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Solutions Manual Chapter11

Oct 23, 2014




CHAPTER 11 PROPERTIES OF SOLUTIONSSolution Review585 g C3 H 7 OH 11.

1 mol C3H 7 OH 60.09 g C3H 7 OH = 9.74 M 1.00 L


0.250 L

0.100 mol 134.00 g = 3.35 g Na2C2O4 L mol


1.00 L

1L 0.040 mol HCl = 0.040 mol HCl; 0.040 mol HCl = 0.16 L L 0.25 mol HCl = 160 mL 1 mol CaCl 2 1L 1000 mL = 19.9 mL 110.98 g CaCl 2 0.580 mol CaCl 2 L3.0 mol Na 2 CO 3 = 0.21 mol Na2CO3 L


1.28 g CaCl2


Mol Na2CO3 = 0.0700 L

Na2CO3(s) 2 Na+(aq) + CO32(aq); mol Na+ = 2(0.21) = 0.42 mol Mol NaHCO3 = 0.0300 L 1.0 mol NaHCO 3 = 0.030 mol NaHCO3 L

NaHCO3(s) Na+(aq) + HCO3(aq); mol Na+ = 0.030 mol

M Na + = 16.

total mol Na + 0.42 mol + 0.030 mol 0.45 mol = = = 4.5 M Na+ total volume 0.0700 L + 0.030 L 0.1000 L b. Na2SO4(s) 2 Na+(aq) + SO42(aq) d. SrBr2(s) Sr2+(aq) + 2 Br(aq) f. NH4Br(s) NH4+(aq) + Br(aq)

a. HNO3(l) H+(aq) + NO3(aq) c. Al(NO3)3(s) Al3+(aq) + 3 NO3(aq) e. KClO4(s) K+(aq) + ClO4(aq) g. NH4NO3(s) NH4+(aq) + NO3(aq) i. NaOH(s) Na+(aq) + OH(aq)

h. CuSO4(s) Cu2+(aq) + SO42(aq)





Questions17. As the temperature increases, the gas molecules will have a greater average kinetic energy. A greater fraction of the gas molecules in solution will have a kinetic energy greater than the attractive forces between the gas molecules and the solvent molecules. More gas molecules are able to escape to the vapor phase, and the solubility of the gas decreases. Henrys law is obeyed most accurately for dilute solutions of gases that do not dissociate in or react with the solvent. NH3 is a weak base and reacts with water by the following reaction: NH3(aq) + H2O(l) NH4+(aq) + OH(aq) O2 will bind to hemoglobin in the blood. Due to these reactions in the solvent, NH3(g) in water and O2(g) in blood do not follow Henrys law. 19. Because the solute is volatile, both the water and solute will transfer back and forth between the two beakers. The volume in each beaker will become constant when the concentrations of solute in the beakers are equal to each other. Because the solute is less volatile than water, one would expect there to be a larger net transfer of water molecules into the right beaker than the net transfer of solute molecules into the left beaker. This results in a larger solution volume in the right beaker when equilibrium is reached, i.e., when the solute concentration is identical in each beaker. Solutions of A and B have vapor pressures less than ideal (see Figure 11.13 of the text), so this plot shows negative deviations from Raults law. Negative deviations occur when the intermolecular forces are stronger in solution than in pure solvent and solute. This results in an exothermic enthalpy of solution. The only statement that is false is e. A substance boils when the vapor pressure equals the external pressure. Because B = 0.6 has a lower vapor pressure at the temperature of the plot than either pure A or pure B, one would expect this solution to require the highest temperature in order for the vapor pressure to reach the external pressure. Therefore, the solution with B = 0.6 will have a higher boiling point than either pure A or pure B. (Note that because PB > PA, B is more volatile than A, and B will have a lower boiling point temperature than A). No, the solution is not ideal. For an ideal solution, the strengths of intermolecular forces in solution are the same as in pure solute and pure solvent. This results in Hsoln = 0 for an ideal solution. Hsoln for methanol-water is not zero. Because Hsoln < 0 (heat is released), this solution shows a negative deviation from Raoults law. The micelles form so that the ionic ends of the detergent molecules, the SO4 ends, are exposed to the polar water molecules on the outside, whereas the nonpolar hydrocarbon chains from the detergent molecules are hidden from the water by pointing toward the inside of the micelle. Dirt, which is basically nonpolar, is stabilized in the nonpolar interior of the micelle and is washed away. See the illustration on the following page.








= detergent molecule = SO4= nonpolar hydrocarbon

= dirt


Normality is the number of equivalents per liter of solution. For an acid or a base, an equivalent is the mass of acid or base that can furnish 1 mole of protons (if an acid) or accept 1 mole of protons (if a base). A proton is an H+ ion. Molarity is defined as the moles of solute per liter of solution. When the number of equivalents equals the number of moles of solute, then normality = molarity. This is true for acids which only have one acidic proton in them and for bases that accept only one proton per formula unit. Examples of acids where equivalents = moles solute are HCl, HNO3, HF, and HC2H3O2. Examples of bases where equivalents = moles solute are NaOH, KOH, and NH3. When equivalents moles solute, then normality molarity. This is true for acids that donate more than one proton (H2SO4, H3PO4, H2CO3, etc.) and for bases that react with more than one proton per formula unit [Ca(OH)2, Ba(OH)2, Sr(OH)2, etc.]. It is true that the sodium chloride lattice must be broken in order to dissolve in water, but a lot of energy is released when the water molecules hydrate the Na+ and Cl ions. These two processes have relatively large values for the amount of energy associated with them, but they are opposite in sign. The end result is they basically cancel each other out resulting in a Hsoln 0. So energy is not the reason why ionic solids like NaCl are so soluble in water. The answer lies in natures tendency toward the higher probability of the mixed state. Processes, in general, are favored that result in an increase in disorder because the disordered state is the easiest (most probable) state to achieve. The tendency of processes to increase disorder will be discussed in Chapter 16 when entropy, S, is introduced. Only statement b is true. A substance freezes when the vapor pressure of the liquid and solid are the same. When a solute is added to water, the vapor pressure of the solution at 0EC is less than the vapor pressure of the solid, and the net result is for any ice present to convert to liquid in order to try to equalize the vapor pressures (which never can occur at 0EC). A lower temperature is needed to equalize the vapor pressure of water and ice, hence, the freezing point is depressed.






For statement a, the vapor pressure of a solution is directly related to the mole fraction of solvent (not solute) by Raoults law. For statement c, colligative properties depend on the number of solute particles present and not on the identity of the solute. For statement d, the boiling point of water is increased because the sugar solute decreases the vapor pressure of the water; a higher temperature is required for the vapor pressure of the solution to equal the external pressure so boiling can occur. 26. This is true if the solute will dissolve in camphor. Camphor has the largest Kb and Kf constants. This means that camphor shows the largest change in boiling point and melting point as a solute is added. The larger the change in T, the more precise the measurement and the more precise the calculated molar mass. However, if the solute wont dissolve in camphor, then camphor is no good and another solvent must be chosen which will dissolve the solute. Isotonic solutions are those which have identical osmotic pressures. Crenation and hemolysis refer to phenomena that occur when red blood cells are bathed in solutions having a mismatch in osmotic pressures inside and outside the cell. When red blood cells are in a solution having a higher osmotic pressure than that of the cells, the cells shrivel as there is a net transfer of water out of the cells. This is called crenation. Hemolysis occurs when the red blood cells are bathed in a solution having lower osmotic pressure than that inside the cell. Here, the cells rupture as there is a net transfer of water to into the red blood cells. Ion pairing is a phenomenon that occurs in solution when oppositely charged ions aggregate and behave as a single particle. For example, when NaCl is dissolved in water, one would expect sodium chloride to exist as separate hydrated Na+ ions and Cl ions. A few ions, however, stay together as NaCl and behave as just one particle. Ion pairing increases in a solution as the ion concentration increases (as the molality increases).



ExercisesSolution Composition29. Because the density of water is 1.00 g/mL, 100.0 mL of water has a mass of 100. g. Density = 10.0 g H 3 PO 4 + 100. g H 2 O mass = = 1.06 g/mL = 1.06 g/cm3 volume 104 mL1 mol = 0.102 mol H3PO4 97.99 g

Mol H3PO4 = 10.0 g Mol H2O = 100. g

1 mol = 5.55 mol H2O 18.02 g0.102 mol H 3 PO 4 = 0.0180 (0.102 + 5.55) mol

Mole fraction of H3PO4 =

H 2O = 1.000 0.0180 = 0.9820

CHAPTER 11Molarity = Molality =

PROPERTIES OF SOLUTIONS0.102 mol H 3 PO 4 = 0.981 mol/L 0.104 L0.102 mol H 3 PO 4 = 1.02 mol/kg 0.100 kg



Molality =

40.0 g EG 1000 g 1 mol EG = 10.7 mol/kg 60.0 g H 2 O kg 62.07 g

Molarity =

40.0 g EG 1.05 g 1000 cm 3 1 mol = 6.77 mol/L 3 100.0 g solution L 62.07 g cm 1 mol 1 mol = 0.644 mol EG; 60.0 g H2O = 3.33 mol H2O 62.07 g 18.02 g

40.0 g EG EG = 31.

0.644 = 0.162 = mole fraction ethylene glycol 3.33 + 0.644

Hydrochloric acid (HCl): molarity = 38 g HCl 1.19 g soln 1000 cm 3 1 mol HCl = 12 mol/L 100. g soln L 36.5 g cm 3 soln38 g HCl 1000 g 1 mol HCl = 17 mol/kg 62 g solvent kg 36.5 g 1 mol 1 mol = 1.0 mol HC

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