1. I. The Nature of Solutions (p. 401 - 410, 425 - 433)
Ch. 13 & 14 - Solutions
I. The Nature of Solutions (p , )
C. Johannesson

2. A. Definitions Solution - homogeneous mixture
Solute - substance being dissolved
Solvent - present in greater amount
C. Johannesson

3. A. Definitions
Solute - KMnO4
Solvent - H2O
C. Johannesson

4. B. Solvation First... Then... Solvation – the process of dissolving
solute particles are surrounded by solvent particles
First...
solute particles are separated and pulled into solution
Then...
C. Johannesson

5. B. Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte+
sugar - +
acetic acid - +
salt
Non-
Electrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules only
solute exists as
ions and
molecules
solute exists as
ions only
DISSOCIATION
IONIZATION
View animation online.
C. Johannesson

6. NaCl(s)  Na+(aq) + Cl–(aq)
B. Solvation
Dissociation
separation of an ionic solid into aqueous ions
NaCl(s)  Na+(aq) + Cl–(aq)
C. Johannesson

7. HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
B. Solvation
Ionization
breaking apart of some polar molecules into aqueous ions
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
C. Johannesson

8. B. Solvation C6H12O6(s)  C6H12O6(aq) Molecular Solvation
molecules stay intact
C6H12O6(s)  C6H12O6(aq)
C. Johannesson

9. B. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR
C. Johannesson

10. B. Solvation Soap/Detergent polar “head” with long nonpolar “tail”
dissolves nonpolar grease in polar water
C. Johannesson

11. C. Solubility UNSATURATED SOLUTION more solute dissolves
no more solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration
C. Johannesson

12. C. Solubility Solubility
maximum grams of solute that will dissolve in 100 g of solvent at a given temperature
varies with temp
based on a saturated soln
C. Johannesson

13. C. Solubility Solubility Curve
shows the dependence of solubility on temperature
C. Johannesson

14. C. Solubility Solids are more soluble at... high temperatures.
Gases are more soluble at...
low temperatures &
high pressures (Henry’s Law).
EX: nitrogen narcosis, the “bends,” soda
C. Johannesson

15. Ch. 13 & 14 - Solutions II. Concentration (p. 412 - 418)
C. Johannesson

16. A. Concentration The amount of solute in a solution.
Describing Concentration
% by mass - medicated creams
% by volume - rubbing alcohol
ppm, ppb - water contaminants
molarity - used by chemists
molality - used by chemists
C. Johannesson

17. SAWS Water Quality Report - June 2000
A. Concentration
SAWS Water Quality Report - June 2000
C. Johannesson

18. B. Molality
mass of solvent only
1 kg water = 1 L water
C. Johannesson

19. B. Molality
Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2
0.25 kg water
= 3.2m MgCl2
C. Johannesson

20. B. Molality
How many grams of NaCl are req’d to make a 1.54m solution using kg of water?
0.500 kg water
1.54 mol NaCl
1 kg water
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl
C. Johannesson

21. C. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.
C. Johannesson

22. C. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
C. Johannesson

23. D. Preparing Solutions 1.54m NaCl in 0.500 kg of water
500 mL of 1.54M NaCl
mass 45.0 g of NaCl
add water until total volume is 500 mL
mass 45.0 g of NaCl
add kg of water
500 mL water
45.0 g NaCl
500 mL
mark
volumetric
flask
C. Johannesson

24. D. Preparing Solutions
Copyright © NT Curriculum Project, UW-Madison
(above: “Filling the volumetric flask”)
C. Johannesson

25. D. Preparing Solutions
Copyright © NT Curriculum Project, UW-Madison
(above: “Using your hand as a stopper”)
C. Johannesson

26. D. Preparing Solutions 250 mL of 6.0M HNO3 by dilution
measure 95 mL of 15.8M HNO3
95 mL of 15.8M HNO3
combine with water until total volume is 250 mL
250 mL mark
Safety: “Do as you oughtta, add the acid to the watta!”
water for safety
C. Johannesson

27. Solution Preparation Lab
Turn in one paper per team.
Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution.
For each of the following solutions:
1) mL of 0.50M NaCl
2) 0.25m NaCl in mL of water
3) mL of 3.0M HCl from 12.1M concentrate.
C. Johannesson

28. III. Colligative Properties (p. 436 - 446)
Ch. 13 & 14 - Solutions
III. Colligative Properties (p )
C. Johannesson

29. A. Definition Colligative Property
property that depends on the concentration of solute particles, not their identity
C. Johannesson

30. B. Types Freezing Point Depression (tf)
f.p. of a solution is lower than f.p. of the pure solvent
Boiling Point Elevation (tb)
b.p. of a solution is higher than b.p. of the pure solvent
C. Johannesson

31. Freezing Point Depression
B. Types
Freezing Point Depression
View Flash animation.
C. Johannesson

32. Boiling Point Elevation
B. Types
Boiling Point Elevation
Solute particles weaken IMF in the solvent.
C. Johannesson

33. B. Types Applications salting icy roads making ice cream antifreeze
cars (-64°C to 136°C)
fish & insects
C. Johannesson

34. C. Calculations t: change in temperature (°C)
t = k · m · n
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles
C. Johannesson

35. C. Calculations # of Particles Nonelectrolytes (covalent)
remain intact when dissolved
1 particle
Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles
C. Johannesson

36. C. Calculations
At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?
GIVEN:
b.p. = ?
tb = ?
kb = 3.60°C·kg/mol
WORK:
m = 0.73mol ÷ 0.225kg
tb = (3.60°C·kg/mol)(3.2m)(1)
tb = 12°C
b.p. = 181.8°C + 12°C
b.p. = 194°C
m = 3.2m
n = 1
tb = kb · m · n
C. Johannesson

37. C. Calculations
Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
tf = 18°C
f.p. = 0.00°C - 18°C
f.p. = -18°C
m = 4.8m
n = 2
tf = kf · m · n
C. Johannesson

38. I. Introduction to Acids & Bases (p. 453 - 473)
Ch. 15 & 16 - Acids & Bases
I. Introduction to Acids & Bases (p )
C. Johannesson

39. A. Properties ACIDS BASES electrolytes  electrolytes sour taste
bitter taste
turn litmus red
turn litmus blue
react with metals to form H2 gas
slippery feel
vinegar, milk, soda, apples, citrus fruits
ammonia, lye, antacid, baking soda
C. Johannesson
ChemASAP

40. Acids form hydronium ions (H3O+)
B. Definitions
Arrhenius - In aqueous solution…
Acids form hydronium ions (H3O+)
HCl + H2O  H3O+ + Cl– H Cl O – +
acid
C. Johannesson

41. Bases form hydroxide ions (OH-)
B. Definitions
Arrhenius - In aqueous solution…
Bases form hydroxide ions (OH-)
NH3 + H2O  NH4+ + OH- H N O – +
base
C. Johannesson

42. HCl + H2O  Cl– + H3O+ B. Definitions acid conjugate base base
Brønsted-Lowry
Acids are proton (H+) donors.
Bases are proton (H+) acceptors.
HCl + H2O  Cl– + H3O+
acid
conjugate base
base
conjugate acid
C. Johannesson

43. B. Definitions
H2O + HNO3  H3O+ + NO3– B A CA CB
C. Johannesson

44. Amphoteric - can be an acid or a base.
B. Definitions
NH3 + H2O  NH4+ + OH- B A CA CB
Amphoteric - can be an acid or a base.
C. Johannesson

45. HF H3PO4 H3O+ F - H2PO4- H2O B. Definitions
Give the conjugate base for each of the following: HF
H3PO4
H3O+
F -
H2PO4-
H2O
Polyprotic - an acid with more than one H+
C. Johannesson

46. Give the conjugate acid for each of the following:
B. Definitions
Give the conjugate acid for each of the following:
Br -
HSO4-
CO32-
HBr
H2SO4
HCO3-
C. Johannesson

47. B. Definitions Lewis Acids are electron pair acceptors.
Bases are electron pair donors.
Lewis base
Lewis acid
C. Johannesson

48. C. Strength Strong Acid/Base 100% ionized in water strong electrolyte- +
HCl
HNO3
H2SO4
HBr HI
HClO4
NaOH
KOH
Ca(OH)2
Ba(OH)2
C. Johannesson

49. does not ionize completely
C. Strength
Weak Acid/Base
does not ionize completely
weak electrolyte - + HF
CH3COOH
H3PO4
H2CO3
HCN
NH3
C. Johannesson

50. Ch. 15 & 16 - Acids & Bases
II. pH (p )
C. Johannesson

51. H2O + H2O H3O+ + OH- Kw = [H3O+][OH-] = 1.0  10-14
A. Ionization of Water
H2O + H2O H3O+ + OH-
Kw = [H3O+][OH-] = 1.0  10-14
C. Johannesson

52. A. Ionization of Water
Find the hydroxide ion concentration of 3.0  10-2 M HCl.
[H3O+][OH-] = 1.0  10-14
[3.0  10-2][OH-] = 1.0  10-14
[OH-] = 3.3  M
Acidic or basic?
Acidic
C. Johannesson

53. pouvoir hydrogène (Fr.)
B. pH Scale 14 7
INCREASING
ACIDITY
INCREASING
BASICITY
NEUTRAL
pH = -log[H3O+]
pouvoir hydrogène (Fr.)
“hydrogen power”
C. Johannesson

54. pH of Common Substances
B. pH Scale
pH of Common Substances
C. Johannesson

55. pH = -log[H3O+] pOH = -log[OH-] pH + pOH = 14
B. pH Scale
pH = -log[H3O+]
pOH = -log[OH-]
pH + pOH = 14
C. Johannesson

56. B. pH Scale What is the pH of 0.050 M HNO3? pH = -log[H3O+]
Acidic or basic?
Acidic
C. Johannesson

57. B. pH Scale
What is the molarity of HBr in a solution that has a pOH of 9.6?
pH + pOH = 14
pH = 14
pH = 4.4
pH = -log[H3O+]
4.4 = -log[H3O+]
-4.4 = log[H3O+]
[H3O+] = 4.0  10-5 M HBr
Acidic
C. Johannesson

58. Ch. 15 & 16 - Acids & Bases III. Titration (p. 493 - 503)
C. Johannesson

59. A. Neutralization Chemical reaction between an acid and a base.
Products are a salt (ionic compound) and water.
C. Johannesson

60. A. Neutralization ACID + BASE  SALT + WATER
HCl + NaOH  NaCl + H2O
strong
strong
neutral
HC2H3O2 + NaOH  NaC2H3O2 + H2O
weak
strong
basic
Salts can be neutral, acidic, or basic.
Neutralization does not mean pH = 7.
C. Johannesson

61. B. Titration
standard solution
unknown solution
Titration
Analytical method in which a standard solution is used to determine the concentration of an unknown solution.
C. Johannesson

62. B. Titration Equivalence point (endpoint)
Point at which equal amounts of H3O+ and OH- have been added.
Determined by…
indicator color change
dramatic change in pH
C. Johannesson

63. n: # of H+ ions in the acid or OH- ions in the base
B. Titration
moles H3O+ = moles OH-
MV n = MV n
M: Molarity
V: volume
n: # of H+ ions in the acid or OH- ions in the base
C. Johannesson

64. B. Titration
42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H2SO4. Find the molarity of H2SO4.
H3O+
M = ?
V = 50.0 mL
n = 2
OH-
M = 1.3M
V = 42.5 mL
n = 1
MV# = MV#
M(50.0mL)(2)
=(1.3M)(42.5mL)(1)
M = 0.55M H2SO4
C. Johannesson