Solutions and their Behavior Chapter 18

1. Identify factors that determine the rate at which a solute dissolves 2. Identify factors that affect the solubility of a solute in solution 3. Calculate the solubility of a gas in a liquid under various pressure conditions Objectives

Factors affecting the rate of dissolving How could you speed up the dissolving of sugar in a glass of iced tea? How could you speed up the dissolving of sugar in a glass of iced tea? Which dissolves faster, table salt or rock salt? Which dissolves faster, table salt or rock salt?

Factors affecting the rate of dissolving Temperature – increasing the temperature speeds up the rate of dissolving Temperature – increasing the temperature speeds up the rate of dissolving Agitation – stirring speeds up the rate of dissolving Agitation – stirring speeds up the rate of dissolving Particle size – smaller particles dissolve faster than large particles (surface area) Particle size – smaller particles dissolve faster than large particles (surface area)

Solubility The solubility of a substance is the amount that dissolves in a given quantity of solvent at a given temperature. The solubility of a substance is the amount that dissolves in a given quantity of solvent at a given temperature.

Solubility A saturated solution contains the maximum amount of solute at a constant temperature. A saturated solution contains the maximum amount of solute at a constant temperature.

Solubility An unsaturated solution contains less solute than a saturated solution. An unsaturated solution contains less solute than a saturated solution. A supersaturated solution contains more solute than a saturated solution. (This occurs when a solution is saturated and then allowed to cool but all of the solid remains dissolved. It is an unstable solution, adding a crystal causes precipitation.) A supersaturated solution contains more solute than a saturated solution. (This occurs when a solution is saturated and then allowed to cool but all of the solid remains dissolved. It is an unstable solution, adding a crystal causes precipitation.)

What would happen… …if you added more sugar to a saturated sugar solution and stirred? …if you added more sugar to a saturated sugar solution and stirred? …if you added more sugar to an unsaturated sugar solution and stirred? …if you added more sugar to an unsaturated sugar solution and stirred?

Factors affecting the solubility of a substance How could you increase the amount of sugar that would eventually dissolve in a glass of tea? How could you increase the amount of sugar that would eventually dissolve in a glass of tea?

Factors affecting the solubility of a substance Only two factors affect the amount of solute that can dissolve. Only two factors affect the amount of solute that can dissolve. 1. Temperature affects solubility of both solids and gases in liquid solvents. 2. Pressure affects solubility of gases in liquid solvents.

Factors affecting the solubility of a substance The solubility of most solid substances increases as the temperature of the solvent increases. The solubility of most solid substances increases as the temperature of the solvent increases. For a few substances, the reverse occurs. For a few substances, the reverse occurs.

The Effect of Temperature on the Solubility of Solids

The Effect of Temperature on Gas Solubility Increasing the temperature of a dissolved gas solution decreases the concentration of the gas. Increasing the temperature of a dissolved gas solution decreases the concentration of the gas. Have you ever tried a hot Dr. Pepper? Heat it in a pan on the stove, pour a cup, and it has no bubbles! Thermal pollution occurs when hot water is added to a lake, the dissolved oxygen levels fall in the water and it kills the fish.

The Effect of Temperature on Gas Solubility

The Effect of Pressure on Gas Solubility Increasing the pressure of a gas over the surface of a solvent increases the solubility of the gas in the solvent. Increasing the pressure of a gas over the surface of a solvent increases the solubility of the gas in the solvent. In a bottled soda, the pressure of CO 2 over the liquid is high and when the cap is opened, the pressure is reduced and bubbles begin to come out of the solution. In a bottled soda, the pressure of CO 2 over the liquid is high and when the cap is opened, the pressure is reduced and bubbles begin to come out of the solution.

The Effect of Pressure on Gas Solubility Henry’s Law: at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. Henry’s Law: at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. S 1 /P 1 = S 2 /P 2

Question: If the solubility of a gas in water is 0.77 g/L at 3.5 atm pressure, what is the solubility (in g/L) at 1.0 atm? If the solubility of a gas in water is 0.77 g/L at 3.5 atm pressure, what is the solubility (in g/L) at 1.0 atm? (The temperature is held constant at 25 o C.) Answer: S 1 /P 1 = S 2 /P 2 Answer: S 1 /P 1 = S 2 /P g/L / 3.5 atm = S 2 / 1.0 atm S 2 = 0.22 g/L

Objectives 1. Solve problems involving the molarity of a solution 2. Describe how to prepare dilute solutions from concentrated solutions of known molarity.

Solution Composition: Molarity Concentration of a solution is the amount of solute in a given volume of solution. Concentration of a solution is the amount of solute in a given volume of solution. A concentrated solution has a high concentration of solute. A concentrated solution has a high concentration of solute. A diluted solution has only a low concentration of solute. A diluted solution has only a low concentration of solute.

A unit of solution concentration A common unit of concentration: A common unit of concentration: Molarity (M) (Notice that the ratio includes liters of total solution, not liters of solvent.)

Solution Composition: Molarity Consider both the moles of solute and the volume of solution to find concentration. Consider both the moles of solute and the volume of solution to find concentration.

Solution Composition: Molarity To find the moles of solute in a given volume of solution of known molarity use the definition of molarity. To find the moles of solute in a given volume of solution of known molarity use the definition of molarity.

Solution Preparation: Molarity To make a molar solution: –Weigh out a sample of solute that contains the necessary moles of solute. –Transfer to a volumetric flask. –Add only enough solvent to mark on flask to get total solution volume.

Problem: Calculate the molarity when 3.0 mol of solute is dissolved in enough water to make 2.0 L of solution. Calculate the molarity when 3.0 mol of solute is dissolved in enough water to make 2.0 L of solution. Answer: Answer: mol solute/L solution = molarity 3.0 mol / 2.0 L = 1.5M

Problem A solution contains 17.0 grams of sodium chloride in 200. mL of solution. What is the molarity of the solution? A solution contains 17.0 grams of sodium chloride in 200. mL of solution. What is the molarity of the solution? Answer: (convert grams to moles) Answer: (convert grams to moles) 17.0 grams. (1 mol/58.5 g) = mol mol solute/L solution = molarity mol / L = molar NaCl

Problem How many grams of sodium chloride are needed to prepare 275 mL of a 0.200M solution of NaCl? How many grams of sodium chloride are needed to prepare 275 mL of a 0.200M solution of NaCl? Answer: Answer: molarity. L solution = mol solute molarity. L solution = mol solute 0.200M L = moles of NaCl mol. (58.5 g/1 mol) =3.22 grams NaCl

Dilution When diluting a solution to a lower concentration, only water is added in the dilution – the amount of solute remains the same. ( remember: mol solute = molarity. L solution) When diluting a solution to a lower concentration, only water is added in the dilution – the amount of solute remains the same. ( remember: mol solute = molarity. L solution) (mol solute) conc = (mol solute) diluted M conc. L conc = M diluted. L diluted M conc. L conc = M diluted. L diluted M conc. V conc = M diluted. V diluted M conc. V conc = M diluted. V diluted M 1. V 1 = M 2. V 2 M 1. V 1 = M 2. V 2

Dilution Diluting a solution Diluting a solution Transfer a measured amount of original solution to a flask containing some water. Transfer a measured amount of original solution to a flask containing some water. Add water to the flask to the mark (with swirling) and mix by inverting the flask. Add water to the flask to the mark (with swirling) and mix by inverting the flask. M 1. V 1 = M 2. V 2 M 1. V 1 = M 2. V 2

Problem How many mL of a concentrated stock solution of 5.0M HCl is needed to prepare 250. mL of 1.75M HCl? How many mL of a concentrated stock solution of 5.0M HCl is needed to prepare 250. mL of 1.75M HCl? Answer: Answer: M 1. V 1 = M 2. V 2 M 1. V 1 = M 2. V 2 5.0M. V 1 = 1.75M. 250 mL V 1 = 87.5 mL (notice: you do not have to convert to liters in dilution problems)

Objectives Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared to the pure solvent Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared to the pure solvent Calculate the molality and mole fraction of a solution Calculate the molality and mole fraction of a solution Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound

Colligative Properties Colligative property – a solution property that depends on the number, but not the type, of solute particles present Colligative property – a solution property that depends on the number, but not the type, of solute particles present There are three colligative properties of solutions: There are three colligative properties of solutions: 1. Vapor pressure lowering 2. Boiling point elevation 3. Freezing point depression

Vapor Pressure Lowering When solute particles are added to a solvent, the solute forms associations with the solvent. Fewer solvent particles are free to vaporize, so the vapor pressure is lower over a solution than over the pure solvent. When solute particles are added to a solvent, the solute forms associations with the solvent. Fewer solvent particles are free to vaporize, so the vapor pressure is lower over a solution than over the pure solvent.

Colligative Properties Adding solute particles causes the liquid range to become wider. The lowering of vapor pressure causes the boiling point to rise and the freezing point to fall. ( interferes with crystal formation) Adding solute particles causes the liquid range to become wider. The lowering of vapor pressure causes the boiling point to rise and the freezing point to fall. ( interferes with crystal formation) Boiling point increases Boiling point increases Freezing point decreases Freezing point decreases

Colligative Properties The higher the number of particles (ions, molecules, etc.) the great the change in the colligative property. The higher the number of particles (ions, molecules, etc.) the great the change in the colligative property. Which would have the greatest elevation in boiling point? Which would have the greatest elevation in boiling point? M C 6 H 12 O M NaCl M CaCl M AlCl 3

Answer: All 4 solutions have the same concentration, but the number of particles that form is not the same: All 4 solutions have the same concentration, but the number of particles that form is not the same: C 6 H 12 O 6(s)  C 6 H 12 O 6(aq) C 6 H 12 O 6(s)  C 6 H 12 O 6(aq) (1 particle) NaCl (s)  Na + (aq) + Cl - (aq) NaCl (s)  Na + (aq) + Cl - (aq) (2 particles) CaCl 2(s)  Ca 2+ (aq) + 2Cl - (aq) CaCl 2(s)  Ca 2+ (aq) + 2Cl - (aq) (3 particles) AlCl 3(s)  Al 3+ (aq) + 3Cl - (aq) AlCl 3(s)  Al 3+ (aq) + 3Cl - (aq) (4 particles, greatest effect) (4 particles, greatest effect) (Remember: ionic compounds dissociate.) (Remember: ionic compounds dissociate.)

Another unit of solution concentration Another common unit of concentration: Another common unit of concentration: Molality (m) (Notice that the ratio includes liters of solvent, not liters of total solution.)

Molality Calculate the molality of a solution prepared by dissolving 45.0 grams of KCl in 500. grams of water. Calculate the molality of a solution prepared by dissolving 45.0 grams of KCl in 500. grams of water. Answer: Answer: 45.0 grams KCl. (74.55 g/1 mol) =0.604 mol m = mol solute/kg solvent mol KCl/.500 kg water = 1.21 molal

Molality How many grams of sodium chloride must be dissolved in grams of water to prepare a 0.75 molal NaCl solution? How many grams of sodium chloride must be dissolved in grams of water to prepare a 0.75 molal NaCl solution? Answer: Answer: mol solute = m. kg solvent mol = 0.75m kg water = 0.19 mol NaCl 0.19 mol NaCl. (58.5 g/1 mol) = 11 grams NaCl

Calculating Solution Boiling Point The boiling point is raised by the addition of a nonvolatile solute. To calculate the change in boiling point, you must consider the concentration of the particles in solution. The boiling point is raised by the addition of a nonvolatile solute. To calculate the change in boiling point, you must consider the concentration of the particles in solution. Remember: water normally boils at 100 o C. Remember: water normally boils at 100 o C.  T = K b i m change in boiling point = (boiling point constant) (number if ions) (molality) (boiling point constant) (number if ions) (molality)

Problem What is the boiling point of 2.75m NaCl (aq) solution? (K b for water is o C/m) What is the boiling point of 2.75m NaCl (aq) solution? (K b for water is o C/m) Answer: Answer:  T = K b i m  T = (0.512 o C/m) (2) (2.75m)  T = 2.82 o C (this is the change in boiling point) Boiling point = 100 o C o C = o C

Calculating Solution Freezing Point You calculate the freezing point of a solution in a similar way. You calculate the freezing point of a solution in a similar way. Remember: water normally freezes at 0.0 o C. Remember: water normally freezes at 0.0 o C.  T = K f i m change in freezing point = (freezing point constant) (number if ions) (molality) (freezing point constant) (number if ions) (molality)

Problem What is the freezing point of a solution that contains 2.50 mol of CaCl 2 in 1250 g of water? What is the freezing point of a solution that contains 2.50 mol of CaCl 2 in 1250 g of water? (K f for water is 1.86 o C/m) Answer:  T = K f i m  T = (1.86 o C/m) (3) (2.50 mol/1.250 kg) = 11.2 o C (this is the freezing point change) Freezing point = 0.0 o C o C = o C

Determining Molar Mass You can use the change in boiling point or freezing point to calculate the molar mass of an unknown solute. You can use the change in boiling point or freezing point to calculate the molar mass of an unknown solute. If you are given a mass of an unknown solute, you can use the boiling point elevation to determine the molality of the solution, then use the volume of solvent to determine the moles of solute. Use the given mass and the moles of solute to determine the molar mass of the solute. If you are given a mass of an unknown solute, you can use the boiling point elevation to determine the molality of the solution, then use the volume of solvent to determine the moles of solute. Use the given mass and the moles of solute to determine the molar mass of the solute.

Problem: The boiling point of water is raised to o C when grams of a nonvolatile molecular solute is dissolved in 500. g of water. Calculate the molar mass of the solute. The boiling point of water is raised to o C when grams of a nonvolatile molecular solute is dissolved in 500. g of water. Calculate the molar mass of the solute. Answer: Answer:  T = K f i m 2.5 o C = (.512 o C/m) (1) (mol solute/.500 kg) mol solute = 2.44 mol grams/2.44 mol = 26.8 g/mol