1. Chapter 18
Solutions
for reference!

2. Section 18.1 Properties of Solutions
Objectives:
Identify the factors that determine the rate at which a solute dissolves
Calculate the solubility of a gas in a liquid under various pressure conditions

3. Definitions Soluble - capable of being dissolved.
Solution - homogeneous mixture of 2 or more substances in a single phase.
Solvent - the dissolving medium in a solution.
Solute - the substance dissolved in a solution.
Phase – a physically distinct & mechanically separable portion of a dispersion or solution.
Combine slides 2 & 3 or fix 3

4. Factors that Increase the Rate of Dissolving
A substance dissolves faster if-
It is stirred or shaken (agitation). This enables fresh solvent to come in contact with the solute.
The particles are made smaller. This increases the surface area.
The temperature is increased. This increases the kinetic energy, hence increases particle movement.

5. Solubility
Solubility- the amount that dissolves in a given quantity of a solvent at a given temperature to produce a saturated solution.

6. Solubility
Saturated solution - Contains the maximum amount of solid dissolved.
Unsaturated solution - Can dissolve more solute.
Supersaturated - A solution that is temporarily holding more than it can, a seed crystal will make it come out.

7. Solubility UNSATURATED SOLUTION more solute dissolves
no more solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration

8. Solubility For liquids dissolved in liquids:
Miscible – Two liquids are said to be miscible if they dissolve in each other.
Ex: Water and ethanol
Immiscible – Liquids that are insoluble in each other.
Ex: Oil and water

9. Factors Affecting Solubility
Solids in liquids - temperature goes up the solubility goes up.
Gases in a liquid - the temperature goes up the solubility goes down.
Gases in a liquid - as the pressure goes up the solubility goes up.

10. Solubility Solids are more soluble at... high temperatures.
Gases are more soluble at...
low temperatures.
high pressure (Henry’s Law)

11. Henry’s Law
(A friend of Dalton’s)
At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P)

12. S2 = 0.2 g/L S1 = P1 S2 P2 0.77 g/L= 350 kPa S2 100 kPa
Ex #1: If the solubility of a gas in water is 0.77 g/L at 350 kPa of pressure, what is its solubility, in g/L at 100 kPa of pressure. (Temperature is constant)
S1 = P1
S2 P2
0.77 g/L= 350 kPa
S kPa
(100 kPa)(0.77 g/L) = (350 kPa)S2
S2 = 0.22 g/L
S2 = 0.2 g/L

13. P2 = 260 kPa S1 = P1 S2 P2 3.6 g/L = 100. kPa 9.5 g/L P2
Ex #2: A gas has a solubility in water at 0°C of 3.6 g/L at a pressure of 100. kPa. What pressure is needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0 °C?
S1 = P1
S2 P2
3.6 g/L = 100. kPa
9.5 g/L P2
(100. kPa)(9.5 g/L) = (3.6 g/L)P2
P2 = kPa
P2 = 260 kPa

14. Section 18.1 Properties of Solutions
Did We Meet Our Objectives?
Identify the factors that determine the rate at which a solute dissolves
Calculate the solubility of a gas in a liquid under various pressure conditions

15. Section 18.2 Concentrations of Solutions
Objectives:
Solve problems involving the molarity of a solution
Describe how to prepare dilute solutions from more concentrated solutions of known molarity
Explain what is meant by percent by volume and percent by mass solutions

16. Molarity
The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent.
A dilute solution is one that contains only a low concentration of solute.
A concentrated solution contains a high concentration of solute.

17. Molarity
Molarity (M) - the number of moles of solute dissolved per liter of solution.
Molarity is also known as molar concentration. It is read as the numerical value followed by the word “molar.”
6 M HCl

18. 0.95 M NaF M = moles Volume (L) 10.0 g NaF x 1 mol NaF = 0.238 moles
Ex #3: Find the molarity of a 250 mL solution containing 10.0 g of NaF.
M = moles
Volume (L)
10.0 g NaF x 1 mol NaF = moles
41.98 g NaF
M = mol
0.25 L
M = 0.952
0.95 M NaF

19. n = 4.5 mol NaCl M = moles Volume (L) 0.75 M NaCl = n . 6.0 L
Ex #4: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?
M = moles
Volume (L)
0.75 M NaCl = n .
6.0 L
n = (0.75 mol)(6.0 L) L
n = 4.5 mol NaCl

20. Ex #5: How many grams of CaCl2 are needed to make 625 mL of a 2
Ex #5: How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution?
M = moles
Volume (L)
2.0 M CaCl2 = n .
0.625 L
n = (2.0 mol)(0.625 L)
L .
n = 1.25 mol CaCl2
1.25 mol CaCl2 x g = g CaCl2
1 mol .
m = 140 g CaCl2

21. Moles before = the moles after
Making Dilutions
The number of moles of solute doesn’t change if you add more solvent.
M = n/V
n = M V
Moles before = the moles after
n1 = n2
M1 x V1 = M2 x V2

22. V1 = 95 mL M1V1 = M2V2 (15.8 M)V1 = (6.0 M)(250 mL)
Ex #6: What volume of 15.8 M HNO3 is required to make 250 mL of a 6.0 M solution?
M1V1 = M2V2
(15.8 M)V1 = (6.0 M)(250 mL)
V1 = mL
V1 = 95 mL

23. 0.46 M solution M1V1 = M2V2 (0.88 M)(2.0 L) = (M2)(3.8 L)
Ex #7: 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity?
M1V1 = M2V2
(0.88 M)(2.0 L) = (M2)(3.8 L)
M2 =
0.46 M solution

24. Percent solutions
Percent by volume %(v/v) – A convenient way to measure if both solute & solvent are liquids
% by volume = Volume of solute x 100% Volume of solution

25. % by volume = Volume of solute x 100%
Ex #8: What is the percent solution if 25 mL of CH3OH is diluted to a volume of 150 mL with water?
% by volume = Volume of solute x 100%
Volume of solution
% by volume = 25 mL x 100% = %
150 mL
17% CH3OH

26. Percent solutions
Percent by mass % (m/v) - is used for solids dissolved in liquids (more common).
% by mass = Mass of solute(g) x 100% Volume of solution(mL)

27. 5.9% NaCl % by mass = Mass of solute(g) x 100% Volume of solution(mL)
Ex #9: 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution?
% by mass = Mass of solute(g) x 100%
Volume of solution(mL)
% by mass = 4.8 g NaCl x 100%
82 mL
% by mass =
5.9% NaCl

28. Ex #10: How many grams of salt are there in 52 mL of a 6.3% solution?
% by mass = Mass of solute(g) x 100%
Volume of solution(mL)
6.3% = m x 100%
52 mL
m = g
m = 3.3 g salt

29. Concentrations Describing Concentration % by mass - medicated creams
% by volume - rubbing alcohol
ppm, ppb - water contaminants
molarity - used by chemists
molality - used by chemists

30. Section 18.2 Concentrations of Solutions
Did We Meet Our Objectives?
Solve problems involving the molarity of a solution
Describe how to prepare dilute solutions from more concentrated solutions of known molarity
Explain what is meant by percent by volume and percent by mass solutions

31. Section 18.3 Colligative Properties of Solutions
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 with the pure solvent.

32. Decrease in Vapor Pressure
Colligative properties depend only on the number of particles dissolved in a given mass of solvent.
Not based on the kind of particle
Three important colligative properties of solutions are:
Vapor pressure lowering
Boiling point elevation
Freezing point depression

33. Number of Dissolved Particles
Add 3 formula units of KMnO4
Add 3 formula units of Na2CO3
Produces 3 K+ ions and 3 MnO4- ions, total of 6 ions
Produces 6 Na+ ions and 3 CO32- ions, total of 9 ions

34. Number of Dissolved Particles
KCl
BeF2
KCl → K+ + Cl-
BeF2 → Be2+ + 2F-
KCl → 2 mols ions
BeF2 → 3 mols ions
MgO
C6H12O
MgO → Mg2+ + O2-
C6H12O → C6H12O
MgO→ 2 mols ions
C6H12O → 1 mol molecule

35. Vapor Pressure
Vapor pressure – the pressure above a substance in a sealed container

36. Vapor Pressure
Vapor pressure is always lowered if a nonvolatile (not easily vaporized) solute is added
The decrease is proportional to the # of particles in solution
NaCl breaks into 2 ions, but sugar stays as 1 molecule, so NaCl will lower the vapor pressure more
The higher the amount of moles of product, the lower the vapor pressure.
When a ions become solvated, they become surrounded by shells of water
Attractions keep molecules from escaping.

37. Boiling Point Elevation
Boiling point – the temperature at which the vapor pressure of a liquid equals the atmospheric pressure.
Adding a solute lowers the vapor pressure
Lower vapor pressure - higher boiling point.
Boiling Point Elevation (tb) – the b.p. of a soln is higher than the b.p. of the pure solvent.

38. Freezing Point Depression
Adding a solute makes the freezing point lower.
Solids form when molecules make an orderly pattern.
The solute molecules break up the orderly pattern.
Freezing Point Depression (tf) – the f.p. of a solution is lower than the f.p. of pure solvent

39. Concept Practice KF → K+ + F- = 2 mols MgF2 → Mg2+ + 2F- = 3 mols
If equal numbers of moles of KF and MgF2 are dissolved in equal volumes of water, state which solution has the highest:
Boiling point
Freezing point
Vapor pressure
KF → K+ + F- = 2 mols
MgF2 → Mg2+ + 2F- = 3 mols

40. Section 18.3 Colligative Properties of Solutions
Did We Meet Our 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 with the pure solvent.

41. Section 18.4 Calculations Involving Colligative Properties
Objectives:
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

42. Molality Another unit for concentration
Molality (m) is the number of moles of solute dissolved in 1 kg of solvent

43. kilograms of solvent vs. liters of solution
Molality vs. Molarity
kilograms of solvent vs. liters of solution

44. 3.2 m MgCl2 m = mol or m = n . mass (kg) m
Ex #11: Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.
m = mol or m = n mass (kg) m
The density of water is 1 g/mL, so 250 mL of water = 250 g
75 g MgCl2 x 1 mol MgCl2 = mol MgCl2
95.21 g MgCl2
m = mol
.250 kg
m = mol/kg
3.2 m MgCl2

45. m = 45.0 g NaCl m = mol or m = n . mass (kg) m 1.54 m = n . 0.500 kg
Ex #12: How many grams of NaCl are required to make a 1.54 m solution using kg of water?
m = mol or m = n mass (kg) m
1.54 m = n
0.500 kg
n = 0.77 mol
0.77 mol NaCl x g NaCl = g NaCl
1 mol NaCl
m = 45.0 g NaCl

46. Mole Fraction
The ratio of the moles of a solute in solution to the total number of moles of solvent and solute is the mole fraction of that solute.
XA = nA XB = nB .
nA + nB nA + nB

47. Ex #13: Compute the mole fraction of each component in a solution of 1
Ex #13: Compute the mole fraction of each component in a solution of 1.25 mol of ethylene glycol (EG) and 4.00 mole water.
1.25 ( ) 1.25 = =

48. Ex #14: What is the mole fraction of each component in a solution made by mixing g of ethanol (C2H5OH) and g of water?
300.0 g C2H5OH x 1 mol C2H5OH = mol C2H5OH g C2H5OH g H2O x 1 mol H2O = mol H2O g H2O XA = nA XB = nB . nA + nB nA + nB Xethanol = Xwater =

49. Nonelectrolytes (covalent)
Determining the Number of Particles in a Covalent and an Ionic Compound
Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles
Nonelectrolytes (covalent)
remain intact when dissolved
1 particle

50. Ex #15: Determine the total number of particles when the following dissolve.
NaCl
BaCl2
CCl4

51. Boiling-Point Elevation
ΔTb = Kb x m
Tb – Boiling-Point Elevation (°C)
Kb – Molal Boiling-Point Elevation Constant (°C/m)
m – Molality (m = moles/kg)
Remember to multiply by the number of particles.

52. Ex #16: What is the boiling point of a 1.50 m NaCl solution?
ΔTb = Kb x m ΔTb = (0.512 °C/m) x (1.50 m x 2) ΔTb =1.55 °C Tsolution = °C °C Tsolution = °C

53. Ex #17: At what temperature will a solution that is composed of 0
Ex #17: At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil (b.p. of phenol 181.7°C)?
mphenol = mol/kg = (0.73 mol/0.225 kg) = 3.2 m ΔTb = Kb x m ΔTb = (3.56 °C/m) x (3.2 m) ΔTb = °C = 11 °C Tsolution = °C + 11 °C = °C Tsolution = 193 °C

54. Freezing-Point Depression
ΔTf = Kf x m
Tf (Freezing-Point Depression - °C) lowers
Kf (Molal Freezing-Point Depression Constant - °C/m)
m (Molality - m = moles/kg)
Remember to multiply by the number of particles.

55. Ex #18: Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
mNaCl = mol/kg = (0.48 mol/0.100 kg) = 4.8 m ΔTf = Kf x m ΔTf = (1.86 °C/m) x (4.8 m x 2) ΔTf = °C = 18 °C Tsolution = 0.0 °C - 18 °C = -18 °C Tsolution = -18 °C

56. Ex #19: What is the boiling point of a solution made by dissolving 1
Ex #19: What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water? What is the freezing point?
ΔTb = Kb x m ΔTb = (0.512 °C/m) x (1.6 m x 2) ΔTb = °C = 1.6 °C Tbp = °C °C = °C Tbp = °C ΔTf = Kf x m ΔTf = (1.86 °C/m) x (1.6 m x 2) ΔTf = °C = 6.0 °C Tfp = 0.0 °C – 6.0 °C = -6.0 °C Tfp = -6.0 °C

57. Molar Mass
One can use the changes in boiling and freezing points to determine the molar mass of a substance.
Molar Mass = Mass of solute
Moles of solute

58. Ex #20: A solution containing 16
Ex #20: A solution containing 16.9 g of a nonvolatile molecular compound in 250 g of water has a freezing point of ˚C. What is the molar mass of the substance?
ΔTf = Kf x m °C = (1.86 °C/m) x (m) m = m m = mol mass (kg) m = mol/(0.250 kg) n = mol Molar Mass = Mass of solute Moles of solute M = 16.9 g = 169 g/mol mol .

60. Section 18.4 Calculations Involving Colligative Properties
Did We Meet Our Objectives?
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