Forming Solutions · Web viewThe saturation of a solution is dependent on the amount of solute present in the solution. Characteristics of solutions What happens when compounds dissolve

Chemistry

Solutions

Unit 9

Learning Objectives Solutions

Essential knowledge and skills:

Compare and contrast the differences between strong, weak, and nonelectrolytes. Perform calculations involving the molarity of a solution, including dilutions. Net Ionic Equations Interpret solubility curves Rates of solution Saturation Examine the polarity of various solutes and solvents in solution formation. Colligative properties

Essential understandings:

Molarity = moles of solute/L of solution. [ ] refers to molar concentration. When solutions are diluted, the moles of solute present initially remain. The saturation of a solution is dependent on the amount of solute present in the solution. Strong electrolytes dissociate completely. Weak electrolytes dissociate partially. Non-electrolytes

do not dissociate. Polar substances dissolve ionic or polar substances; nonpolar substances dissolve nonpolar

substances. The number of solute particles changes the freezing point and boiling point of a pure substance.

A liquid’s boiling point and freezing point are affected by changes in atmospheric pressure. A liquid’s boiling point and freezing point are affected by the presence of certain solutes.

When solutions are diluted, the moles of solute present initially remain. The saturation of a solution is dependent on the amount of solute present in the solution.

Characteristics of solutions

What happens when compounds dissolveI. Forming Solutions

a. Solution Homogeneous mixture

1. Components are uniformly intermingled Solvent + Solute

1. Solventa. Substance present in the largest amountb. Dissolves the solute

2. Solutea. Substance(s) present in the lesser amountb. Dissolved in the solvent

b. Aqueous Solution Solvent = Water Will focus on these types of solutions

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II. Solubilitya. Like Dissolves Like, a given solvent can dissolve solutes that have similar polarities

Solvent = water 1. Water is polar

Solute1. Ionic Substances

a. Have a dipoleb. Ions are attracted to the polar water moleculesc. Can dissolve in water

2. Polar substancesa. Have a dipoleb. Charged poles are attracted to the polar water moleculesc. Can dissolve in water

3. Nonpolar substancesa. Do not have a dipoleb. Cannot dissolve in water

Solvation – Process of surrounding solute particles with solvent particles to form a solution

***solvation in water = also called hydration

Solvent particles surround the solute particles. The attractive forces between the solvent particles and solute particles are greater than the forces holding the solute particles together. The solvent particles pull the solute particles apart and pull them into the solution.

Electrolytes/Non-electrolytes

Discussion of Electrolytes and Nonelectrolytes electrolytes solutions that conduct an electric current due to formation of hydrates ions in

solution; 1 mol of solute produces more than 1 mol of dissolved particles Acids, bases, and salts dissolve in water to form conducting solutions Non-electrolytes solutions that do not conduct an electric current

Solution Composition

Saturated Solution

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The solution contains as much solute as will dissolve at that temperature If more solute is added it will not dissolve

Unsaturated Solution The solution contains less solute that can dissolve If more solute is added it will dissolve

Supersaturated Solution The solution contains more solute than a saturated solution will hold at a given temperature Formed by increasing the temperature, adding more solute, then letting it cool Very unstable Can add a crystal more of the solute to cause the excess solute to precipitate out until the solution

is saturated - Seed crystal

Describing a Solution

Concentrated A large amount of solute is dissolved A Measure of the amount of solute dissolved in a given volume of solution

Dilute A small amount of solute is dissolved

Factors Affecting the Rate of Dissolving: Surface Area

More surface area = Faster rate

Stirring Increases the rate Removes newly dissolved particles from the solid surface to allow the solute to become exposed

to fresh solvent

Temperature High temperature = Faster rate Cause solvent particles to move faster Most solids are more soluble at high temperatures Most gases are more soluble at low temperatures

Calculations of concentrations of solutions

Parts per million (ppm)

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One ppm is equivalent to 1 milligram of something per dm3 of water (mg/dm3) or 1 milligram of something per kilogram (mg/kg).

=   0 .000001 = 1 ppm

Parts per million (ppm): It is often used to express the concentration of very dilute solutions.

1 ppm =

grams of solutegrams of solution

×106

OR

mg of solutekg of solution

Since the amount of solute relative to the amount of solvent is typically very small, the density of the solution is to a first approximation the same as the density of the solvent. For this reason, parts per million may also be expressed as:

Cppm= mg of soluteL of solution

Parts per billion (ppb)

This concentration unit used for very dilute solutions. The "technical" definition is as follows:

1 ppb =

grams of solutegrams of solution

×109

OR

ppm and ppb concentration calculations (1L of water weighs 1kg due to the density of water)

1. 0.00025 grams of a chemical is dissolved in 75 grams of water.a. What is the concentration of the chemical in parts per million (ppm)?

b. What is the concentration of the chemical in parts per billion (ppb)?

2. Suppose 0.0017 grams of sucrose is dissolved in 183 grams of water. What is the concentration of sucrose in ppm?

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3. 3.5 x 10-7 grams of ethanol is dissolved in 115 grams of water. What is the concentration of ethanol in parts per billion (ppb)?

4. Change 50 ppm to ppb.

5. The solubility (the maximum amount that can dissolve) of NaCl is 284 grams in 100 grams of water. What is this concentration in ppb?

6. The solubility of AgCl is 0.008 grams in 100 grams of water. What is this concentration in ppm?

7. A certain pesticide has a toxic solubility of 5.0 grams for every 1 kg of body weight. What is this solubility in ppm?

Mass percentage

Solution Chemistry - Percent Mass

Mass percentage denotes the mass of a substance in a mixture as a percentage of the mass of the entire mixture. For instance: if a bottle contains 40 grams of ethanol and 60 grams of water, then it contains 40% ethanol by mass. Commercial concentrated aqueous reagents such as acids and bases are often labeled in concentrations of weight percentage with the specific gravity also listed. In older texts and references this is sometimes referred to as weight-weight percentage (abbreviated as w/w). In water pollution chemistry, a common term of measuring total mass percentage of dissolved solids in an aqueous medium is total dissolved solids.

% mass = mass of solute (g) / mass of solution (g) x 100%

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Mass-volume percentage

Solution Chemistry - Percent Mass VolumeMass-volume percentage, (sometimes referred to as weight-volume percentage or percent weight per volume and often abbreviated as % m/v or % w/v) describes the mass of the solute in g per 100 mL of the resulting solution. Mass-volume percentage is often used for solutions made from a solid solute dissolved in a liquid. For example, a 40% w/v sugar solution contains 40 g of sugar per 100 mL of resulting solution.

Mass/volume % = mass of solute (g) / volume of solution (ml) x 100%

Volume-volume percentage

Solution Chemistry - Percent VolumeVolume-volume percentage (sometimes referred to as percent volume per volume and abbreviated as % v/v) describes the volume of the solute in mL per 100 mL of the resulting solution. This is most useful when a liquid - liquid solution is being prepared. For example, a 40% v/v ethanol solution contains 40 mL ethanol per 100 mL total volume.

Volume/volume % = volume of solute (ml) / volume of solution x 100%

Worksheet on using % mass and volume in solution

1. Glucose is a sugar found abundantly in nature. What is the percent by mass of a solution made by dissolving 163 g of glucose in 755 g of water? Do you need to know the formula of glucose? Why or why not?

2. What is the mass percent sucrose in a solution obtained by mixing 225 g of an aqueous solution that is 6.25% sucrose by mass with 135 g of an aqueous solution that is 8.20% sucrose by mass?

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3. Determine the volume percent of toluene in a solution made by mixing 40.0 mL toluene with 75.0 mL of benzene.

4. What is the mass percent of solute when 4.12 g is dissolved in 100.0 g of water?

5. What is the volume percent of 10.00 g of acetone (d = 0.789 g/mL) in 1.55 L of an acetone-water solution?

6. What is the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 g of water?

7. A solution is made by adding 25 mL of benzene to 80 mL of toluene. What is the percent by volume of benzene?

8. A solution is formed by adding 35 g of ammonium nitrate to 250 g of water. What is the percent by mass of ammonium nitrate?

9. What is the percent by volume of a solution formed by mixing 25 mL of isopropanol with 45 mL of water?

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10. What is the mass percent of each component in the mixture formed by adding 12 g of calcium sulfate, 18 g of sodium nitrate, and 25 g of potassium chloride to 500 g of water?

11. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by mass?

12. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?

13. An experiment requires a solution that is 45% methyl alcohol by volume. What volume of methyl alcohol should be added to 200 mL of water to make the solution.

Reading Solubility Curves

Interpreting Solubility CurvesThe curve shows the amount of solute in grams of a saturated solution containing 100 mL or 100 g of water at a certain temperature.

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Any amount of solute below the line indicates the solution is unsaturated at a certain temperature Any amount of solute above the line in which all of the solute has dissolved shows the solution is

supersaturated. If the amount of solute is above the line but has not all dissolved, the solution is saturated and the

grams of solute settled on the bottom of the container = total mass in solution – mass of a saturated solution at that temperature. (according to the curve)

Solutes whose curves move upward w/ increased temperature are typically solids b/c the solubility of solids increases w/ increased temperature.

Solutes whose curves move downward w/ increased temperature are typically gases b/c the solubility of gases decreases with increased temperature.

Solubility Problems (worked example)

Calculate how much NaNO3 will dissolve in 60 mL of water @10ºC?

From the solubility curve:

At 10oC, 80 g of NaNO3 will dissolve in 100 mL (a saturated solution)

To find the number of grams needed to saturate a solution when the volume is NOT 100 mL use the following strategy to find answer:

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Start with known volume x solubility/100mL at set temp. = amount of Solute needed to saturate

Ex. 60 mL H2O x 80 g NaNO3 = 48 g NaNO3 needed to saturate solution 100 mL H2O

or if the chart is in units of 100 g of H2O then use the density of water conversion 1mL H2O = 1 g H2O

Ex. 60 mL H2O x 1 g H2O x 80 g NaNO3 = 48 g NaNO3

1 mL H2O 100 g H2O

Solubility Curve Question Worksheet 1

1. Which of the salts shown on the graph is the least soluble in water at 10oC?

2. Which of the salts shown on the graph has the greatest increase in solubility as the temperature increases from 30ºC to 60ºC?

3. Which of the salts has its solubility affected the least by a change in temperature?

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4. At 20oC, a saturated solution of sodium nitrate contains 100 grams of solute in 100 ml of water. How many grams of sodium chlorate must be added to saturate the solution at 50oC?

5. At what temperature do saturated solutions of potassium nitrate and sodium nitrate contain the same weight of solute per 100 mL of water?

6. What two salts have the same amount of solubility at approximately 19oC?

7. How many grams of potassium chlorate must be added to 1 liter of water to produce a saturated solution at 50oC?

8. A saturated solution of potassium nitrate is prepared at 60oC using 100.mL of water. How many grams of solute will precipitate out of solution if the temperature is suddenly cooled to 30oC?

9. What is the average rate of increase for the solubility of KNO3 in grams per 100 mL per degree Celsius in the temperature range of 60oC to 70oC?

10. If 50. mL of water that is saturated with KClO3 at 25oC is slowly evaporated to dryness, how many grams of the dry salt would be recovered?

11. Thirty grams of KCl are dissolved in 100 mL of water at 45oC. How many additional grams of KCl are needed to make the solution saturated at 80oC?

12. What is the smallest volume of water, in mL, required to completely dissolve 39 grams of KNO3 at 10oC?

13. What is the lowest temperature at which 30. grams of KCl can be dissolved in 100 mL of water?

14. Are the following solutions saturated, unsaturated or supersaturated (assume that all three could form supersaturated solutions)

40. g of KCl in 100 mL of water at 80oC120. g of KNO3 in 100 mL of water at 60oC80. g of NaNO3 in 100 mL of water at 10oC

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15. Assume that a solubility curve for a gas such as ammonia, at one atmosphere of pressure, was plotted on the solubility curve graph. Reading from left to right, would this curve would _____

a. slope upward b. slope downward c. go straight across

Answers: KClO3, KNO3, NaCl, 15 g, 72 3oC, KNO3 and KCl, 200 g, 55 3g, 2.7 g/oC, 5 g, 20 g, 175 mL, 10oC, a) unsat., b) supersat., c) sat, b

Solubility Curve Questions Worksheet 2

You'll notice that for most substances, solubility increases as temperature increases. As discussed earlier in solutions involving liquids and solids typically more solute can be dissolved at higher temperatures. Can you find any exceptions on the graph? ____________________

Directions: Use the graph to answer the following questions. REMEMBER UNITS!

1) What mass of solute will dissolve in 100mL of water at the following temperatures?

a. KNO3 at 70°C = ____________

b. NaCl at 100°C= ____________

c. NH4Cl at 90°C= ____________

d. Which of the above three substances is most soluble in water at 15°C

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2) Types of Solutions (Saturated or Unsaturated)

On a solubility curve, the lines indicate the concentration of a __________________ solution - the maximum amount of solute that will dissolve at that specific temperature.

Values on the graph ____________ a curve represent unsaturated solutions - more solute could be dissolved at that temperature.

Label the following solutions as saturated or unsaturated. If unsaturated, write how much more solute can be dissolved in the solution.

Solution Saturated or Unsaturated? If unsaturated: How much more solute can dissolve in the solution?

a solution that contains 70g of NaNO3 at 30°C (in 100 mL H2O)a solution that contains 50g of NH4Cl at 50°C (in 100 mL H2O)a solution that contains 20g of KClO3 at 50°C (in 100 mL H2O)a solution that contains 70g of KI at 0°C (in 100 mL H2O)

1. a. What is the solubility of KCl at 5C? _______

b. What is the solubility of KCl at 25C? _______

c. What is the solubility of Ce2(SO4)3 at 10C? _______

d. What is the solubility of Ce2(SO4)3 at 50C? _______

2. At 90C, you dissolved 10 g of KCl in 100. g of water. Is this solution saturated or unsaturated?

3. A mass of 100 g of NaNO3 is dissolved in 100 g of water at 80ºC.

a) Is the solution saturated or unsaturated? ______________________________

b) As the solution is cooled, at what temperature should solid first appear in the solution? Explain.

4. Use the graph to answer the following two questions:Which compound is most soluble at 20 ºC? ________Which is the least soluble at 40 ºC? ________

5. Which substance on the graph is least soluble at 10C? __________

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6. A mass of 80 g of KNO3 is dissolved in 100 g of water at 50 ºC. The solution is heated to 70ºC. How many more grams of potassium nitrate must be added to make the solution saturated? Explain your reasoning.

Molarity (M)

Molarity ExplainedThe concentration of a solution is defined as the amount of solute in a given volume of solution. The most commonly used expression of concentration is molarity (M). The molarity of a solution describes the number of moles of solute in one liter of solution. Molarity can be written as follows:

Molarity ( M )=moles of solute (mol)volume (L)

For instance, 1.0 M means that the solution has a concentration of 1.0 molar, or contains 1.0 moles of solute in a liter of solution.

As an example, determine the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. First, you must covert grams of NaOH to moles.

11.5 g NaOH x 1mol NaOH40.0 g NaOH = 0.288 mol.

Then you can solve for molarity:

Molarity ( M )=0.288 mol .1.50 L

=0.192 mol L−1∨0.192 M

Example A: If you are given a 0.20 M Sodium Chloride solution, what does this mean?

This means there are 0.20 moles of salt dissolved in 1.0 litres of solution

Also expressed as 0.20 molL-1 or 0.20 mol/L

Example B. If you wanted to make 3.0 liters of a 0.20M Sodium Chloride solution, how many grams of sodium chloride would you need?

Step 1: Figure out how many moles of NaCl you would need

0.2 M= xmoles of NaCl3.0 L solution

=0.6 molesof NaCl

Step 2: Convert moles to mass. Calculate molar mass of solute

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58.44 g/mol x 0.6 moles NaCl = 35.06 g of NaCl

Using dimensional analysis:

Example C: How many liters of a 5.7 M NaCl solution will contain 5.2 grams of salt?

Step 1: Figure out how many moles of NaCl you have (mass to moles)

5.2 g ÷ 58.44 g/mol = 0.09 moles

Step 2: Divide moles by molarity to calculate volume of solution

0.09 moles

5.7 molesL

=0.02 L

Using dimensional analysis:

Molarity Worksheet

1. Sea water contains roughly 28.0 g of NaCl per liter. What is the molarity of sodium chloride in sea water?

2. What is the molarity of 245.0 g of H2SO4 dissolved in 1.00 L of solution?

3. What is the molarity of 5.30 g of Na2CO3 dissolved in 400.0 mL solution?

4. What is the molarity of 5.00 g of NaOH in 750.0 mL of solution?

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5. How many moles of Na2CO3 are there in 10.0 L of 2.0 M solution?

6. How many moles of Na2CO3 are in 10.0 mL of a 2.0 M solution?

7. How many moles of NaCl are contained in 100.0 mL of a 0.20 M solution?

8. What mass (in grams) of NaCl would be contained in problem 7?

9. What mass (in grams) of H2SO4 would be needed to make 750.0 mL of 2.00 M solution?

10. What volume (in mL) of 18.0 M H2SO4 is needed to contain 2.45 g H2SO4?

11. What volume (in mL) of 12.0 M HCl is needed to contain 3.00 moles of HCl?

12. How many grams of Ca(OH)2 are needed to make 100.0 mL of 0.250 M solution?

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13. What is the molarity of a solution made by dissolving 20.0 g of H3PO4 in 50.0 mL of solution?

14. What weight (in grams) of KCl is there in 2.50 liters of 0.50 M KCl solution?

15. What is the molarity of a solution containing 12.0 g of NaOH in 250.0 mL of solution?

16. Determine the molarity of these solutions:

a) 4.67 moles of Li2SO3 dissolved to make 2.04 liters of solution.

b) 0.629 moles of Al2O3 to make 1.500 liters of solution.

c) 4.783 grams of Na2CO3 to make 10.00 liters of solution.

d) 0.897 grams of (NH4)2CO3 to make 250 mL of solution.

e) 0.0348 grams of PbCl2 to form 45.0 mL of solution.

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17. Determine the number of moles of solute to prepare these solutions:

a) 2.35 liters of a 2.00 M Cu(NO3)2 solution.

b) 16.00 mL of a 0.415-molar Pb(NO3)2 solution.

c) 3.00 L of a 0.500 M MgCO3 solution.

d) 6.20 L of a 3.76-molar Na2O solution.

18. Determine the grams of solute to prepare these solutions:

a) 0.289 liters of a 0.00300 M Cu(NO3)2 solution.

b) 16.00 milliliters of a 5.90-molar Pb(NO3)2 solution.

c) 508 mL of a 2.75-molar NaF solution.

d) 6.20 L of a 3.76-molar Na2O solution.

e) 0.500 L of a 1.00 M KCl solution.

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f) 4.35 L of a 3.50 M CaCl2 solution.

19. Determine the final volume of these solutions:

a) 4.67 moles of Li2SO3 dissolved to make a 3.89 M solution.

b) 4.907 moles of Al2O3 to make a 0.500 M solution.

c) 0.783 grams of Na2CO3 to make a 0.348 M solution.

d) 8.97 grams of (NH4)2CO3 to make a 0.250-molar solution.

e) 48.00 grams of PbCl2 to form a 5.0-molar solution.

Answers

Work required for answers…

1. (x) (1.00 L) = 28.0 g / 58.45 g mol¯1; x = 0.479 M

2. (x) (1.00 L) = 245.0 g / 98.08 g mol¯1; x = 2.498 M

3. (x) (0.4000 L) = 5.30 g / 106.0 g mol¯1; x = 0.125 M

4. (x) (0.7500 L) = 5.00 g / 40.00 g mol¯1; x = 0.167 M

5. 2.0 M = x / 10.0 L 20 mol

6. 2.0 M = x / 0.0100 L 0.02 mol

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7. 0.20 M = x / 0.1000 L 0.02 mol

8. (0.20 mol L¯1) (0.100 L) = x / 58.45 g mol¯1 1.17g

9. (2.00 mol L¯1) (0.7500 L) = x / 98.08 g mol¯1 147 g

10. (18.0 mol L¯1) (x) = 2.45 g / 98.08 g mol¯1 1.39 mL

This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value.

11. 12.0 M = 3.00 mol / x 250 mL

This calculates the volume in liters. Multiplying the answer by 1000 provides the required mL value.

12. (0.250 mol L¯1) (0.100 L) = x / 74.1 g mol¯1 1.9 g

13. (x) (0.050 L) = 20.0 g / 97.99 g mol¯1 4.08 M

14. (0.50 mol L¯1) (2.50 L) = x / 74.55 g mol¯1 93 g

15. (x) (0.2500 L) = 12.0 g / 40.00 g mol¯1 1.20 M

16. Determine the molarity of these solutions:

a) x = 4.67 mol / 2.04 L 2.29 Mb) x = 0.629 mol / 1.500 L 0.419 Mc) (x) (10.00 L) = 4.783 g / 106.0 g mol¯1 0.0045Md) (x) (0.250 L) = 0.897 g / 96.09 g mol¯1 0.037Me) (x) (0.0450 L) = 0.0348 g / 278.1 g mol¯1 0.0028 M

17. Determine the number of moles of solute to prepare these solutions:

a) x = (2.00 mol L¯1) (2.35 L) 4.70 molb) x = (0.415 mol L¯1) (0.01600 L) 0.00664 molc) x = (0.500 mol L¯1) (3.00 L) 1.5 mold) x = (3.76 mol L¯1) (6.20 L) 23.3 mol

18. Determine the grams of solute to prepare these solutions:

a) (0.00300 mol L¯1) (0.289 L) = x / 187.56 g mol¯1 0.163 gb) (5.90 mol L¯1) (0.01600 L) = x / 331.2 g mol¯1 31.3 gc) (2.75 mol L¯1) (0.508 L) = x / 41.99 g mol¯1 58.7 gd) (3.76 mol L¯1) (6.20 L) = x / 61.98 g mol¯1 1440 ge) (1.00 mol L¯1) (0.500 L) = x / 74.55 g mol¯1 37.3 gf) (3.50 mol L¯1) (4.35 L) = x / 110.99 g mol¯1 1690 g

19. Determine the final volume of these solutions:

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a) x = 4.67 mol / 3.89 mol L¯1 1.2 Lb) x = 4.907 mol / 0.500 mol L¯1 9.81 Lc) (0.348 mol L¯1) (x) = 0.783 g / 105.99 g mol¯1 0.0212 Ld) (0.250 mol L¯1) (x) = 8.97 g / 96.01 g mol¯1 0.374 Le) (5.00 mol L¯1) (x) = 48.0 g / 278.1 g mol¯1 0.0345 L

Making a Standard solution

A solution of known molar concentration is called a standard solution. Its concentration is determined by a process known as standardization.

A primary standard is a compound which is very pure, stable, nonhygroscopic, and with a high molecular weight. You can prepare a standard solution simply by dissolving a known amount of the compound in a known volume of liquid.

Use a wash bottle to wash all solid into the flask. Then rinse down the beaker, the funnel, and the inside neck of the volumetric flask.

Add enough water to the flask to dissolve the solid. After dissolution, add distilled water to just below the calibration line. A dropper should be used to make the volume to the calibration mark itself.

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Molarity by Dilution

Video tutorialDilution: To dilute a solution means to add more solvent without the addition of more solute. Of course, the resulting solution is thoroughly mixed as to ensure that all parts of the solution are identical. The fact that the solute amount stays constant allows us to develop calculation techniques so that:

moles before dilution = moles after dilution

From the definition of molarity, we know that the moles of solute equals the molarity times the volume. So we can substitute MV (molarity x volume) into the above equation, like this:

M1V1 = M2V2

where the left side of the equation is before dilution and the right side after dilution.

1. What is the concentration (molarity) of a solution of NaCl if 40. mL of a 2.5 M NaCl solution is diluted to a total volume of 500. mL?

2. A stock solution of 1.00 M NaCl is available. How many mL are needed to make 100.0 mL of 0.750 M

3. What volume of 0.250 M KCl is needed to make 100.0 mL of 0.100 M solution?

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Now shake the flask several time to ensure that the solution is completely uniform (homogenous)

4. To properly dispose of acid its concentration must be less than 1.00 x 10-5 M. How much water must be added to 25 mL of 6.0 M HCl before it can be disposed of safely? (This is why we don’t pour it away).

5. 500. mL of a 6.00 M stock solution of NaCl is added to 2.00 L of water. How much of the solution must you pour away and replace with water to get exactly 2.00 L of 1.00 M NaCl?

6. How much solvent must be added to 200. mL 1.50 M NaNO3 to make a solution with a concentration of 0.800 M NaNO3 ?

7. Suppose you have just received a shipment of sodium carbonate, Na2CO3. You weigh out 50.00 g of the material, dissolve it in water, and dilute the solution to 1.000 L. You remove 10.00 mL from the solution and dilute it to 50.00 mL. By measuring the amount of a second substance that reacts with Na2CO3, you determine that the concentration of sodium carbonate in the diluted solution is 0.0890 M. Calculate the percentage of Na2CO3 in the original batch of material. 

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Molality (m)

Molality Problems ExplainedDifference between Molarity and MolalityWhat is the point of MolalityThe unit molality is an additional way in which the concentration of a solution is expressed.

Molality, m, is the number of moles of solute dissolved in 1 kilogram (1000 g) of solvent (mol/kg). Remember that the density of H2O is 1 g/ml. Molality is also called molal concentration and can be expressed as:

Molality (m)=molesof solute (mol)Mass of solvent (kg)

1. Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.

2. 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in 1.00 kg of water. Calculate the molality.

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3. Calculate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.

4. 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality?

5. 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

6. Find the molality of the following solutions;

3.40 moles of NH3 dissolved in 4500. grams of water

0.500 moles of HNO3 dissolved in 1.25 kilograms of water

3.00 liters of 1.50M KBr solution

17.0 grams of LiF in 34.0 moles of water

7. What is the molality of a solution made by combining 1.25 x 10-3 moles of glucose in 500.0 ml of water?

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8. What is the molality of a solution made by dissolving 6.75 grams of CaCl2 in 2125 grams of water?

9. What is the molality of a solution made by dissolving 2.98 grams of NaCl in water if the mass of the resulting solution is 976.5 grams?

10. How many grams of potassium nitrate need to be added to 2.00 kg of water to make a 0.150 m solution?

Colligative Properties of Solutions

Colligative Properties – physical properties of solutions that are affected by the number of solute particles but not the identity of dissolved solute particles

I. Vapor Pressure Lowering

A Vapor Pressure – the pressure exerted in a closed container by liquid particles that have escaped the liquid’s surface & entered the gaseous state

B Adding a nonvolatile solute (one that has little tendency to become a gas) to a solvent lowers the solvent’s vapor pressure

In pure water – all the water (solvent) molecules occupy the surface area, which allows many water molecules to evaporate

In a mix of water & solute particles – less water molecules occupy the surface area – fewer enter the gaseous state – lower vapor pressure. The greater the number of solute particles in a solvent, the lower the resulting vapor pressure.

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II. Boiling Point Elevation

Boiling Point – temperature at which the vapor pressure equals the atmospheric pressure in order for the solvent to escape the liquid state

Boiling Point Elevation - the temperature difference between a solution’s boiling point and a pure solvent’s boiling point

The boiling point elevation is directly proportional to the solutions molality – the greater the number of solute particles in solution, the greater the boiling point elevation.

Freezing Point Depression

Freezing Point Temperature of a Solution: particles no longer have sufficient kinetic energy to overcome the intermolecular attractive forces → particles form into a more organized structure in the solid state

In solution: solute particles interfere with the attractive forces among the solvent particles → prevents solution from entering solid state at normal freezing point → lowers freezing point

Freezing Point Depression – difference in temperature between its freezing point and the freezing point of its pure solvent If we denote the increase in BP as Tb, the molality of the solution as m, then there is a molal boiling-

point-elevation constant, Kb, that is the constant of proportionality between the two:Tb = iKbm

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Some typical values for Kb are found in the table below. Once again, it is important to recognize that the concentration of solute particles in solution can vary depending on whether or not the solute is an electrolyte.

Similar explanation can be given for the depression of the freezing point, so that the following relationship holds:

Tf = iKfm

where Tf is the change in the freezing point, and Kf is the molal freezing-point-depression constant, i= dissociation factor (non-electrolytes =1, electrolytes = number of ions in solution)

For example, table salt, NaCl dissolves in water to form ions NaCl(s) Na+(aq) + Cl-

(aq). Each mole of NaCl makes two moles of ions, so if we have a 1 molal solution of NaCl in water, the ion concentration is 2 molal. Remember, nonelectrolyte solutes do not produce any ions.

Molal freezing point and boiling point constant table:

Explaination of Colligative Properties and calculation

Worked example2.40 moles of sucrose are dissolved in 320. mL of distilled water. (i = 1 for a non-electrolyte solid.) What is the boiling point of this solution?

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Freezing Boiling

Solvent Point, C Kf C/m point, C Kb, C/m

Acetic acid 16.60 3.90 117.9 3.07

Benzene 5.50 4.90 80.10 2.53

Camphor 179.8 39.7 207.42 5.61

Ethyl ether 34.51 2.02

Nitrobenzene 5.7 7.00 210.8 5.24

Phenol 40.90 7.40 181.75 3.56

Water 0.00 1.86 100.00 0.512

1. 56.8g of NaCl is dissolved in 560. mL distilled water. What is the freezing point of this solution?

2. 89.7g of Ba(NO3)2 is dissolved in 1.25L of distilled water. What is the boiling point of this solution?

3. An aqueous sodium chloride solution boils at 381K. What is the molal concentration of the solution?

4. An aqueous barium fluoride solution freezes at –4.20C. What is the molal concentration of the solution?

Colligative Properties Worksheet 2

1. What is the boiling point of a solution of 25.00 g of MgCl2 in 100.0 g of water? The kb of water is 0.52°C/m.

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2. A water tank contains 6.50 kg of water. Will the addition of 1.20 kg of ethylene glycol (C2H6O2 – a covalent compound) be sufficient to prevent the solution from freezing if the temperature drops to -25.0°C? What is the freezing temperature of this solution?

3. NaCl, CaCl2, and ethylene glycol (C2H6O2 – a covalent compound) are three common road deicers. For each deicer, calculate the freezing point of a solution of 100.0 g of the solute in 0.250 kg of water. The kf of water is 1.86 °C/m. Which one of the three makes the best deicer (i.e. lowers the freezing point of water the most)?

Colligative Properties Worksheet 3

1) If I add 45 grams of sodium chloride to 500 grams of water, what will be the melting and boiling points of the resulting solution?

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2) Which solution will have a higher boiling point: A solution containing 105.5 g of sucrose (C12H22O11) in 500.0 grams of water or a solution containing 35.0 g of sodium chloride in 500. grams of water?

3) 5.0 grams of salt (NaCl) is added to 170.0 mL of water. What are the new freezing and boiling points?

4) What is the change in freezing point of a solution containing 132.0 g C12H22O11 and 250.0 g of H2O?

Solubility Rules.

1. All salts containing ions of Group IA, NH4+, NO3

-, C2H3O2- (acetate ion), or CIO3

- ARE SOLUBLE.

2. Salts of Cl-, Br-, or I- ARE SOLUBLE, EXCEPT with Ag+, Cu+, Hg22+, or Pb2+.

3. Salts of SO42- ARE SOLUBLE, EXCEPT with Pb2+, Ca2+, Sr2+, Ba2+, or Ra2+.

4. Salts of OH- and S2- ARE NOT SOLUBLE, EXCEPT for Rule 1, or Ca2+, Sr2+, Ba2+ or Ra2+.

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5. Salts of O2-, CO32-, PO4

3-, SO32-, CrO4

2-, and SiO4-2 ARE, NOT SOLUBLE,, EXCEPT for Rule 1.

1. Based upon the above rules, predict whether each of the compounds listed will be Soluble or Insoluble.

a. K2CO3 ____ b. CaSO4 ______ c. AgNO3 ______ d. MnPO4 _____ e. PbI2 ______ f. (NH4)2SO3 _____

g. Ca(OH)2 _____ h. Pb(C2H3O2)2 _____ i. AgI _____ j. Na2SiO4 _____ k. FeSO4 _____ m. WC14 _____

Complete and Net Ionic Equations

How to write Net Ionic Equations

Precipitation Reactions:

Pb(NO3)2(aq) + NaI(aq)

Step 1: Write the balanced general reaction including the reactants and the products

Pb(NO3)2(aq) + 2 NaI(aq) PbI2 + 2 NaNO3

Step 2: Look up the products on your solubility chart and insert the subscripts (for solid, and aq for aqueous) If something is insoluble you use the solid subscript, if it is soluble you use the aq subscript)

Pb(NO3)2(aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

Step 3: Break all reactants and products into ions, (you can not break up pure liquids, solids, or gases). This is the COMPLETE IONIC EQUATION. If you have polyatomic ions such as nitrate (NO3

-) that is an ion, do not break it up further

Pb2+(aq) + 2 NO3

-(aq) + 2 Na+

(aq) + 2 I-(aq) PbI2(s) + 2 Na+

(aq) + 2 NO3-(aq)

Step 4: Cancel out any spectator ions. Spectator ions are ions that appear in the reactant side and the product side. (they have to be exactly the same, you could cancel out NO3

- & NO3- but not Fe+ & Fe3+, the

oxidation # of iron has changed therefore it is not the exact same on the reactant side as the product side.

Pb2+(aq) + 2 NO3

-(aq) + 2 Na+

(aq) + 2 I-(aq) PbI2(s) + 2 Na+

(aq) + 2 NO3 -(aq)

Step 5: Re-write the chemical reaction without the spectator ions. This is the NET IONIC EQUATION

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Pb2+(aq) + 2 I-

(aq) 2 PbI2(s)

Step 6: Check yourself. The only chemicals on the reactants side should be what makes up the solid product. In this case PbI2 is the product so only Pb2+ and I- should be on the reactant side.

Net Ionic Equation Worksheet: Write balanced, complete and net ionic equations for the following

1. AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

2. strontium bromide (aq) + potassium sulfate (aq) strontium sulfate + potassium bromide

3. Ni(NO3)3 (aq) + KBr (aq)

4. cobalt (III) bromide + potassium sulfide  

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