Solutions for Chapter 6 End-of-Chapter Problemsquantum.bu.edu/courses/ch102-summer-2010/text/... · ACS Chemistry Chapter 6 suggested solutions 1 Solutions for Chapter 6 End-of-Chapter

ACS Chemistry Chapter 6 suggested solutions 1

Solutions for Chapter 6 End-of-Chapter Problems

Problem 6.1. (a) When water is boiled in a teakettle, no chemical reaction occurs. There is a physical change from H2O(l) to H2O(g). (b) When boiling water is poured into a bowl containing a package of instant oatmeal and stirred to make a hot cereal breakfast, various components of the cereal dissolve or are re-hydrated (the oatmeal had been previously prepared and the chemical changes that make oatmeal from oat flakes had occurred and then water was removed to produce the instant product). The changes occurring when the product is re-hydrated are probably reasonably classified as physical, not chemical changes. (c) When a glass of a carbonated soft drink is left overnight and tastes “flat” the next morning, CO2(g) dissolved in the soft drink as CO2(aq) has escaped to the atmosphere. You might think of this as a physical reaction, but it can also be considered a chemical reaction, if some of the dissolved carbon dioxide was present as carbonic acid, (HO)2CO(aq), which is a different chemical species. (d) When a glass of ice cubes and water is left overnight and there are no ice cubes in the water the next morning, no chemical reaction has occurred. The ice cubes melted, H2O(s) → H2O(l), but no new chemical species formed.

Problem 6.2. (a) When a match is dropped on the floor, no chemical change (reaction) occurs. The match is still the match. (b) When a match is struck and used to start a barbeque, several chemical reactions occur. There is a chemical reaction that uses the energy of the friction of the match head to initiate the reaction that lights the match (which continues to burn because the energy released in the burning sustains the reactions). Once the match is burning, the flame is used to start the chemical reaction in the charcoal, propane, or other fuel used in the barbeque. In both cases, the products of the change, carbon dioxide, water, ashes, and so on, are different chemical species than were present in the match and fuel. (c) When a piece of paper is folded to make a paper airplane, no chemical change (reaction) occurs. The paper is still the paper. (d) When a piece of paper is torn into many small pieces to make confetti, you can argue that no chemical change (reaction) has occurred because the little pieces are still paper. On the other hand, tearing the paper apart breaks a large number on intermolecular (not covalent) bonds between the cellulose polymer molecules that make up the paper, so many of these molecules are not making the same bonding interactions they were previously. The distinction between chemical change and physical change is sometimes quite fuzzy and probably not worth making.

Problem 6.3. (a) When an acorn buried and forgotten by a squirrel grows into an oak tree, a very large number of chemical reactions have occurred. Many compounds not originally present in the acorn are now present in the oak tree. Other compounds that were originally present in the acorn (glucose, for example) are also present in the oak tree, but they are not the same molecules as were in the acorn and there are, of course, a great many more of them in the tree.

Chapter 6 Chemical Reactions

2 ACS Chemistry Chapter 6 suggested solutions

(b) When a bottle of milk left too long in the refrigerator turns sour, chemical reactions have occurred. New compounds (some of which taste and/or smell bad) have been formed, so souring of milk is a chemical reaction. (c) When equal volumes of solutions of blue food coloring and yellow food coloring are mixed and the resulting solution is green, no chemical reaction has occurred to produce the color change. Mixing two different colored solutions results in a mixed solution that retains the characteristics of the molecules in the original solutions. In this case, we know that a mixture of blue and yellow colors produces green. In principle, we could use physical methods (that is, ones that do not required changing the colored molecules to different molecules) to separate the mixture into its original components. (d) When a few drops of bromocresol green solution are added to 20 mL of a colorless 0.1 M solution of sodium acetate, the resulting solution is blue. When 10 mL of a colorless solution of 0.1 M hydrochloric acid are added, the mixture is green. When a further 10 mL of the hydrochloric acid are added, the mixture is yellow. Chemical reactions occurring in these solutions are responsible for the color changes. Bromocresol green is an acid-base indicator, a substance whose color depends upon the pH of the solution it is in. It is blue in a higher pH solution, yellow in a low pH solution, and, in a solution at some intermediate pH, both the blue and yellow forms are present, yielding a green color [compare with the dyes in part (c)]. The acetate ion in sodium acetate solution is a base that gives the solution high enough pH to put the indicator entirely in its blue form. Added hydrochloric acid reacts with some of the acetate and reduces the pH to an intermediate value where the indicator is present in both its blue and yellow forms and the solution appears green. Further addition of acid reacts with all the acetate and lowers the pH still further, which changes the indicator entirely to its yellow form. A chemical reaction between the hydronium ion in the solution and the indicator is responsible for the different colors of the indicator.

Problem 6.4. When a few small pieces of dry ice (solid carbon dioxide, –78 °C) are dropped into a tall glass cylinder containing a red solution of dilute ammonia to which has been added a few drops of phenol red acid-base indicator solution, bubbles of gas are rapidly evolved, a white fog forms above the solution, and, after a short time, the solution in the cylinder turns yellow. A number of physical and chemical reactions are occurring in this system. Solid carbon dioxide is subliming, that is going from solid to gas, CO2(g) → CO2(g), which accounts for the bubbles that are formed. This process is usually classified as a physical change. When the cold CO2(g) reaches the surface and leaves the liquid, it cools the gaseous molecules, including the water vapor above the surface. Some of the water vapor condenses to tiny droplets that we observe as a fog above the surface and spilling out of the cylinder and falling toward the ground as the fog is carried down by the cold dense CO2(g) (and the cooled air) spilling out of the cylinder as well. Some of the CO2(g) dissolves in the solution forming some carbonic acid, (HO)2CO(aq), which can react with the base, ammonia: (HO)2CO(aq) + NH3(aq) → HOCO2

–(aq) + NH4+(aq). This

reaction lowers the pH of the solution changing the phenol red acid-base indicator from its red (basic) form to its yellow (acidic) form. These physical and chemical changes explain the observations described in the first sentence.

Problem 6.5. (a) NaCl is a water-soluble ionic compound composed of 1+ and 1– ions, Na+ and Cl–.

Chemical Reactions Chapter 6


(b) CaCO3 is a water insoluble ionic compound because both the cation, Ca2+, and anion, CO32–,

are multiply charged. (c) Na2CO3 is probably a water-soluble ionic compound, because alkali metal compounds are almost all soluble, even if the anions are multiply charged. (d) BaCl2 is probably a water-soluble ionic compound, because most chloride compounds are soluble, even if the cations are multiply charged. (e) BaSO4 is a water insoluble ionic compound because both the cation, Ba2+, and anion, SO4

2–, are multiply charged.

Problem 6.6. Mixing 25.0 mL of 0.100 M sodium sulfate, Na2SO4, solution with 25.0 mL of 0.200 M barium chloride, BaCl2 solution results in the formation of a white precipitate. The precipitate must be, BaSO4(s), since it is the only possible ionic product with a multiply charged cation and anion. The net ionic reaction equation for its formation is: Ba2+(aq) + SO4

2–(aq) ⇔ BaSO4(s) This equation shows that equal numbers of moles of Ba2+(aq) and SO4

2–(aq) react to give the product. To find the number of moles of product formed (and from that, the mass of product formed), we need to calculate the number of moles of each ion present in the original mixture: mol Ba2+(aq) = (0.0250 L)(0.200 M) = 0.00500 mol Ba2+(aq) mol SO4

2–(aq) = (0.0250 L)(0.100 M) = 0.00250 mol SO42–(aq)

Since there are fewer moles of the sulfate ion, SO42–(aq) is the limiting reactant and we are

limited, by the stoichiometry of the reaction equation to formation of 0.00250 mol BaSO4(s). The molar mass of BaSO4 is 233.4 g, so the mass of product is:

mass BaSO4(s) = (0.00250 mol BaSO4)233.4 g

1 mol BaSO4

⎛⎝⎜

⎞⎠⎟

= 0.584 g BaSO4(s)

Problem 6.7. (a) All of the iron in the 0.264 g of precipitated Fe2O3 came from the 2.998 sample of powdered tablets, which was only part of the original 22.131 g of 20 powdered tablets. To determine the average mass of iron in one tablet, we need to find out how much iron is in the sample of 20 tablets and divide this amount by 20 to find the average per tablet. To find the mass of iron in the 20-tablet sample, we first calculate the mass of iron in the 2.998 g sample that was analyzed:

0.264 g Fe2O3 = (0.264 g Fe2O3)1 mol Fe2O3

159.7 g Fe2O3

⎛

⎝⎜⎞

⎠⎟2 mol Fe

1 mol Fe2O3

⎛

⎝⎜⎞

⎠⎟55.85 g Fe1 mol Fe

⎛⎝⎜

⎞⎠⎟

= 0.185 g Fe Next we need to find the mass of Fe in the total mass of powdered tablets and divide this mass by 20 to find the average mass of Fe per tablet. We get the total mass of Fe (in 20 tablets) using the factor by which the total sample is larger than the analyzed sample.

mass Fe per tablet = (0.185 g Fe)22.131 g for 20 tablets

2.998 g analyzed sample⎛⎝⎜

⎞⎠⎟

120 tablets

⎛⎝⎜

⎞⎠⎟

= 0.0683 g·tablet–1



(b) The mass percent of Fe in a tablet is equal to the mass percent of iron in any of the samples, so we can go back to the analyzed sample that contained 0.185 g Fe in 2.998 g of sample:

mass % Fe = 0.185 g Fe

2.998 g tablets⎛⎝⎜

⎞⎠⎟

100% = 6.17% Fe

(c) In this analysis, we assume that all chemical conversions are quantitative, that is all iron is converted to Fe3+, which is all precipitated, and which all appears in the final product, Fe2O3. We assume that the stoichiometry of the product is correct and that it is really only this product with no leftover water or other contaminants. We assume that the 20 tablets chosen accurately represent the entire batch of tablets (which may be many thousands or millions) being tested, that the tablets are relatively uniform in composition (so the average is meaningful), and that the sample taken for analysis has the same composition as the bulk sample from which it came. Sampling problems are some of the most important in analyses like this, often adding more uncertainly than the chemistry itself.

Problem 6.8. We are to analyze a mixture containing only potassium sulfate, K2SO4, and ammonium sulfate, (NH4)2SO4, to determine the mass percent of K2SO4 in the mixture. A 0.649 g sample of the mixture is dissolved in water and then treated with excess barium nitrate, Ba(NO3)2, solution to precipitate all of the sulfate ion as barium sulfate, BaSO4(s). (a) The net ionic equation for the precipitation reaction is: Ba2+(aq) + SO4

2–(aq) → BaSO4(s) (b) The mass of precipitate formed is 0.977 g, so the number of moles of sulfate ion in the original sample is:

0.977 g BaSO4(s) = (0.977 g BaSO4(s)) 1 mol BaSO4

233.4 g BaSO4

⎛

⎝⎜⎞

⎠⎟1 mol SO4

2–

1 mol BaSO4

⎛

⎝⎜⎞

⎠⎟

= 0.00419 mol SO42–

(c) To find the mass percent of K2SO4 in the original sample, we first find the mass of K2SO4 in the original sample. To do this, let w = mass of K2SO4 in the sample. Then we can write the mass of (NH4)2SO4 in the sample as (0.649 – w) g. Use these masses to express the number of moles of sulfate present in each compound and set their sum equal to the total number of moles of sulfate from part (b). There is one mole of sulfate in one mole of each compound, so we have: mol K2SO4 + mol (NH4)2SO4 = total mol SO4

2– = 0.00419 mol

(w g)1 mol K2SO4

174.3 g⎛⎝⎜

⎞⎠⎟

+ [(0.649 – w) g]1 mol (NH4 )2 SO4

132.1 g⎛⎝⎜

⎞⎠⎟

= 0.00419 mol

(0.00574)·w mol + 0.00491 mol – (0.00757)·w mol = 0.00419 mol – (0.00183)·w mol = – 0.00072 mol w = 0.393 g K2SO4 The mass percent K2SO4 in the sample is:

mass percent K2SO4 = 0.393 g K2SO4

0.649 g sample⎛⎝⎜

⎞⎠⎟

100% = 60.6 %



Problem 6.9. (a) The total volume of each sample in Figure 6.1 (six units of volume) is proportional to the total number of moles of cation plus anion, because the cation- and anion-containing solutions have the same molarity. The mole fraction of cation in each solution is the ratio of the volume of cation added to the total volume. For the five samples, the mole fractions of cation are: 0.17, 0.33, 0.50, 0.67, and 0.83. (b) Count the number of cation-anion pairs at the bottom of each container in Figure 6.1 to get a measure that is proportional to the amount of precipitate formed. For the five samples, the results are 2, 4, 6, 4, and 2. The plot of amount of precipitate as a function of the mole fraction of cation is:

(c) The plot in part (b) has obvious straight-line segments (shown on the graph) that meet at the point corresponding to the largest amount of precipitate formed. This intersection point occurs when the mole fraction of cation is 0.5. The corresponding mole fraction of anion is also 0.5. That is, the straight lines intersect where the mole fractions of the reactants are in the same ratio as the stoichiometry of their reaction with one another. (d) For an unknown precipitation stoichiometry, you could perform a similar series of reactions in which you measure the mass of precipitate formed as you keep the total number of moles of the cation and anion constant while varying their ratio. A plot of these data will probably yield a graph similar to the one in part (b) with the intersection of the two straight-line portions of the data occurring at the mole fraction of cation corresponding to the fraction of cations in the precipitate. For example, in a 1:2 ionic compound, such as PbI2, the fraction of cations in the solid is one-third (one in three of the ions in the compound is a cation), so the lines would intersect at about 0.33 mole fraction of cation.

Problem 6.10. The mole fractions of nickel cation in the six samples in Check This 6.11 are obtained by dividing the number of moles of nickel cation by the total number of moles of nickel plus dimethylglyoxime in each sample. The results are shown in this table (where all molar quantities are mol × 104)



Sample 1 2 3 4 5 6 mol Ni2+(aq) 2.55 4.89 5.98 7.50 9.89 12.6 mol dmg 18.2 15.8 14.8 13.1 10.9 8.19 total mol 20.8 20.7 20.8 20.6 20.8 20.8 mol fraction Ni2+(aq) 0.123 0.236 0.288 0.364 0.475 0.606 mass ppt, g 0.074 0.141 0.173 0.189 0.158 0.118

The plot of the mass of precipitate formed as a function of the cation mole fractions is:

We see that the two straight-line segments intersect at a nickel cation mole fraction of 0.33; one third of the ions in the solid ionic compound are nickel cation, Ni2+. The other two thirds are dimethylglyoxime anions, so the ionic compound must be Ni(dmg)2. Note that this result also tells us that each dimethylglyoxime anion must have a charge of –1 in order to balance the cationic charge. When you analyzed the data in Check This 6.11, you probably focused on the sample that produced the largest amount of precipitate and found that the molar ratio of nickel to dimethylglyoxime in this sample solution is 7.50:13.1 = 1.00:1.75 and concluded (assuming simple small whole numbers of each ion) that the cation to anion ratio in the solid product is 1:2. The graphical method confirms this analysis and provides even more convincing evidence for this ratio.

Problem 6.11. Lewis (or partial Lewis) structures are drawn here for several ions or molecules, in order to help predict which will be Lewis bases. To act as a Lewis base, an ion or molecule must that a nonbonding pair of electrons that can bond to a proton. (a) Ammonia, :NH3, has a nonbonding electron pair on the nitrogen atomic core and is a Lewis base. (b) A proton (hydrogen atomic core), H+, has no nonbonding electron pairs and cannot be a Lewis base. (c) The ammonium ion, NH4

+, has no nonbonding electron pairs and cannot be a Lewis base.

(d) The hydroxide ion, OH , has three nonbonding electron pairs on the oxygen atomic core and is a Lewis base.



(e) The cyanide ion, C N , has a nonbonding electron pair on both a carbon and nitrogen atomic core and is a Lewis base. Because the carbon is less electronegative, it has the greater ability to share its electron pair, so a proton will bond there to form HCN.

(f) Methyl amine, CH3NH2 , like ammonia, has a nonbonding electron pair on the nitrogen atomic core and is a Lewis base. (g) Although the oxygen atomic core in the hydronium ion, H3O+, has an unshared electron pair, the species has an overall positive charge, which will repel another proton. The hydronium ion shows no Lewis base properties.

(h) The methide ion, CH3 , has a nonbonding electron pair on a carbon atomic core and is a Lewis base.

Problem 6.12. (a) All Lewis bases have a pair of non-bonding electrons capable of forming a covalent bond with a Lewis acid. (b) Lewis acids can share a pair of non-bonding electrons from a Lewis base to form a covalent bond. Protons are Lewis acids (just as they are Brønsted-Lowry acids), but the array of ions and molecules that act as Lewis acids is much broader than this, as this chapter tries of show.

Problem 6.13. From the point of view of the Brønsted-Lowry acid-base model, we might write the reaction in this problem like this:

O HH

H N HH

HOH

H NH

HH

H

The red color of one of the hydrogen atomic cores draws attention to proton transfer from the Brønsted-Lowry acid, H3O+ (hydronium ion), to the Brønsted-Lowry base, NH3 (ammonia). As in all acid-base reactions, there are two acid-base pairs, H3O+/H2O and NH4

+/NH3, that differ by the presence of a hydrogen atom core in the acid that is not present in the base.

From the point of view of the Lewis acid-base model, we might write the reaction like this:

O HH

H N HH

HOH

H NH

HH

H

Here, the color focuses our attention on electron pairs that bond a hydrogen core to another atomic core in the Lewis acids and that are nonbonding electron pairs in the Lewis bases. The identification of the Lewis acids and bases and conjugate Lewis acid-base pairs is identical to that for the Brønsted-Lowry model above, as it always is when it is a proton that is transferred to the Lewis acid.

Problem 6.14. The reaction of lactic acid, CH3CH(OH)C(O)OH, transferring its acidic proton to water in a Brønsted-Lowry acid-base reaction, with the acid-base pairs denoted, is:



CH3CH(OH)C(O)OH + H2O ⇔ CH3CH(OH)C(O)O– + H3O+ acid 1 base 2 base 1 acid 2 We write the reaction with reversible equilibrium arrows, because we are told that the only a fraction of the lactic acid molecules transfer their protons to water. Evidently the reaction can go in either direction and comes to a balance point (equilibrium) when there are substantial amounts of all the species present in the solution.

Problem 6.15. The concentration of hydronium ion in a solution of pH 1.3 is: [H3O+(aq)] = 10–pH = 10–1.3 = 0.05 M Since the pH of a 0.05 M aqueous solution of perchloric acid, HOClO3, is 1.3, it looks as though the acid has transferred its proton completely to the water: HOClO3(aq) + H2O(aq) → H3O+(aq) + ClO4

–(aq) The perchlorate ion, ClO4

–(aq), must be quite a weak base (similar to the chloride ion), since it gives up all its protons to water. Therefore, if sodium perchlorate, NaClO4, is dissolved in water, the solution pH will be the same as the water before the salt was added, because the weak perchlorate base does not react with water to produce hydroxide ion and perchloric acid. Note the position of the perchlorate ion in Table 6.2.

Problem 6.16. A 0.1 M solution of sodium cyanide, NaCN, in water has a pH of 11, so the cyanide ion, CN–

(aq), must be reacting as a base to produce a basic solution containing some OH–(aq): CN–(aq) + H2O(aq) → HCN(aq) + OH–(aq) The concentration of hydronium ion in a solution of pH 11 is: [H3O+(aq)] = 10–pH = 10–11 M Rearranging equation 6.14 and substituting this value for [H3O+(aq)], we find:

[OH–(aq)] = 10–14 M2

[H3O+(aq)]

= 10–14 M2

10–11 M = 10–3 M

If the chemical reaction written above were to go to completion, a 0.1 M solution of sodium cyanide would have [OH–(aq)] = 10–1 M. It appears that only about 1% of the cyanide ion reacts as shown, so we should write the equation with double equilibrium arrows to signify that the reaction reaches a balance point (equilibrium) when there are substantial amounts of all the species present in the solution. This conclusion tells us that hydrogen cyanide gas, HCN, dissolved in water should transfer some of its protons to the water molecules forming some H3O+(aq). Thus, the pH of a 0.1 M aqueous solution of HCN, will be lower, more acidic, than the water before the HCN was dissolved. Since, some HCN(aq) forms in aqueous cyanide solutions, we know that the acid is not a strong acid that transfers all its protons to water. The pH will be higher than 1 (the value for a 0.1 M solution of a strong acid), but lower than 7 (the pH of pure water). An intermediate value like 4 would be reasonable estimate based on the acid-base concepts developed so far. [The mathematical analyses from Chapter 9 give a pH of 5.]



Problem 6.17. (a) From its position in Table 6.2, we see that 2,4-dinitrophenol is only a little stronger acid than ethanoic (acetic) acid. Thus, the anionic (more water soluble), basic form of 2,4-dinitrophenol readily accepts protons from hydronium ion to produce its electrically neutral (less water soluble), acidic form. This is just the property we want in a molecule added to interrupt the formation of ATP (by reducing the pH difference across membranes), so we can study how this interruption affects other metabolic processes. (b) Our bodies need a continuous supply of ATP to keep the processes of life going. If the rate of production of ATP goes down, we compensate by metabolizing more carbohydrate and fat molecules to try to maintain the difference in pH across membranes that is required for ATP synthesis. The rate of production of ATP can be slowed down by the presence of 2,4-dinitrophenol, which reduces the pH difference across membranes. In this case, the body metabolizes more carbohydrate and fat than it usually would and the result is the loss of weight, since these are the fuel molecules the body uses to store energy. This treatment can obviously be carried too far with too large a decrease in ATP production and inability of the body to sustain itself.

Problem 6.18. (a) The reaction between boric acid and water with the Lewis structures written out is:

BOH

HOOH

+ O HH

2 BOH

HOOH

OH + O HH

H

The boron atom is only surrounded by three electron pairs, so it can share an electron pair from a water molecule. Boric acid is the Lewis acid, and water is the Lewis base on the left-hand side of this equation. Note that you might think of this reaction occurring in two steps:

BOH

HOOH

+ O HH

BOH

HOOH

OH

H

+ O HH

BOH

HOOH

OH + O HH

HBOH

HOOH

OH

H

The first step is the Lewis acid-base reaction between the electron deficient boron atomic center (see Problem 5.58), a Lewis acid, and one of the non-bonding electron pairs on a water molecule to form a species that has three bonds to an oxygen core. This species would have properties similar to a hydronium ion, in particular, it could transfer a proton to water to form a hydronium cation and the B(OH)4

– anion. These reactions are reversible and solutions of boric acid are only weakly acidic (which is why you may be familiar with its use in dilute solutions as an eyewash solution), so the equilibrium balance must lie far to the reactant side. (b) Writing the formula for boric acid as H3BO3 is misleading, because it might suggest that the three hydrogen atomic centers are directly bonded to the boron atomic center. The formula,



B(OH)3, makes the connectivity explicit and shows that this structure is related to the oxyacids (although its acidic reaction is quite different). Conventionally, oxyacids, like sulfuric, (HO)2SO2, are written with the acidic hydrogen centers on the left by analogy with simpler acids like hydrochloric, HCl. Writing boric acid as B(OH)3 is a subtle way of indicating that it does not act like other oxyacids, that is, that the hydrogen cores on this species are not acidic. (c) Polyprotic acids, like sulfuric, (HO)2SO2, and phosphoric, (HO)3PO, are all oxyacids that can donate more than one proton to a Lewis (or Brønsted-Lowry) base, because they have more than one –OH group attached to the central atomic core. Boric acid, B(OH)3, is an interesting apparent exception to this observation. The difference is that the central boron atomic core is electron deficient and itself acts as a Lewis acid [see part (a)], to add an OH– group from a water molecule as the proton from the water molecule is transferred to another water molecule to form a single hydronium ion for each B(OH)3 that reacts, that is, boric acid is monoprotic.

Problem 6.19. Based on these experimental observations, we are to rank these bases from weakest to strongest: CH3C(O)O– (ethanoate or acetate), NO3

– (nitrate), OH– (hydroxide). (i) 0.1 M CH3C(O)OH has pH = 2.9 (ii) 0.1 M HNO3 has a pH = 1.0 (iii)0.1 M NaNO3 has a pH = water in which it was dissolved (iv) 0.1 M CH3C(O)O–Na+ (sodium ethanoate) has a pH = 8.9 (v) 0.1 M NaOH has a pH = 13.0 The ranking is (weakest) NO3

– < CH3C(O)O– < OH– (strongest) Observation (ii) shows that nitric acid, HNO3(aq), transfers its protons completely to water, since the pH corresponds to [H3O+(aq)] = 0.1 M, which is the concentration of the nitric acid solution. In addition, observation (iii) shows that nitrate anion does not attract enough protons from water to form HNO3(aq) and OH–(aq) and change the pH of the water in which NaNO3 is dissolved. Thus, nitrate anion, NO3

–(aq), must be a very weak base. Observation (i) shows that ethanoic acid, CH3C(O)OH(aq), does not transfers all its protons completely to water, since the pH corresponds to [H3O+(aq)] = 10–2.9 ≈ 0.001 M, which is less than the concentration of the ethanoic acid solution. In addition, observation (iv) shows that ethanoate anion attracts enough protons from water to form CH3C(O)OH(aq) and OH–(aq) and makes the pH of the water in which CH3COO-Na+ is dissolved somewhat basic, pH 8.9. Thus, ethanoate anion, , must be a stronger base than nitrate ion, which has no effect on the pH. To compare the ethanoate and hydroxide anions, we consider the reaction that makes the solution basic in observation (iv): CH3COO-(aq) + H2O(aq) ⇔ CH3C(O)OH(aq) + OH–(aq)

If this reaction went to completion as written, the solution would have [OH–(aq)] = 0.1 M, with a pH of 13. But the pH is only 8.9, showing that this reaction precedes only a small amount as written. Since the weaker base predominates in equilibria like this, the ethanoate anion is a weaker base than the hydroxide anion, which completes the ordering we were asked to find,

Problem 6.20. (a) NH3 is a stronger Lewis base than PH3. The central atomic center in each molecule is in group V of the periodic table. We know that basicity decreases going down a column of the



table, because the size of the central atomic core increases and the charge density of the nonbonding electrons becomes less concentrated. (b) HP2– is a stronger Lewis base than S2–. The atomic center in each ion is in the third period of the periodic table, so these atomic cores are about the same size. P is less electronegative than S, so holds its nonbonding electrons less strongly, which makes them are more available to bond with a Lewis acid and makes HP2– the stronger Lewis base. (c) It’s difficult to predict whether HP2– or O2– is the stronger Lewis base. Because of size considerations, we know that O2– is a stronger Lewis base than S2–. But from part (b) we know that HP2– is a stronger Lewis base than S2–. Thus, we see that O2– and HP2– are both stronger bases than S2–, but our directional models don’t give us any quantitative information about how much stronger each one is, so our models are inadequate to make this prediction. (d) CH3O– is a stronger Lewis base than CH3S–. The nonbonding electrons in each ion are on atomic centers that are in group VI of the periodic table. We know that basicity decreases going down a column of the table, because the size of the central atomic core increases and the charge density of the nonbonding electrons becomes less concentrated.

Problem 6.21. When equal volumes of 0.1 M aqueous solutions of sodium hydroxide, NaOH, and ammonium chloride, NH4Cl, are mixed this net acid-base reaction can occur: NH4

+(aq) + OH–(aq) ⇔ NH3(aq) + H2O(aq) Since equal volumes of solutions of the same molarity are mixed, the number of moles of the reactants, NH4

+(aq) and OH–(aq), are stoichiometrically equivalent and they would react completely to use one another up, if this reaction were to go to completion. The Lewis bases in this reaction are OH–(aq) and NH3(aq). The strongest base that can exist in aqueous solution is the hydroxide ion, so NH3(aq) is the weaker base and it will be the predominant Lewis base in the solution, that is, the reaction will lie far toward the product side. Water is a Lewis acid in this reaction system and it will be the predominant Lewis acid in the solution. As we have said in the text, it is a bit unfair to count water as the predominant Lewis acid (or base) in aqueous solution, because its concentration overwhelms all other species, regardless of whether it is the weaker acid or base. In this case, the comparison is somewhat valid, because water is on the product side and it is the weaker acid compared to the ammonium ion, NH4

+(aq.

Problem 6.22. The reactions that describe the solution that results when equal volumes of 0.10 M NH3 and 0.10 M HCl are mixed can be written as: HCl(aq) + H2O(aq) → H3O+(aq) + Cl– (aq) H3O+(aq) + NH3(aq) ⇔ NH4

+(aq) + H2O(aq) We know that the chloride anion is such a weak base that HCl(aq) completely transfers its proton to water, as shown in the first equation. Since equal volumes of solutions of the same molarity are mixed, the number of moles of the reactants, H3O+(aq) and NH3(aq), are stoichiometrically equivalent and they would react completely to use one another up, if the second reaction were to go to completion. The Lewis acids in this reaction are H3O+(aq) and NH4

+(aq). The strongest acid that can exist in aqueous solution is the hydronium ion, so NH4

+(aq) is the weaker acid and it will be the predominant Lewis acid in the solution, that is, the reaction will lie far toward the product side. Thus, the predominant ions in the solution will



be the Cl– (aq) anion from the first acid-base reaction and the NH4+(aq) cation from the second.

There will, of course, be hydronium and hydroxide ions in the solution, but their concentrations will be much lower.

Problem 6.23. For two oxyanions with the same amount of electron delocalization, such as (HO)2AsO2

– and (HO)2PO2

–, the anion with the larger central atom is more stable (the ion is larger and the electrons can occupy a larger volume) and does not accept protons as well. Since atomic size increases down a family of the periodic table, we expect the dihydrogen arsenate anion to be a somewhat weaker base than the dihydrogen phosphate anion. Conversely, we expect arsenic acid, (HO)3AsO, to be a somewhat stronger acid (more transfer of a proton to water) than phosphoric acid, (HO)3PO. Experimentally, we find that these two bases (acids) are almost equal in their basicity (acidity); the size difference between a phosphorus atom and an arsenic atom is small.

Problem 6.24. (a) The Lewis structures for CH3C(O)O– and CH3C(O)S– are:

C CO

O

HH

HC C

O

O

HH

Hand

C CS

O

HH

HC C

O

S

HH

Hand

These ions have delocalized pi electrons spread over the three atoms of the carboxyl, O–C–O, and thiocarboxyl, S–C–O, groups. (b) The CH3C(O)O– (ethanoate) anion is likely to be the stronger base, because the delocalization of the electrons is likely to be enhanced by the somewhat larger sulfur atomic center in the CH3C(O)S– (thioethanoate) anion, which would stabilize thioethanoate more than ethanoate. This added stability would make CH3C(O)S– less likely to react with a proton donor and hence make it the weaker base. [Quantitatively, CH3C(O)S– (pKb = 10.67) is about 26 times less basic than CH3C(O)O– (pKb = 9.25).]

Problem 6.25. 6.25. (a) A 0.1 M solution of potassium amide, KNH2, in water has a pH of 13. Write a chemical equation for the acid-base reaction that occurs in this solution. What is the predominant Brønsted-Lowry acid and base in the solution? How would you characterize the base strengths of the bases in this solution. Explain your responses. (b) A 0.1 M solution of ethylamine, CH3CH2NH2, in water has a pH of about 11. Write a chemical equation for the acid-base reaction that occurs in this solution. What is the predominant Brønsted-Lowry acid and base in the solution? How would you characterize the base strengths of the bases in this solution. Explain your responses. (c) How are the reactions you write and your interpretations of the experimental evidence in parts (a) and (b) the same and different? Explain clearly. Answer to 6.25: (a) A 0.1 M solution of potassium amide, KNH2, in water has a pH of 13. In order to form a strongly basic, pH 13, solution, the amide ion must react with water to accept a proton: NH2

–(aq) + H2O(l) ⇔ NH3(aq) + OH–(aq)



In Check This 6.18, you calculated that the concentration of hydroxide ion in this solution is 0.1 M, which is the concentration of amide ion that would be present if reaction had not occurred. We can conclude that essentially all of the amide ion has reacted and that reaction equation (6.12), which shows the reaction going to completion, better represents the situation in this solution: NH2

–(aq) + H2O(l) → NH3(aq) + OH–(aq) The predominant acid and base in this solution are ammonia, NH3(aq), and hydroxide ion, OH–

(aq), respectively. Since the weaker acid and base predominate in a solution, we can conclude that the amide ion, NH2

–(aq), is a stronger base than the hydroxide ion. (b) A 0.1 M solution of ethylamine, CH3CH2NH2, in water has a pH of about 11. In order to form a basic, pH 11, solution, the ethylamine must react with water to accept a proton: CH3CH2NH2(aq) + H2O(l) ⇔ CH3CH2NH3

+(aq) + OH–(aq) In this pH 11 solution, [H3O+(aq)] = 10–11 M, and, from equation (6.14), we get:

[OH–(aq)] = 10–14 M2

[H3O+(aq)]

= 10–14 M2

10–11 M = 10–3 M

This concentration of hydroxide ion is 100 times less than the concentration of ethylamine in the solution. We can conclude that the reaction we wrote occurs to only a small extent, so the predominant acid and base in this solution are water, H2O(l), and ethylamine, CH3CH2NH2(aq), respectively. Since the weaker acid and base predominate in a solution, we can conclude that the hydroxide ion is a stronger base than ethylamine. (c) The reactions we wrote in parts (a) and (b) are essentially the same, a base accepts a proton transferred from water to form a solution with excess hydroxide ion compared to pure water. The great difference is in the extent of the proton transfer. In part (a), the transfer is “complete,” with virtually none of the amide ion left unreacted; the amide ion is a strong base in water. In part (b), the transfer is incomplete and a great deal of the ethylamine is left unreacted; ethylamine is a weak base in water.

Problem 6.26. (a) Consider the Lewis acid-base reactions occurring in these reaction steps for the formation of carbonic acid, H2CO3 [(HO)2CO], from water and carbon dioxide:

OH

H

C

O

O

C

O

O

O

H

H C

O

O

O

H

H

The water is acting as an electron-pair donor, so it is the Lewis base. The C=O bonds in carbon dioxide are polarized with the more electronegative oxygen atomic cores pulling electron density away from the carbon. Thus, a positive center is created at the carbon core, which acts as an electron-pair acceptor, so it is the Lewis acid (as represented in the reaction by the arrow showing the attraction of the electron pair to this atomic center). As the new C–O bond forms, there has to be a redistribution of electrons, because only four electron pairs can be accommodated around the carbon core. Therefore, as the small red arrow indicates, one of the pairs of electrons forming the pi bond between carbon and oxygen moves to an oxygen. This oxygen now has three nonbonding electron pairs, which gives it properties similar to the oxygen

Dan Dill

Highlight

Dan Dill

Highlight



core in hydroxide ion, OH–, a strong base. The oxygen core just bonded to the carbon has three sigma bonds to other atomic cores, just as in the hydronium ion, H3O+, a strong acid. Thus, the intermediate molecular structure is set up for the electron pair on the basic oxygen to donate a pair of electrons to the acidic hydrogen atomic core on the other, as represented by the small red arrow on the center structure. The result is the formation of the carbonic acid structure at the right, in which all atomic centers have their conventional number of bonding and nonbonding sigma electron pairs. (b) We have previously written the overall chemical equation for this reaction as: H2O(l) + CO2(g) ⇔ H2CO3(aq) The structural equation in part (a) gives more information about electron arrangement, making it easier to identify the electron pair donors and acceptors. The formula equation does give not give information about which atoms are linked to which nor any indication which are the Lewis acids and bases in this reaction.

Problem 6.27. (a) The chemical equation H2O + HCO3

– ⇔ H3O+ + CO32– can be rewritten using the

representations given in Table 6.2 and identifying all Lewis acids and bases in the reaction and the conjugate pairs of Lewis acids and bases (red is used to label nonbonding electron pairs in Lewis bases and corresponding bonding pairs in Lewis bases):

H2O OCO2ŠH H2O H+ OCO2

2Š

Lewis base Lewis acid

conjugate pair

Lewis baseLewis acid

conjugate pair

(b) The hydronium ion, H3O+(aq), is the strongest acid that can exist in aqueous solution, so the weaker acid, HOCO2

2–(aq), will be the predominant acid in this reaction. The equilibrium will lie mostly toward the reactant side where the weaker Lewis acid (and weaker Lewis base) is found. The same conclusion can be drawn comparing the Lewis bases, water and carbonate anion. Sodium carbonate, Na2CO3 (often called washing soda), dissolves in water to form a basic solution, showing that the carbonate anion wins the competition with water molecules for protons and is the stronger base and will not predominate in the reaction we wrote.

Problem 6.28. Since aluminum is in the boron family of elements, group III, its bonding characteristics will probably be similar to boron’s. In boron compounds, the boron atomic center is often electron deficient, because the atom has only three valence electrons, and, in Problem 6.18, we found that boric acid, B(OH)3, reacts with water to form the B(OH)4

– anion (and hydronium). It seems likely that Al(OH)3(s) might dissolve in basic solution by reacting with a hydroxide ion to form a soluble anion: Al(OH)3(s) + OH–(aq) → Al(OH)4

–(aq)



Problem 6.29. (a) The independent variable in a continuous variations analysis is the mole fraction one of the reactants is of the total number of moles of reactants added. If we suspect that reactant A will react with more than one mole of reactant B, then, the graphs are easier to interpret if we plot the mole fraction of reactant B. For the copper-ethylenediamine complex, it’s likely that more than one ethylenediamine (en) will react with each copper (Cu) ion, so we’ll use the mole fraction of en in each sample as the independent variable. Since the molarities of the reactant solutions are equal, there is the same total number of moles of Cu plus en in each sample. The fraction that is en is just the ratio of the volume of en added divided by the total volume, 12 mL in each case. This table gives the mole fraction of en in each sample and their absorbances at 640 nm (red light) and 560 nm (green light). The graph shows the two absorbances, A, plotted as a function of the mole fraction of en with blue squares for A640 and magenta circles for A560. [These are actual data from an experiment on the formation of Cu(II)-ethylenediamine complexes for use as an analytical method for determining Cu(II) concentrations.]

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

0.000 0.200 0.400 0.600 0.800 1.000

Abs

orba

nce

mol fraction en

The maximum absorbance at 640 nm is reached at a lower mole fraction of en than for the absorbance at 560 nm. The problem states that copper (II) solutions initially become a deep blue when en is added. A solution that absorbs red light (640 nm) will appear blue, since blue light is being transmitted. Note that there is some light absorption with no en added and this is consistent with the original light blue copper (II) solution. At higher mole fractions of en, the absorbance at 640 nm decreases, but the absorbance at 560 nm (green light) continues to grow. If green light is absorbed by the solution, the colors that are transmitted will be violet and red, which accounts for the magenta color of solutions with more en added. At even higher mole fractions of en, the absorbance at 560 nm also decreases. If we extrapolate both the increasing and decreasing parts of the A640 plot, it looks like the lines would meet at about a mole fraction of 0.5, at an absorbance of about 0.35. At a mole fraction of 0.5, the number of moles of Cu and en are equal. We conclude that a 1:1 complex, Cu(en)2+(aq), is a deep blue.

mol fraction en A640 A560 0.000 0.044 0.001 0.083 0.095 0.026 0.167 0.150 0.050 0.250 0.200 0.075 0.333 0.256 0.107 0.417 0301 0.146 0.500 0.583 0.261 0.307 0.667 0.187 0.411 0.750 0.312 0.833 0.208 0.917 0.105



The peak in the 560 nm absorbance is at an en fraction of 0.67, that is two-thirds en and one-third copper ion [twice as much en as Cu(II)]. We can conclude that increasing amounts of en produced a second complex with a 1:2 ratio of copper (II) to en, Cu(en)2

2+(aq). When more en is added, the absorbance at 560 nm decreases linearly with decreasing number of moles of copper(II) in solution. This is exactly what you expect, if no complexes with more than two en are formed. Once there are two or more en molecules for every copper(II), all the copper(II) two is complexed and the absorbance of the solution depends only on how many moles of copper(II) are present. When the fraction of en is 1.00 (no copper(II) in the solution), the absorbance is zero, as the extrapolation of the plot shows.

(b) The color (and the absorbance) of these solutions are only dependent on the ratio of en to Cu. Thus, if the 6:6 solution had been mixed to make it uniform before it was spilled, spilling some would make no difference to the absorbance. The students could have used the sample (as long as there was enough left to measure in their spectrophotometer). The estimated 640 nm absorbance of this sample was given above. You can get the estimated 560 nm absorbance by interpolating between the points on either side of the missing point. Drawing a smooth curve through the points gives an estimated absorbance of 0.22.

(c) A model of the 1:1 complex has one ethylenediamine attached through the lone electron pairs on its two nitrogen atomic cores at two positions on the metal ion. The 2:1 complex has two ligands (ethylenediamine molecules) bonded by their nitrogen lone electron pairs at four positions around the central metal ion. The geometry of the nitrogen atomic cores around the Cu(II) central atomic center might be either square planar or tetrahedral. [The evidence is that the complex has the nitrogen atomic cores arranged in a square about the Cu(II) ion with water molecules occupying positions above and below the plane of the square, so that the bonding about the Cu(II) core is octahedral.]

Problem 6.30. The light blue solution first formed when a 4.0 M aqueous solution of ethylenediamine, en, is added slowly to a 200 mL of a stirred 0.10 M aqueous solution of nickel(II) ion is probably the color of a complex between the Ni(II) ion and one en, Ni(en)2+. As more en solution is added the solution changes from light blue to purple, which suggests that further Ni(II)-en complexes are formed with two or more en ligands per metal ion. The color stops changing after 15 mL of the en solution has been added. We can calculate the ratio of moles of en to moles of Ni(II) when the color stops changing to get an indication how many en ligand molecules complex to the Ni(II). The number of moles of en added is: mol en = (4.0 M)(0.015 L) = 0.060 mol The number of moles of Ni(II) in the original solution is: Mol Ni(II) = (0.10 M)(0.200 L) = 0.020 mol The ratio en:Ni(II) is 3:1 when the color stops changing, so we can conclude that the purple complex is probably Ni(en)3

2+.

Problem 6.31. (a) This is a representation of the structure of BAL (British AntiLewisite with the H atoms omitted) that focuses on the relationship of the three group VI atomic cores (each of which has two pairs of nonbonding electrons) to one another. All three of these Lewis base centers are “reaching” toward you from the plane defined by the three carbon cores.



CCC

OS

S

(b) The nonbonding electron pairs in the molecule (and your model) are in position to form bonds to a single Lewis acid central metal ion, such as Pb2+. Just as ethylenediamine can form two bonds in a metal ion complex and EDTA can form six, BAL can form three, so two of these molecules could complex with and successfully sequester potentially toxic Pb2+ ions, preventing them from being incorporated into bone and blood. The sulfur atomic centers are weaker bases than the oxygen center and probably will transfer their protons to water (forming BAL2–) as they bind to the metal ion. A likely complex with Pb(II) would be Pb(BAL)2–, a soluble complex that normal excretion processes taking place in the kidney will eliminate from the body.

Problem 6.32. (a) Brønsted-Lowry acid-base reactions occur when a positively polarized proton donor (B-L acid) and a negatively polarized proton acceptor (B-L base) are attracted to one another and get close enough for the proton to be transferred from the acid to the base. The polarized molecules charges may carry full ionic charges or partial charges caused by electronegativity differences within the molecules. Lewis acid-base reactions shift the focus from proton exchange to electron-pair donors (Lewis bases), which are negatively polarized, and electron-pair acceptors (Lewis acids), which are positively polarized. When these positive and negative centers are attracted close enough, an electron pair from the Lewis base forms an electron-pair bond with the positive center on the Lewis acid. (b) Brønsted-Lowry acid-base definitions are particularly straightforward to apply to aqueous proton transfer systems. The particular usefulness of Lewis acid-base definitions is extension of the concept of acid-base reactions to many organic reactions and to complex-ion reactions.

Problem 6.33. (a) If we analyze this possible reaction, CH3C(O)OH + NH3 ⇔ CH3C(O)NH2 + H2O, as an electrophile-nucleophile reaction, the electrophile is CH3C(O)OH. The carbonyl carbon is positively polarized, because the two electronegative oxygen atoms draw bonding electron density away from the carbon. Thus, the carbonyl carbon attracts electron density from another molecule and is an electrophile. (b) The ammonia molecule, NH3, has an unshared pair of electrons that can interact with a positive center in another molecule, so it is a nucleophile. (c) These reactants, a carboxylic acid and a base (ammonia) can undergo an acid-base reaction, CH3C(O)OH + NH3 ⇔ CH3C(O)O– + NH4

+, instead of an electrophile-nucleophile reaction. The marginal note on page 391 suggests that electrophile-nucleophile reactions are often slow and we have observed that acid-base reactions (proton transfers) are quite rapid. Thus, it seems most likely that the acid-base reaction will be observed. [This is true.]

Problem 6.34. (a) The structural formula of the tripeptide that is formed when three glycine molecules, H2NCH2C(O)OH, are linked together by amide bonds structure is:

N C CO

N C CO

N C CO

OHH

H

H

H

HH H

H

HH



(b) The reaction equation for formation of this tripeptide is:

N C CO

N C CO

N C CO

OH2

HH

H

H

HH H

H

HHN C C

O

OHH

H

HH3 2 H2O

Problem 6.35. The animation in the Web Companion, Chapter 6, Section 6.7, page 2, shows the (somewhat hypothetical) formation of an amide from a carboxylic acid and an amine that we can compare to the reaction sequence shown in equation (6.34) for formation of an ester from a carboxylic acid and an alcohol. The animation represents the attraction of the electrophilic carbonyl carbon for the electron pair on the nucleophilic nitrogen of the amine as a developing bond when the molecules approach one another. This is the attraction represented in equation (6.34) by the long curved arrow from the electron pair on the alcohol oxygen toward carbonyl carbon in the acid. (By convention, these kinds of arrows go from electrons to the atomic core where they end up. This is because electrons are moving rapidly and the much heavier cores move quite slowly, so rearrangements are likely to involve mostly electronic movement.) As the molecules get closer and the new bond in the animation gets more prominent, note that the double bond from the carbon to the carbonyl oxygen fades to become a single bond. This is represented in equation (6.34) by the small curved arrow in the left-hand structure that shows one of the pairs of electrons in the C=O double bond moving onto the oxygen. The result in the animation is a structure like the central structure in the equation: a nitrogen in the animation replaces the oxygen that comes from the alcohol in the equation. The animation next shows the transfer of a proton from the nitrogen (now bonded to the original carbonyl carbon) to the oxygen that was the –OH oxygen in the carboxylic acid. The proton transfer occurs from an ammonium-like nitrogen species (sigma bonded to four other atoms) to the oxygen. Similarly, equation (6.34) represents such a proton transfer by the two lower curved arrows in the central structure, which show a proton transfer from a hydronium-like oxygen species (sigma bonded to three other atoms) to the oxygen that was the –OH oxygen in the carboxylic acid. The animation follows this transfer with the loss of a water molecule, which takes with it the electron pair that was bonding it to the original carbonyl carbon. As the carbon-oxygen bond extends and breaks, note that the double bond from reforms between the carbonyl carbon and the originally double-bonded oxygen. This concerted bond breaking and formation are represented in equation (6.34) by the upper pair of curved arrows in the central structure. The reaction is now complete with the formation of an amide (in the animation) or an ester (in the equation) and a molecule of water in both cases. These representations are quite parallel and hopefully reinforce and clarify the molecular changes that occur in these electrophile-nucleophile reactions. Both representations are probably incorrect in some details, but they provide pictures that help us understand how these reactions can occur.

Problem 6.36. The nucleophile, the electrophile, and any spectator ions are identified in each of these balanced equations. (a) NaOH + CH3–I → NaI + CH3–OH The oxygen (electron pair) of hydroxide (HO–) is the nucleophile. The positively polarized carbon atom of methyl iodide (CH3—I) is the electrophile. The sodium cation (Na+) is a spectator ion.



(b) (CH3)3N + CH3–I → (CH3)4N+ + I– The nitrogen (electron pair) of trimethyl amine ((CH3)3N) is the nucleophile. The positively polarized carbon atom of methyl iodide (CH3—I) is the electrophile. There are no spectator ions. (c) (CH3)3C–Br + CH3CH2OH → (CH3)3C–O–CH2CH3 + HBr The oxygen (electron pair) of ethanol (CH3CH2OH) is the nucleophile. The central (tertiary) carbon positively polarized of tertiary butyl bromide ((CH3)3C—Br) is the electrophile. There are no spectator ions. (d) NH3 + CH3C(O)OCH2CH3 → CH3C(O)NH2 + CH3CH2OH The nitrogen (electron pair) of ammonia (NH3) is the nucleophile. The positively polarized carbonyl carbon of ethyl acetate (CH3C(O)OCH2CH3) is the electrophile. There are no spectator ions. (e) CH3NH2 + CH3C(O)Cl → CH3C(O)NHCH3 + HCl The nitrogen (electron pair) of methylamine (CH3NH2) is the nucleophile. The positively polarized carbonyl carbon of acetyl chloride (CH3C(O)Cl) is the electrophile. There are no spectator ions.

Problem 6.37. The formulas of the alcohol and the acid from which each ester is made are shown:

ester alcohol acid (a) CH3C(O)OCH2CH2CH(CH3)2 HOCH2CH2CH(CH3)2 CH3C(O)OH (b) CH3CH2CH2C(O)OCH2CH3 HOCH2CH3 CH3CH2CH2C(O)OH (c) CH3C(O)O(CH2)7CH3 HO(CH2)7CH3 CH3C(O)OH (d) C6H5C(O)OCH3 HOCH3 C6H5C(O)OH (e) CH3C(O)OCH2C6H5 HOCH2C6H5 CH3C(O)OH

Problem 6.38. The formula of the ester that is formed by reaction of each of these acid and alcohol combinations is shown. (a) CH3(CH2)2C(O)OH + CH3(CH2)4OH → CH3(CH2)2C(O)O(CH2)4CH3 + H2O (b) CH3C(O)OH + CH3CH2OH → CH3C(O)OCH2CH3 + H2O (c) CH3C(O)OH + (CH2)3CH(CH2)2OH → CH3C(O)O(CH2)2CH(CH3)2 + H2O (d) CH3CH2CH2C(O)OH + CH3OH → CH3CH2CH2C(O)OCH3 + H2O (e) HC(O)OH + CH3CH2OH → HC(O)OHCH2CH3 + H2O

Problem 6.39. The reactions of an alcohol with a carboxylic acid, reaction (6.34), and an alcohol with an aldehyde, this problem, are shown here for comparison.



C C

O

O

HH

HH

+ O CH

C

H

H H

HH

C C

HH

H

O

O HO

CH

HC

HH

H

C C

HH

H

O

OC

CH H

H

H HH

+ H OH

C

O

O

H

H3C C

O

OH3C

H

H3C O

H

CCH2CH3

O

HH

CH2CH3

H

CH2CH3

In both cases, an electrophilic carbon atom doubly-bonded to oxygen reacts with a nucleophilic nonbonding electron pair on the alcohol oxygen atom to form a new sigma bond between the carbon and oxygen atoms. The electron rearrangement accompanying this interaction displaces one of the electron pairs in the carbon-oxygen double bond onto the oxygen. In the intermediates shown, the oxygen atom with three nonbonding electron pairs has a –1 formal charge and the oxygen atom bonded to three other atoms has a +1 formal charge. In the second reaction, these formal charges are reduced to zero by transfer of a proton from the oxygen with the positive formal charge and an acidic proton to the oxygen with the negative formal charge acting as a base to accept the proton. The result is the hemiacetal product shown. An exactly analogous rearrangement can take place in the intermediate in the first reaction. The result would be:

CH3C

OH

OH

OCH2CH3

Experimentally, we find that molecules (intermediates) with carbon atoms sigma bonded to four other groups, two of which are –OH groups, are unstable and readily lose water to form a carbonyl double bond between the carbon atom and the remaining oxygen atom. These are the products shown in the first reaction, which may actually proceed through this second intermediate to the final product. Thus, the difference between reactions of alcohols with carboxylic acids compared to those with aldehydes and ketones is a result of the intermediate structure with its two –OH groups from the carboxylic acid. This reaction goes on to eliminate a water molecule, a fate that is not possible for the product from the aldehyde or ketone reaction.

Problem 6.40. (a) This intramolecular reaction sequence between an alcohol and aldehyde functional group in the same molecule is modeled after the intermolecular sequence shown in Problem 6.39. To simplify the pictures, we use skeletal structures with only the atoms that enter into the reaction shown explicitly.



CO

HOH

CO O

H

H

CO O

H

H

(b) Skeletal structures for the product from part (a) and for glucose are:

OOH

O

OH

HO

HO OH

OH

The similarities between these structures are: (1) the six-membered ring with an oxygen atom as one of the ring atoms, (2) the –OH group bonded to one of the ring carbon atoms adjacent to the ring oxygen atom, and (3) the non-ring, sixth carbon bonded to the other ring carbon adjacent to the ring oxygen.

Problem 6.41. (a) The glucose ring carbon atom denoted by the arrow in this structure is bonded to two electronegative oxygen atoms:

O

OH

HO

HO OH

OH

This carbon atom acts very much like a carbon atom doubly bonded to an oxygen atom (which it is in the linear form of the glucose molecule). Thus, this electrophilic carbon (Lewis acid) is likely to be the site of reaction between this glucose molecule and an electron pair from another. (b) One of the nonbonding electron pairs on the oxygen atom that is highlighted in the reaction shown will be the pair to interact with the electrophilic center (carbon) in the other molecule to form the product shown in the problem. The highlighted oxygen ends up bonded to this electrophilic carbon:

O

OH

HO

HO OH

OHO

OH

HO

HO OH

OHO

OH

HO

HO OH

OO

OH

HO OH

OH H2O

The oxygen atomic core (and all its valence electrons) originally bonded to the electrophilic carbon ends up in the water product. Loss of this oxygen and formation of water is represented very schematically by the dashed ellipse.

Problem 6.42. (a) Lewis structures for nitrite anion and isoamyl nitrite are:



NOO

O

NO

H2C

CH2

CHCH3

CH3

(b) The reaction between the nucleophilic nitrite anion and the electrophilic carbon center (bonded to the electronegative bromine atom) is:

ON

O

H2C

CH2

CHCH3

CH3N

OO CH2

H2C

CH2

CHCH3

CH3

BrNa+ Na+ BrŠ

nucleophile

electrophilespectator ion

(c) The nitrogen atomic center, with its unshared electron pair, as well as the oxygen atomic centers of the nitrite anion, are possible nucleophiles. (d) Reaction of the other nucleophilic site in the nitrite anion with 3-methyl-1-bromobutane is represented by this reaction that forms 3-methyl-1-nitrobutane.

ON

H2C

CH2

CHCH3

CH3N

OO CH2

H2C

CH2

CHCH3

CH3

BrNa+ Na+ BrŠ

O

(e) There is reason to expect the oxygen atoms in the nitrite ion to be less nucleophilic than the oxygen atom in the hydroxide. In the Lewis structure shown above, the right-hand oxygen core is sigma bonded to another atomic core and has three pairs of unshared electrons, just like the hydroxide ion. However, the pi electrons in the nitrite anion are delocalized and spread over all three atomic centers:

NOO

NO O

Delocalization gives each of the oxygen atomic centers an electron density intermediate between the extremes represented by either Lewis structure, thus making the nitrite anion oxygen centers less nucleophilic than the oxygen in the hydroxide ion.

Problem 6.43. (a) The stoichiometry of the reaction of glycerol, OHCH2CH(OH)CH2OH, two carboxylic acids, R1C(O)OH and R2C(O)OH, choline, HOCH2CH2N(CH3)3

+, and phosphoric acid to produce phosphatidyl choline can be represented as:



R1C

O OP

ON

O

OC

O

R2

OO

R1C

OH

O

R2C

OH

OHO OH

OHN

HOHOP

OH

OO

4H2O

(b) All four of the bonds formed between the units are ester bonds, bonds formed by loss of water between and an acid and an alcohol in an electrophile-nucleophile reaction. The electrophilic carbon atoms in the two carboxylic acids and the electrophilic phosphorus atom in the dihydrogen phosphate ion (which reacts to form a phosphodiester) are highlighted in blue. The nucleophilic alcohol oxygen atoms are highlighted in red.

Problem 6.44. (a) When the four pairs of sigma bonding electrons in NH4

+ are divided equally between the atomic cores they bond, the N has four valence electrons and a core charge of +5, so its formal charge is +1. Each of the H atomic cores has one valence electron and a core charge of +1, so its formal charge is 0. The sum of the formal charges, +1 + 0 + 0 + 0 + 0 = +1, is the charge on the ion, as it must be. (b) Breaking apart a Lewis structure for NO3

– helps to find the formal charges on each atom:

NO

OO

O

OON

–

The topmost O core is surrounded by six valence electrons and has a 6+ core charge, so its formal charge is 0. Both of the lower O cores are surrounded by seven valence electrons and have a 6+ core charge, so they each have a formal charge of –1. The N core is surrounded by four valence electrons and has a 5+ core charge, so its formal charge is +1. The sum of the formal charges, 0 + 1– 1– 1 = –1, is the charge on the ion. [Recall that there are three energy equivalent Lewis structures for the nitrate ion:

NOO

ONO

O

ON

O

OO

Note that it does not matter which of the three equivalent Lewis structures we use to find the formal charges. The result is that two O atoms have –1 formal charge and the third has 0 formal charge. Since all the Os are equivalent, we can “average” the formal charge on each to give all three a –2/3 formal charge. The sum of the formal charges is then 1 – 2/3 – 2/3 – 2/3 = –1, as it



must be. This is a more realistic way to think about the formal charge in a molecule or ion with delocalized electrons.] (c) Breaking apart the Lewis structure for NH3 gives:

N HHH

NH H

H

The N core is surrounded by five valence electrons and has a 5+ core charge, so its formal charge is 0. Each of the H atomic cores has one valence electron and a core charge of +1, so its formal charge is 0. The sum of the formal charges is 0 for this electrically neutral molecule. (d) Breaking apart the Lewis structure for H2O gives:

OHH

OH

H

The O core is surrounded by six valence electrons and has a 6+ core charge, so its formal charge is 0. Each of the H atomic cores has one valence electron and a core charge of +1, so its formal charge is 0. The sum of the formal charges is 0 for this electrically neutral molecule.

Problem 6.45. Lewis structures with formal charges for each atom are shown for five anions:

[NOTE: Most of the solution for Problem 6.46 is also included here, so as to pull together the ideas of formal charge and electron delocalization that spreads formal charge more uniformly over a molecule or ion.] (a) Three energy equivalent Lewis structures are possible for the carbonate anion, CO3

2–:

COO

OCO

O

OC

O

OO

2Š 2Š 2Š-1

-1

00

The formal charge on each atom in any one of these structures is like that shown for the first one. However, as we argued in the solution to Problem 6.44(b), these may be a bit misleading, because delocalization of the pi electrons makes all the O atoms equivalent and each will have a formal charge of –2/3. (In two out of three of the structures, a particular O has a –1 formal charge and in the other structure it has a 0 formal charge, thus averaging to –2/3.) The sum of the formal charges, 0 – 2/3 – 2/3 – 2/3 = –2, is the charge on the anion. (b) Two energy equivalent Lewis structures are possible for the nitrite anion, NO2

–:

NOO

NO O

00 –1 – –

The formal charge on each atom in either of these structures is like that shown for the first one. However, delocalization of the pi electrons makes both the O atoms equivalent and each will have a formal charge of –1/2. (In half of the structures, a particular O has a –1 formal charge and



in the other structure it has a 0 formal charge, thus averaging to –1/2.) The sum of the formal charges, 0 – 1/2 – 1/2 = –1, is the charge on the anion. (c) Four energy equivalent Lewis structures are possible for the phosphate anion, PO4

3–: 3Š

P OO

OO

P OO

OO

PO

OO

O PO

OO

O

3Š 3Š 3Š-1 -1

-1

0

0

The formal charge on each atom in any one of these structures is like that shown for the first one. However, delocalization of the pi electrons makes all the O atoms equivalent and each will have a formal charge of –3/4. (In three out of four of the structures, a particular O has a –1 formal charge and in the fourth structure it has a 0 formal charge, thus averaging to –3/4.) The sum of the formal charges, 0 – 3/4 – 3/4 – 3/4 – 3/4 = –3, is the charge on the anion. (d) Three energy equivalent Lewis structures are possible for the hydrogen phosphate anion, HPO4

2–:

P OO

OOH

POH

OO

O POH

OO

O

2Š 2Š 2Š-1

-1

0

0 0

The formal charge on each atom in any one of these structures is like that shown for the first one. (The formal charge on the H is 0, but, for simplicity, is not shown.) However, delocalization of the pi electrons makes all the O atoms not bonded to hydrogen equivalent and each will have a formal charge of –2/3. (In two out of three of the structures, a particular O has a –1 formal charge and in the other structure it has a 0 formal charge, thus averaging to –2/3.) The sum of the formal charges, 0 + 0 + 0 – 2/3 – 2/3 – 2/3 = –2, is the charge on the anion [just as in the case of the carbonate anion in part (a)]. (e) Two energy equivalent Lewis structures are possible for the dihydrogen phosphate anion, H2PO4

–:

P OO

HOOH

POH

OO

HO

Š Š0

-1

0

0 0

The formal charge on each atom in either of these structures is like that shown for the first one. (The formal charge on each H is 0, but, for simplicity, is not shown.) However, delocalization of the pi electrons makes both the O atoms not bonded to hydrogen equivalent and each will have a formal charge of –1/2. (In half of the structures, a particular O has a –1 formal charge and in the other structure it has a 0 formal charge, thus averaging to –1/2.) The sum of the formal charges, 0 + 0 + 0 + 0 + 0 – 1/2 – 1/2 = –1, is the charge on the anion [just as in the case of the nitrite anion in part (b)].

Problem 6.46. [NOTE: This problem asks if there is more than one possible way to write Lewis structures for any of the anions in the preceding Problem 6.45, and if so, calculate the formal charges for each atom in each structure and explain which structure(s) are most favored. In the solution for



Problem 6.45, energy-equivalent structures for each anion are given for the Lewis structure with the least possible formal charge on the atoms (which, according to our rule in the textbook, is the structure of lowest energy). The consequence of electron delocalization, the spreading of formal charge more uniformly over the anions, is explained as well. The remainder of a solution to the problem here is to examine other Lewis structures that might be possible and explain why they are not favored.]

If we assume that the best Lewis structures for molecules with only second period atoms have octets of electrons (four pairs) on each second row atom, then the only way to write Lewis structures for CO3

2– and NO2– are the ones in Problem 6.45. In each case, energy equivalent

structures were written with the charges on different oxygen atoms, but the structures are the same as far as formal charges are concerned. The oxygen atoms with three nonbonding electron pairs have a formal charge of –1 and all the other atoms have a formal charge of zero. This is the best we can do for formal charge; some atoms in an ion must bear formal charges and the minimum number of formal charges is present in the structures written. If we do not assume octets on all second period atoms, we could write these Lewis structures for CO3

2– and NO2–:

COO

O2Š

-1

-1

+1

-1

N

OO

+1-1 –1 –

In each case, the oxygen atoms with three nonbonding electron pairs have a formal charge of –1 and the central atoms have a +1 formal charge. The net charge on each ion is still what it should be, but the amount of formal charge on the atoms is greater than for the structures in Problem 6.45. In addition, these are the only Lewis structures possible with these electron arrangements, so there is no electron delocalization. These new structures are energetically unfavorable compared to the ones written previously. For the ions with a central phosphate atom, we are not restricted to an octet of electrons on the phosphate atom, as you saw in the structures written in the solution for Problem 6.45. In each case, energy equivalent structures were written with the charges on different oxygen atoms, but the structures are the same as far as formal charges are concerned. The oxygen atoms with three nonbonding electron pairs have a formal charge of –1 and all the other atoms have a formal charge of zero. This is the best we can do for formal charge; some atoms in an ion must bear formal charges and the minimum number of formal charges is present in the structures written. If we assume that the phosphorus atoms should have octets, we can draw these structures:

P OO

OO

3Š-1 -1

-1

-1

+1

POH

OO

O

2Š-1

-1

0

+1

-1

POH

OO

HO

Š0

-1

0

+1

-1

In each case, all the oxygen atoms with three nonbonding electron pairs have a formal charge of –1 and the central atoms have a +1 formal charge. The net charge on each ion is still what it should be, but the amount of formal charge on the atoms is greater than for the structures in Problem 6.45. In addition, these are the only Lewis structures possible with these electron arrangements, so there is no electron delocalization. These new structures are energetically unfavorable compared to the ones written previously.



Problem 6.47. Lewis structures for ozone and nitric acid can be broken apart to show how the formal charges are determined.

OO

O OO

O0 +1

-1

The core charge on O is 6+, which is balanced by the six valence electrons on the left-hand O for a formal charge of 0. The central O has only five valence electrons and a formal charge of +1. The right-hand O has seven valence electrons and a formal charge of –1. The pi electrons are delocalized in the ozone molecule, so the end O atoms each have an average formal charge of –1/2. (See the solutions to Problems 6.44 and 6.45.)

ON

O

O

H ON

O

O

H0

0

0

+1–1

The core charge on O is 6+, which is balanced by the six valence electrons on the left-hand and center O for a formal charge of 0 on these atoms. The right-hand O has seven valence electrons and a formal charge of –1. The N with a core charge of +5 and four valence electrons has a formal charge of +1. The H with a core charge of +1 and one valence electron has a formal charge of 0. The pi electrons are delocalized in the nitric acid molecule, so the two O atoms not bonded to H each have an average formal charge of –1/2.

Problem 6.48. (a) You can tell that a reaction occurs by observing that a solid forms when solutions of Sn2+(aq) and Bi3+(aq) are mixed: 3Sn2+(aq) + 2Bi3+(aq) → 3Sn4+(aq) + 2Bi3+(s) If you determine that it is a metallic solid, Bi(s), that precipitates, you can conclude that the reaction is a reduction-oxidation reaction, because the only way to get the metal from the cation is by reduction (addition of electrons). (b) The element Sn, is oxidized from Sn2+ to Sn4+. Each Sn2+ cation loses two electrons in the reaction. (c) The element Bi is reduced from Bi3+ to Bi0. Each Bi3+ cation gains three electrons in the reaction. (d) Each Sn2+ cation loses two electrons in the reaction. (e) Each Bi3+ cation gains three electrons in the reaction.

Problem 6.49. (a) The evolution of hydrogen gas and the disappearance of the solid metallic sodium are indications that a chemical reaction occurs when a piece of sodium metal, Na(s), is placed in water. The resulting solution is basic, since it turns pink when phenolphthalein, an acid-base indicator is added. Since addition of phenolphthalein to pure water does not produce a pink color, we can conclude that another product of the reaction is hydroxide anion, OH–(aq). Some cation must be formed to balance the charge from OH–(aq), and a logical candidate is the sodium cation, Na+(aq), which also accounts for the disappearance of the sodium metal. The



transformation of Na(s) to Na+(aq, is an oxidation reaction, Na0 → Na+. Some other element in the system must be reduced and it is logical to think of the H in water molecules, which has an oxidation number +1, being reduced to the H in molecular hydrogen, H2(g), which has an oxidation number of 0. (b) The net ionic equation for the reaction described and analyzed in part (a) is: 2Na(s) + 2H2O(l) → 2Na+(aq) + H2(g) + 2OH–(aq)

(c) The element sodium, Na(s), is the reactant that is oxidized, as shown in part (a). (d) Hydrogen atomic cores from two water molecules are reduced to H2(g), (also producing two hydroxide anions).

Problem 6.50. (a) The change Zn → Zn2+ is an oxidation, because the oxidation number increases going from Zn (the element with oxidation number 0) to Zn2+ (monatomic ion with oxidation number +2). (b) The change S2– → S is a reduction, because the oxidation number decreases going from S2– (monatomic ion with oxidation number –2) to S (the element with oxidation number 0). (c) The change Fe3+ → Fe2+ is a reduction, because the oxidation number decreases going from Fe3+ (monatomic ion with oxidation number +3) to Fe2+ (monatomic ion with oxidation number +2). (d) The change Ag+ → Ag is a reduction, because the oxidation number decreases going from Ag+ (monatomic ion with oxidation number +1) to Ag (the element with oxidation number 0).

Problem 6.51. (a) Balanced oxidation-reduction reactions are written for each of these observations:

(i) Mg metal reacts with Zn2+(aq) to yield Mg2+(aq) and Zn metal. Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)

(ii) Zn metal reacts with Cu2+(aq) to yield Zn2+(aq) and Cu metal. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

(iii) Mg metal reacts with Cu2+(aq) to yield Mg2+(aq) and Cu metal. Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

(b) Mg(s) is oxidized in reactions with both Zn2+(aq) and Cu2+(aq), observations (i) and (iii). Thus, we conclude that Mg(s) is the most likely metal (among these three) to undergo an oxidation, (c) Cu2+(aq) is reduced in reactions with both Zn(s) and Mg(s), observations (ii) and (iii). Thus, we conclude that Cu2+(aq)) is the most likely metal cation (among these three) to undergo a reduction,

Problem 6.52. For these reactions we (i) assign oxidation numbers to each atom, (ii) identify the element that is oxidized, (iii) identify the element that is reduced. (a) V2O5 + 2H2 → V2O3 + 2H2O

(i) In the reactants, V = +5, O = –2, and H = 0. In the products, V = +3, O = –2, H = +1, and O = –2



(ii) H goes from 0 → +1, an increase in oxidation number, so it is oxidized. (iii) V goes from +5 → +3, a decrease in oxidation number, so it is reduced.

(b) 2K + Br2 → 2K+ + 2Br- (i) In the reactants, K = 0, and Br = 0. In the products, K = +1 and Br = –1. (ii) K goes from 0 → +1, an increase in oxidation number, so it is oxidized. (iii) Br goes from 0 → –1, a decrease in oxidation number, so it is reduced.

(c) N2 + 3H2 → 2NH3 (i) In the reactants, N = 0, and H = 0. In the products, N = –3 and H = +1. (ii) H goes from 0 → +1, an increase in oxidation number, so it is oxidized. (iii) N goes from 0 → –3, a decrease in oxidation number, so it is reduced.

Problem 6.53. (a) The oxidation number for the sulfur atom in sulfur trioxide, SO3, is +6. as we can see by breaking apart the molecule and assigning all bonding electrons to the most electronegative atom and then finding the charge on the atomic core plus valence electrons for each core:

SO O

OS

O O

O Š2

Š2Š2 +6

Note that the structure here has formal charges of 0 on all the atoms. It is possible to write a Lewis structure with an octet of electrons on the S, but this would give formal charges of –1 on two O atoms and +2 on the S atom, which is unfavorable relative to this one. No matter which structure you start with, all the valence electrons end up on the O atoms, just as here, when you apply the rules to get the oxidation numbers in the molecule. (b) The oxidation number for the sulfur atom in the sulfite ion, SO3

2– is +4. Let’s apply the alternative set of rules for assigning oxidation numbers to get this value. The ON for each of the three O atoms is –2. The total for the ONs has to sum to the charge on the ion, so we have: ion charge = –2 = 3(–2) + ON(sulfur); ∴ ON(sulfur) = +4 The oxidation number for the sulfur atom in the sulfate ion, SO4

2– is +6: ion charge = –2 = 4(–2) + ON(sulfur); ∴ ON(sulfur) = +6

(c) The dissolution of sulfur trioxide in water to give an acidic solution is not a redox reaction. This means that the S in the acid produced has to have the same oxidation number, +6, as the S in sulfur trioxide. The S in the sulfate ion has an oxidation number of +6, so the acid produced must be sulfuric: SO3(g) + H2O(l) → H2SO4(aq)

Problem 6.54. (a) The oxidation number is –3 for the N in NH3. N is more electronegative than H, so all the bonding electrons (plus its nonbonding pair) are assigned to N, which gives it eight electrons (charge –8) and core charge of +5. Thus, the oxidation number is –3 (= 5 – 8). (b) The oxidation number is +2 for the N in NO. Because this molecule has an odd number of valence electrons (11), no Lewis structure can be written with all the electrons paired and in



octets on these second row atoms. Since O is more electronegative than N, it will take all the electrons it can (eight of the 11) in the imaginary separation of the atoms we use to assign oxidation numbers. This leaves N with three valence electrons (charge –3) and core charge of +5, for an oxidation number of +2. The same result is obtained if you apply the alternative method for assigning oxidation numbers. (c) The oxidation number is +3 for the N in NO2

–. The Lewis structure(s) for the nitrite anion is shown in the solutions for Problems 6.42, 6.45, and 6.46. In all the structures, the N atom has a nonbonding pair of electrons and the others are shared with the O atoms. In the imaginary separation of he atoms we use to assign oxidation numbers, all the electrons, except the unshared pair on N will go to O. This leaves N with two valence electrons (charge –2) and core charge of +5, for an oxidation number of +3. The same result is obtained if you apply the alternative method for assigning oxidation numbers. (d) The oxidation number is +5 for the N in NO3

–. The Lewis structure(s) for the nitrate ion is shown in the solution for Problem 6.44. All the valence electrons on the N are shared with O atoms, so in the imaginary separation of he atoms we use to assign oxidation numbers, all the electrons go to O. This leaves N with no valence electrons and core charge of +5, for an oxidation number of +5. The same result is obtained if you apply the alternative method for assigning oxidation numbers. (e) The oxidation number for N in some common molecules and ions, parts (a) through (d), ranges from –3 to +5. The location of nitrogen near the middle of the second period of the periodic table helps to explain this wide range of possible oxidation numbers. The atomic core can be assigned a maximum of eight electrons (the maximum for a second-period element) when it is bonded to less electronegative atoms in molecules. This gives N an oxidation number of –3. When an N atomic core shares all its valence electrons with more electronegative atoms in a molecule, all of the valence electrons can be assigned to the other atoms, leaving N with none and an oxidation number of +5. Intermediate values are possible, as you see in parts (b) and (c). [What is the oxidation number of N in H2NOH, NO2, and H2NNH2, a few more common N-containing molecules.]

Problem 6.55. To find the oxidation number of the metal in a metal-ion complex, we take advantage of the fact that metals (Lewis acids) have low electronegativities and, in the ligands (Lewis bases), the atom bonded to the metal has a high electronegativity. Thus, when the bonding electrons are partitioned between the metal and the ligand to assign oxidation numbers, the ligand will get all the electrons. The ligands will end up with the charge they have as free molecules or ions and the oxidation number of the metal will be whatever is required to give the correct charge on the complex. (a) The free ligands in PtCl4

2– are chloride ions, Cl–, so Pt has to have an oxidation number of +2 to give a complex with a 2– charge. (b) The free ligands in Cu(NH3)4

2+ are ammonia molecules, NH3, so Cu has to have an oxidation number of +2 to give a complex with a 2+ charge. (c) The free ligands in Fe(CN)6

3– are cyanide anions, CN–, so Fe has to have an oxidation number of +3 to give a complex with a 3– charge. (d) The free ligands in MoS4

2– are sulfide anions, S2–, so Mo has to have an oxidation number of +6 to give a complex with a 2– charge.



(e) The free ligands in Zn(OH)42– are hydroxide anions, OH–, so Zn has to have an oxidation

number of +2 to give a complex with a 2– charge.

Problem 6.56. [NOTE: Both the oxidation-number and half-reaction methods for balancing oxidation-reduction reactions are presented in the text with Worked Examples and Check This problems for practice. For the reactions here, we give the reasoning for determining the reactant that is reduced and the one that is oxidized, and then give the balanced redox reaction but do not show the balancing in detail.] (a) MnO4

–(aq) + SO32–(aq) + H+(aq) → Mn2+(aq) + SO4

2–(aq) + H2O Mn in the MnO4

–(aq) reactant has ON = +7 and Mn in the Mn2+(aq) product has ON = +2, so the Mn is reduced and gains 5 electrons in the reaction. S in the SO3

2–(aq) reactant has ON = +4 and S in the SO4

2–(aq) product has ON = +6, so the S is oxidized and loses 2 electrons in the reaction. We require two Mn reactants for every five S reactants, in order to balance the electron transfer. The balanced redox equation is: 2MnO4

-(aq) + 5SO32-(aq) + 6H+(aq) → 2Mn2+(aq) + 5SO4

2-(aq) + 3H2O

(b) NO3–(aq) + Zn(s) + H+(aq) → NH4

+(aq) + Zn2+(aq) + H2O N in the NO3

–(aq) reactant has ON = +5 and N in the NH4+(aq) product has ON = –3, so the N is

reduced and gains 8 electrons in the reaction. Zn in the Zn(s) reactant has ON = 0 and Zn in the Zn2+(aq) product has ON = +2, so the Zn is oxidized and loses 2 electrons in the reaction. We require one N reactant for every four Zn reactants, in order to balance the electron transfer. The balanced redox equation is: NO3

-(aq) + 4Zn(s) + 10H+(aq) → NH4+(aq) + 4Zn2+(aq) + 3H2O

(c) Cl2(g) + OH–(aq) → ClO3–(aq) + Cl–(aq) + H2O

This reaction is a bit complicated to analyze because some of the reactant chlorine, Cl2(g), is oxidized and some of it is reduced, [This is called disproportionation.] To make reaction easier to analyze, we will rewrite it with the two chlorine molecules shown as reactants. We will assume that the first is reduced and the second oxidized in the reaction: Cl2(g) + Cl2(g) + OH–(aq) → ClO3

–(aq) + Cl–(aq) + H2O Cl in the first Cl2(g) reactant has ON = 0 and Cl in the Cl–(aq) product has ON = –1, so the Cl is reduced and gains 1 electron in the reaction. Cl in the second Cl2(g) reactant has ON = 0 and Cl in the ClO3

–(aq) product has ON = +5, so the Cl is oxidized and loses 5 electrons in the reaction. Since each of the Cl2(g) reactants has two Cl atoms to gain or lose electrons, we might consider multiplying the electron gains and losses by two, but this is not necessary in this case, because it would not change the ratio of gain to loss. We require five of the first Cl reactant for every one of the second Cl reactant, in order to balance the electron transfer. The balanced redox equation is: 5Cl2(g) + Cl2(g) + 12OH–(aq) → 2ClO3

–(aq) + 10Cl–(aq) + 6H2O Now we can combine the Cl2(g) reactants, since there is not really any way to differentiate them, and then divide through the entire equation by two, to reduce the coefficients to their lowest possible integer values (as we conventionally show balanced equations). 3Cl2(g) + 6OH-(aq) → ClO3

-(aq) + 5Cl-(aq) + 3H2O



Problem 6.57. (a) In a zinc-air cell, oxygen gas, O2, is converted in aqueous solution to hydroxide anion, OH– and zinc metal is converted to zinc oxide, ZnO(s), during discharge of the cell. In this cell, zinc metal is being oxidized, going from an oxidation state of 0 in Zn(s) to +2 in ZnO(s). Oxygen gas is being reduced, going from an oxidation state of zero in O2(g) to –2 in ZnO(s) (and OH–). (b) The net (overall) reaction in this cell is 2Zn(s) + O2(g) → 2ZnO(s). You might, however, wonder what happened to the hydroxide anion that is said to have been formed. We can break the process into two half reactions, whose sum is the net reaction just written: O2(g) + 2H2O(l) + 4e– → 4OH–(aq) 2Zn(s) + 4HO–(aq) → 2ZnO(s) + 2H2O(l) + 4e–

(c) For each mole of zinc undergoing reaction, two moles of electrons have to be transferred from the zinc to oxygen in order to oxidize the zinc metal to zinc(II) ion. You can also see this stoichiometry in the half reactions where four moles of electrons are transferred for every two moles of zinc metal reacted, or, again, two moles of electrons per mole of zinc.

Problem 6.58. [NOTE: Both the oxidation-number and half-reaction methods for balancing oxidation-reduction reactions are presented in the text with Worked Examples and Check This problems for practice. For the reactions here, we give the reasoning for determining the reactant that is reduced and the one that is oxidized, and then give the balanced redox reaction but do not show the balancing in detail. Both methods give the same result, as they must. Preference for one or the other is a matter of taste.]

(a) S2–(aq) + NO3–(aq) → NO2(g) + S8(s) (S8 rings are the common form of elemental

sulfur.) N in the NO3

–(aq) reactant has ON = +5 and N in the NO2(g) product has ON = +4, so the N is reduced and gains 1 electron in the reaction. S in the S2–(aq) reactant has ON = –2 and S in the S8(s) product has ON = 0, so the S is oxidized and loses 2 electrons per sulfur atom in the reaction. Since eight sulfide anions have to be oxidized to form one S8(s), we require 16 N reactants for every eight S reactants, in order to balance the electron transfer. The balanced redox equation is: 8S2–(aq) + 16NO3

–(aq) + 32H+(aq) → 8NO2(g) + S8(s) + 16H2O

(b) Hg2+(aq) + NO2–(aq) → Hg(l) + NO3

–(aq) Hg in the Hg2+(aq) reactant has ON = +2 and Hg in the Hg(l) product has ON = 0, so the Hg is reduced and gains 2 electrons in the reaction. N in the NO2

– (aq) reactant has ON = +3 and N in the NO3

–(aq) product has ON = +5, so the N is oxidized and loses 2 electrons in the reaction. We require one Hg reactant for every N reactant, in order to balance the electron transfer. The balanced redox equation is: Hg2+(aq) + NO2

–(aq) + H2O → Hg(l) + NO3–(aq) + 2H+(aq)

(c) Fe2+(aq) + NO3–(aq) → Fe3+(aq) + NO(g)

N in the NO3–(aq) reactant has ON = +5 and N in the NO(g) product has ON = +2, so the N is

reduced and gains 3 electrons in the reaction. Fe in the Fe2+(aq) reactant has ON = +2 and Fe in the Fe3+(aq) product has ON = +3, so the Fe is oxidized and loses 1 electron in the reaction. We



require one N reactant for every three Fe reactants, in order to balance the electron transfer. The balanced redox equation is: 3Fe2+(aq) + NO3

–(aq) + 4H+(aq) → 3Fe3+(aq) + NO(g) + 2H2O

Problem 6.59. [NOTE: For the reactions here, we give the reasoning for determining the reactant that is reduced and the one that is oxidized, and then give the balanced redox reaction in acidic solution followed by the balanced reaction in basic solution but do not show the balancing or addition of OH–(aq) in detail.] (a) S2–(aq) + I2(s) → SO4

2–(aq) + I–(aq) I in the I2(s) reactant has ON = 0 and I in the I–(aq) product has ON = –1, so the I is reduced and gains 2 electrons per I2(s) in the reaction. S in the S2–(aq) reactant has ON = –2 and Fe in the SO4

2–(aq) product has ON = +6, so the S is oxidized and loses 8 electrons in the reaction. We require four I2(s) reactant for every S2–(aq) reactant, in order to balance the electron transfer. The balanced redox equations (in acid —imagined— and in base) are: S2–(aq) + 4I2(s) + 4H2O(l) → SO4

2–(aq) + 8I–(aq) + 8H+(aq) S2–(aq) + 4I2(s) + 8OH-(aq) → SO4

2–(aq) + 8I–(aq) + 4H2O(l)

(b) MnO4–(aq) + C2O4

2–(aq) → MnO2(s) + CO2(g) Mn in the MnO4

–(aq) reactant has ON = +7 and Mn in the MnO2(s) product has ON = +4, so the Mn is reduced and gains 3 electrons in the reaction. C in the C2O4

2–(aq) reactant has ON = +3 and C in the CO2(g) product has ON = +4, so the C is oxidized and loses 2 electrons per C2O4

2-(aq) in the reaction. We require two MnO4–(aq) reactants for every three C2O4

2–(aq) reactants, in order to balance the electron transfer. The balanced redox equations (in acid —imagined— and in base) are: 2MnO4

–(aq) + 3C2O42–(aq) + 8H+(aq) → 2MnO2(s) + 6CO2(g) + 4H2O(l)

2MnO4–(aq) + 3C2O4

2–(aq) + 4H2O(l) → 2MnO2(s) + 6CO2(g) + 8OH-(aq)

(c) Bi(OH)3(s) + Sn(OH)3–(aq) → Bi(s) + Sn(OH)6

2–(aq) Bi in the Bi(OH)3(s) reactant has ON = +3 and Bi in the Bi(s) product has ON = 0, so the Bi is reduced and gains 3 electrons in the reaction. Sn in the Sn(OH)3

–(aq) reactant has ON = +2 and Sn in the Sn(OH)6

2–(aq) product has ON = +4, so the Sn is oxidized and loses 2 electrons in the reaction. We require two Bi(OH)3(s) reactants for every three Sn(OH)3

–(aq) reactants, in order to balance the electron transfer. The balanced redox equations (in acid —imagined— and in base) are: 2Bi(OH)3(s) + 3Sn(OH)3

–(aq) + 3H2O(l) → 2Bi(s) + 3Sn(OH)62–(aq) + 3H+(aq)

2Bi(OH)3(s) + 3Sn(OH)3–(aq) + 3OH-(aq) → 2Bi(s) + 3Sn(OH)6

2–(aq)

Problem 6.60. (a) The reaction that occurs when solid cobalt(II) sulfide dissolves in nitric acid can be represented as: CoS(s) + NO3

–(aq) → Co2+(aq) + NO(g) + S8(s) (S8 rings are the common form of elemental sulfur.)



N in the NO3–(aq) reactant has ON = +5 and N in the NO(g) product has ON = +2, so the N is

reduced and gains 3 electrons in the reaction. S in the CoS(s) reactant has ON = –2 and S in the S8(s) product has ON = 0, so the S is oxidized and loses 2 electrons per sulfur atom in the reaction. (b) If we imagine that elemental sulfur is S(s), then we require 2 NO3

–(aq) reactants for every three CoS(s) reactants, in order to balance the electron transfer. The balanced redox equation and its eight-fold multiple to account for S8(s) are: 3CoS(s) + 2NO3

–(aq) + 8H+(aq) → 3Co2+(aq) + 2NO(g) + 3S(s) + 4H2O 24CoS(s) + 16NO3

–(aq) + 64H+(aq) → 24Co2+(aq) + 16NO(g) + 3S8(s) + 32H2O

Problem 6.61. To determine whether the conversion of lactic acid, CH3CH(OH)COOH, to pyruvic acid, CH3C(O)COOH, is an oxidation or a reduction reaction we need to focus attention on the part of the molecule that changes, the bonding about the central carbon atomic core. In lactic acid, this carbon is bonded to two other carbon atoms, a hydrogen atom, and an oxygen atom. Using our rules for oxidation numbers, we assign ON = +1 to this carbon. In pyruvic acid, this carbon is bonded to two other carbons and doubly bonded to an oxygen atom. Using our rules for oxidation numbers, we assign ON = +2 to this carbon. Since the oxidation number has increased, this change is an oxidation.

Problem 6.62. The reaction of interest, ethanol in hard cider reacting with oxygen from the air to form ethanoic acid, is an oxidation of ethanol to ethanoic acid: CH3CH2OH(aq) → CH3C(O)OH(aq) The ethanol has to gain an oxygen and lose two hydrogen atomic cores to undergo this oxidation. If an oxide ion is added on the left and two hydronium ions are added on the right, we can write the reaction as a half reaction: CH3CH2OH(aq) + O2–(aq) → CH3C(O)OH(aq) + H+(aq) + 4e– The problem states that the oxidant is oxygen from the air, which is reduced in the reaction, as we can represent in another half reaction: O2(g) + 4e– → 2O2–(aq) The sum of this oxidation and reduction (after canceling an oxide anion from each side is) is: CH3CH2OH(aq) + O2(g) → CH3C(O)OH(aq) + 2H+(aq) + O2–(aq) The two hydronium ions and the oxide ion will, of course, react to form water. The balanced net reaction for making cider vinegar is: CH3CH2OH(aq) + O2(g) → CH3C(O)OH(aq) + 2H2O(aq)

Problem 6.63. (a) Three separate equations showing the formation of carbon dioxide, carbon monoxide, and free carbon particles (plus water) when glucose in wood burns with oxygen gas from the air are: C6H12O6 + 6O2 → 6CO2 + 6H2O C6H12O6 + 3O2 → 6CO + 6H2O C6H12O6 → 6C + 6H2O



(b) All three reactions can take place at the same time. The first reaction produces nontoxic carbon dioxide and water. The second reaction uses less oxygen, and produces toxic carbon monoxide. In poorly ventilated areas the reaction can use up oxygen faster than it is replaced, thus tipping the balance of the reactions toward those that require less oxygen. Thus, more carbon monoxide (and soot) would be produced and the lack of ventilation would permit its concentration to build up, creating a dangerous situation.

Problem 6.64. [NOTE: Based on the rules for oxidation numbers that we have adopted in the text, we identify each of these reactions (incomplete equations) as an oxidation, a reduction, or neither with respect to the carbon atom on the right in each structure. This identification is not universally accepted. Another convention that assigns and oxidation number of +1 to all H atoms and –2 to all O atoms in a carbon-containing molecule leads to different interpretations in some cases. This convention is sometimes useful for looking at the overall oxidation state of a molecule, but often not useful for looking at the oxidation state of single atoms in a molecule.]

(a)

H2C CH2 H3C CO

OH The C in the reactant is bonded to two H atoms and doubly bonded to another carbon atom (which counts as bonding to two C atoms in our convention), which gives ON = 0. The C in the product is singly bonded to another C and an O atom, and doubly bonded to an O atom (which counts as bonding to two O atoms), which gives ON = +3. This is an oxidation of this carbon. The alternative convention can be applied to each carbon individually. In the reactant, both carbons are the same, =CH2. If the H atoms are assigned ON = +1, the C must have ON = –2. In the product, the right-hand C is part of a –C(O)OH group in which the C has to have ON = +3 in order to give the group zero charge. This is an oxidation of this carbon, just as in the previous paragraph. If we take account of the other C in the reactant and product, sum of the C oxidation numbers in the reactant is –4. In the product, the left-hand C is part of a –CH3 group in which the C has an ON = –3. The sum of the C oxidation numbers for the product is 0 [= (=3) + (–3)]. Again, we conclude that the reaction is an oxidation, but the focus has shifted from an individual carbon atom to the whole molecule.

(b)

H2C CH2 H2C CH2

HO

OH The C in the reactant is bonded to two H atoms, a C atom, and O atom, which gives ON = +1. The C in the product is singly bonded to two H atoms and doubly bonded to another C atom (which counts as bonding to two C atoms), which gives ON = 0. This is a reduction of this carbon.

(c)

H3C C H2C CH2H

O

The C in the reactant is bonded to an H atom, a C atom, and doubly bonded to an O atom (which counts as bonding to two O atoms), which gives ON = +2. The C in the product is singly bonded to two H atoms and doubly bonded to another C atom (which counts as bonding to two C atoms), which gives ON = 0. This is a reduction of this carbon.



(d) H3C CH2H2C CH2

OH

The C in the reactant is bonded to two H atoms and doubly bonded to another carbon atom (which counts as bonding to two C atoms in our convention), which gives ON = 0. The C in the product is singly bonded to another C, two H atoms, and an O atom, which gives ON = +1. This is an oxidation of this carbon.

Problem 6.65. (a) The oxidation number of manganese in MnO4

–(aq), Mn2+(aq), MnO2(s), and MnO42–(aq) is,

respectively, +7, +2, +4, and +6. If MnO4–(aq) is used to oxidize other substances, it must itself

be reduced as electrons are transferred from the substance oxidized. The latter three manganese species are product of these reactions under various conditions and, in each case, the oxidation number of the manganese goes down, that is, the manganese is reduced, as it must be when permanganate, ON = +7 is used as an oxidant. (b) The reduction half-reaction for each of the reactions of permanganate anion acting as an oxidizing agent is: MnO4

–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(aq) MnO4

–(aq) + 4H2O(aq) + 3e– → MnO2(s) + 4OH–(aq) MnO4

–(aq) + e– → MnO42–(aq)

(c) The color change from intense pink of the MnO4–(aq) anion to colorless (or almost colorless)

Mn2+(aq) cation (formed as the reduction product in the titrated solution) can used as a visual indicator of the endpoint of a permanganate oxidation in acidic solutions. Generally the reaction proceeds until one drop of the intense pink titrant [MnO4

–(aq)] changes the colorless solution being titrated to pale pink. In neutral or slightly basic solution, formation of brown solid MnO2 could serve to signal that oxidation is taking place, although it is difficult to determine when the solid stops forming and the brown solid obscures the disappearance of the intense pink color. The color change in highly alkaline solutions may provide an endpoint, as a drop of the intense pink titrant changes the pale green color of the solution being titrated [pale green because of the reduction product MnO4

2–(aq)]. Of course, these visual color methods will not work if the material being oxidized undergoes a color change itself.

Problem 6.66. (a) Each chromium atom in the dichromate anion has ON = +6 each.

Cr O Cr OOO

O

O

O

Oxygen is more electronegative than any metal so, all the electrons will be assigned to the O atoms when oxidation numbers are assigned to the atoms in dichromate. Each oxygen will have 8 electrons for an oxidation number of –2 (= 6 – 8). One way to get the oxidation number of the Cr atoms is to set the sum of the oxidation numbers to the charge on the ion: ion charge = –2 = 7(–2) + 2ONCr; ∴ ONCr = +6 Another way to find ONCr is to use the information in the problem that the chromium atom has six valence electrons, which gives the atom a core charge of +6. If, when we assign electrons to



get the oxidation numbers, the Cr atoms are left with no valence electrons (the O atoms have taken them all), each has an oxidation number of +6 (= 6 – 0). (b) Each Cr in the dichromate anion has an ON = +6 and ends up as a Cr3+ cation, with an ON = +3. Each Cr gains three electrons in the reaction. The dichromate must gain six electrons, in order to reduce both Cr atoms. To balance this gain of electrons, the iron, oxidizing from ON = +2 to ON = +3, must lose six electrons; so there must be six irons in the balanced equation:

Cr2O72Š(aq) + 6Fe2+(aq) 2Cr3+(aq) + 6Fe3+(aq)

gain 6 electrons

lose 6 electrons

This equation is balanced in electron transfers, but not in atoms. Seven oxygen atoms are needed on the right; add seven water molecules. This also adds 14 hydrogen atoms on the right; add 14 hydronium cations on the left to balance the hydrogen: Cr2O7

2–(aq) + 6Fe2+(aq) + 14H+(aq) → 2Cr3+(aq) + 6Fe3+(aq) + 7H2O(aq)

(c) This balanced equation gives the stoichiometric relationship between dichromate and iron(II): 1 mole of dichromate reacts with 6 moles of iron(II). The titration used 0.03457 L (34.57 mL) of 0.0100 M dichromate; the number of moles of dichromate is: (0.03457 L)(0.0100 mol·L–1) = 3.457 × 10–4 mol dichromate The number of moles of iron(II) in the sample is:

(3.457 × 10–4 mol dichromate)6 mol iron(II)

1 mol dichromate⎛ ⎝

⎞ ⎠ = 2.07 × 10–3 mol iron(II)

The mass of iron that is equivalent to this number of moles is:

(2.07 × 10–3 mol iron(II))55.85 g iron1 mol iron

⎛ ⎝

⎞ ⎠ = 0.116 g iron

The mass percent of iron in the ore sample is:

0.116 g iron0.178 g ore

⎛⎝⎜

⎞⎠⎟

100% = 65.2% iron in the ore

Problem 6.67. Apparently, in the presence of hydroxide ion, OH–(aq), dichromate ion reacts to produce chromate ion: Cr2O7

2–(aq) + OH–(aq) → CrO42–(aq)

orange yellow The oxidation number of the chromium atoms in dichromate is +6 (See Problem 6.66). Likewise, the oxidation number of chromium in chromate ion is +6. This reaction is not an oxidation-reduction reaction. To balance the reaction, two chromate ions are produced for each dichromate that reacts: Cr2O7

2–(aq) + OH–(aq) → 2CrO42–(aq) + H+(aq)

The hydronium ion product is required to balance the equation in terms of hydrogen. This equation is not, however, a good representation of the chemistry of the system. In basic solution,



the hydronium immediately reacts with hydroxide to produce water. To balance the equation “chemically,” add a hydroxide on each side and let it react with the hydronium on the right: Cr2O7

2–(aq) + 2OH–(aq) → 2CrO42–(aq) + H2O(l)

This reaction is not an oxidation-reduction and certainly not a precipitation. It must be a Lewis acid-base reaction. The Lewis base (nucleophilic nonbonding electron pair on hydroxide) attracts the Lewis acid (electrophilic chromium in dichromate):

Cr O Cr OOO

O

O

O (aq) (aq)(aq)O H Cr O Cr OO

O

O

O

O (aq)OH

In basic solution, the hydrogen (proton) on the hydrogen chromate ion will react with base (another Lewis acid-base reaction) to form water. Adding that reaction to this one gives the previous balanced reaction equation. The reaction is readily reversed by adding acid, which converts the chromate to dichromate.

Problem 6.68. The responses here are based on the data in this table:

Metal Ion Ratio of Ionic Charge/Ionic Radius

Li+ 1.5 Ca2+ 2.1 Mg2+ 3.1 Al3+ 6.7

(a) The process of a metal ion interacting with water involves the metal ion acting as a Lewis acid and water acting as a Lewis base. The strength of this interaction varies with concentration of the charge on the Lewis acid. If the charge is spread out over a large ion, the interaction is decreased. Metal ions that are strong Lewis acids have a high charge to radius ratio because this means the charge will be concentrated on a small ion. In this list, the Al3+ ion has the highest charge to radius ratio and will be the strongest Lewis acid. Indeed, Al3+ compounds are often used in reactions in which a strong Lewis acid is required, for example, as a catalyst, (b) Magnesium and calcium are both members of group II on the periodic table. Both ions have the same charge, 2+. However, the calcium ion (period 4), is larger than the magnesium ion (period 3), which reduces the ratio of charge to radius, as shown in the table. (c) Sodium ion, Na+, does not act as a Lewis acid with water. This means the cation must have a low charge density, low charge to radius ratio. Sodium and lithium are both in group I on the periodic table and both form +1 ions. Since Na is in period 3 and Li in period 2, Na+ will have a larger radius and a lower charge to radius ratio than 1.5, the value for Li+ in the above table. We might predict a value close to 1. (The actual value given in the literature is 1.0.)

Problem 6.69. (a) As molten nylon polymer flows through the very fine holes in the spinnerets shown in the chapter opening illustration the molecules represented in the Web Companion, Chapter 6, Section 6.7, page 2, are forced to come out more or less lined up so the long molecules can get through the opening (rather like a thread passing through the eye of a needle). Thus the



molecules are lying side-by-side and hydrogen bonds are readily formed between adjacent molecules, which help to account for the strength of nylon fibers. (b) Spiders produce several different relatively simple protein polymers that are mixed in various ratios to produce more complex polymers with different properties, because this is a more efficient use of their DNA and it allows them to store these ingredients in easily usable liquid suspensions. If each complex polymer had to be coded separately by its DNA, the spider would require many very long sequences of DNA that were highly repetitive from one coding gene to another. By synthesizing several smaller polymers that can be mixed and matched to make the final filament, a much smaller amount of DNA is required. Also, if much longer polymers were synthesized, they would probably be a good deal less soluble and would probably have to be made as needed (a relatively slow process), instead of stored for quick use.

Solutions for Chapter 6 End-of-Chapter Problemsquantum.bu.edu/courses/ch102-summer-2010/text/... · ACS Chemistry Chapter 6 suggested solutions 1 Solutions for Chapter 6 End-of-Chapter

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