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

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

  
temperature speeds up the rate of dissolving – increasing the
– stirring speeds up the rate of dissolving
– 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.

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

Solubility
 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.)

 …if you added more sugar to a saturated 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?

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

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

 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.

 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 out of the solution.
2 over the liquid is high and when the cap is opened, the pressure is reduced and bubbles begin to come


the liquid.
: at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above S 1 /P 1 = S 2 /P 2

 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.) 
: S 1 /P 1 0.77 g/L = S 2 /P / 3.5 atm = S 2 2 S 2 = 0.22 g/L / 1.0 atm

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

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

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

 Consider both the moles of solute and the volume of solution to find concentration.

 To find the moles of solute in a given volume of solution of known molarity use the definition of 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.

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

 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) 17.0 grams
.
(1 mol/58.5 g) = 0.2906 mol mol solute/L solution = molarity 0.2906 mol / 0.200 L = 0.145 molar NaCl

 How many grams of sodium chloride are needed to prepare 275 mL of a 0.200M solution of NaCl?
 Answer: molarity
.
L solution = mol solute 0.200M
.
0.275 L = 0.0550 moles of NaCl 0.0550 mol
.
(58.5 g/1 mol) =3.22 grams NaCl

 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
M
conc . L conc
M
conc . V
M
1
.
conc V 1 = (mol solute) diluted = M diluted . L diluted = M = M 2
.
diluted V 2 . V diluted

 Diluting a solution  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. 
M
1
.
V 1 = M 2
.
V 2

 How many mL of a concentrated stock solution of 5.0M HCl is needed to prepare 250. mL of 1.75M HCl?
 Answer: 
M
1
.
V 1 = M 2
.
V 2 5.0M V 1
.
V 1 = 1.75M = 87.5 mL
.
250 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 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 molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound

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

 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.

   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 Freezing point decreases

1.
2.
3.
4.
  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?
1.0M C 6 H 12 O 6 1.0M NaCl 1.0M CaCl 2 1.0M AlCl 3

     
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) (1 particle) NaCl (s)  Na + (aq) + Cl (aq) (2 particles) CaCl 2(s)  Ca 2+ (aq) + 2Cl (aq) (3 particles) AlCl 3(s)  Al 3+ (aq) + 3Cl (aq) (4 particles, greatest effect) (Remember: ionic compounds dissociate.)

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

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

 How many grams of sodium chloride must be dissolved in 250.0 grams of water to prepare a 0.75 molal NaCl solution?
 Answer: mol solute = m mol = 0.75m
.
.
kg solvent 0.250 kg water = 0.19 mol NaCl 0.19 mol NaCl
.
(58.5 g/1 mol) = 11 grams NaCl

  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.
D T = K b
i m
change in boiling point = (boiling point constant) (number if ions) (molality)

 What is the boiling point of 2.75m NaCl (aq) solution? (K b for water is 0.512
o C/m)  Answer: D T = K b
i m
D T = (0.512
o C/m) (2) (2.75m) D T = 2.82
o C (this is the change in boiling point) Boiling point = 100 o C + 2.82
o C = 102.82
o C

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

 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: D T = K f
i m
D 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 - 11.2
o C = -11.2
o C

 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.

 The boiling point of water is raised to 102.5
o C when 65.50 grams of a nonvolatile molecular solute is dissolved in 500. g of water. Calculate the molar mass of the solute.
 Answer: D T = K f
i m
2.5
o C = (.512
o C/m) (1) (mol solute/.500 kg) mol solute = 2.44 mol 65.50 grams/2.44 mol = 26.8 g/mol