Name: _______________________________________________ Date: ____________ Period: ________
AP Chemistry Unit 10 Review Sheet
Part 1. Choose the best answer to each question below. You should not need a
calculator for this part of the review sheet.
1) The calculation of concentration and pH for weak acids is more complex than for
strong acids due to:
a) the incomplete ionization of weak acids.
b) the low Ka value for strong acids.
c) the more complex atomic structures of strong acids.
d) the low percent ionization of strong acids.
e) the inconsistent Kb value for strong acids.
2) The general reaction of an acid dissolving in water may be shown as:
HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)
A conjugate acid base pair for this reaction is:
a) HA and H2O
b) HA and A-
c) H2O and A-
d) H3O+ and A-
e) HA and H3O+
3) Strong acids are those which
a) have an equilibrium lying far to the left.
b) yield a weak conjugate base when reacting with water.
c) have a conjugate base which is a stronger base than water.
d) readily remove the H+ ions from water.
e) are only slightly dissociated (ionized) at equilibrium.
4) When calculating the pOH of a hydrofluoric acid solution (Ka = 7.2 x 10-4) from its
concentration, the contribution of water ionizing (Kw = 1.0 x 10-14) is usually
a) hydrofluoric acid is such a weak acid.
b) hydrofluoric acid can dissolve glass.
c) the ionization of pure water provides relatively few H+ ions.
d) the [OH-] for pure water is unknown.
e) the conjugate base of HF is such a strong base.
5. The percent dissociation (percent ionization) for weak acids:
a) is always the same for a given acid, no matter what the concentration.
b) usually increases as the acid becomes more concentrated.
c) compares the amount of acid that has dissolved at equilibrium with the initial
concentration of the acid.
d) may only be used to express the dissociation of weak acids.
e) has no meaning for polyprotic acids.
6. The [OH-] of a certain aqueous solution is 1.0 x 10-5 M. The pH of this same
solution must be:
a) 1.0 x 10-14
7. In many calculations for the pH of a weak acid from the concentration of the acid,
an assumption is made that often takes the form of [HA]0 – x = [HA]0.
a) This is valid because x is very small compared to the initial concentration of the
b) This is valid because the concentration of the acid changes by such large
c) This is valid because the actual value of x cannot be known.
d) This is valid because pH is not dependent upon the concentration of the weak
e) The approximation is always valid and does not need to be checked.
8. HA is a weak acid which is 4.0% dissociated at 0.100 M. Determine the Ka for this
9. Sulfur trioxide is an acidic oxide due to the
a) high electronegativity of sulfur in the O-S bond, forming strong covalent bonds.
b) low electronegativity of sulfur in the O-S bond, forming strong covalent bonds.
c) high electronegativity of sulfur in the O-S bond, forming strong ionic bonds.
d) low electronegativity of sulfur in the O-S bond, forming strong ionic bonds.
e) attraction for the H-O bonds by the H+ ion in water.
10. Ionic substances known as salts can form acidic, basic, and neutral solutions
when dissolved in water. When dissolved in water,
a) KNO3 forms a basic solution.
b) NaCl forms an acidic solution.
c) NaNO3 forms an acidic solution.
d) NaF forms a basic solution.
e) KClO4 forms an acidic solution.
11. The net ionic equation representing the equilibrium established when potassium
sulfide dissolves in water is shown. Label the species present as either acid or
base according to the Bronsted-Lowry theory.
S2- + H 2O ↔ HS- OH-
a) acid base acid base
b) base acid acid base
c) acid base base acid
d) base acid base acid
e) acid acid base base
12. Which one of the following is a Lewis acid, but not a Bronsted-Lowry acid?
13. Which of the following metal ions would you expect to show the greatest acidic
properties in water?
14. Which of the following binary hydrides is the most basic?
Part 2. Answer the questions below. Show all work you did to arrive at your answer.
15) Methylamine, CH3NH2, is a weak base. A 0.065 M solution of methylamine was
prepared by dissolving the substance in water. The solution has a pH of 11.70.
a) Write the reaction that occurs when methylamine is dissolved in water.
b) What is the value of Kb for methylamine?
16) a) What are the equilibrium concentrations of [H3O+], [CH3CO2-], and [CH3CO2H]
in a 0.20 M aqueous solution of CH3CO2H? (Ka for CH3CO2H is 1.8 x 10-5).
17) Lactic acid (C3H5O3H) is found in sour milk and in muscles after activity. The Ka
for lactic acid is 1.4 x 10-4.
a) What is the pH of a 0.50 M lactic acid solution?
18) The Kw for water at 25OC is 1.0 x 10-14, but is 1.0 x 10-13 at 60OC.
a) Give the chemical equation for the autoionization of water.
b) Determine the [OH-] for water at 60OC.
c) Determine the pH of water at 60OC.
d) Is the reaction of the autoionization of water endothermic or exothermic? Support
your answer with data and explanation.
19) Phosphoric acid, H3PO4, is a triprotic acid.
a) Show the three equations involved in the dissociation of this substance.
b) Illustrate how these three equations might be combined to show the complete
dissociation of phosphoric acid.
c) If a 7.0 M H3PO4 solution dissociated, calculate the pH of the solution.
Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13
d) Determine the concentration for the ions H2PO4-, HPO42-, and PO43- (aq) in the
dissociated 7.0 M H3PO4.
e) Determine the pOH for this same 7.0 M H3PO4.
1) A 2)B 3)B 4)C 5)C 6)D 7)A 8)B 9)A 10)D 11)B 12)B 13)E 14)E
15) a) CH3NH2 (aq) + H2O (l) ↔ CH3NH3+ (aq) + OH- (aq)
b) Kb = 4.2 x 10-4
16) [CH3CO2H] = 0.20 M, [CH3CO2-] = [H3O+] = 1.9 x 10-3 M
17) pH = 2.08
18) a) H2O (l) + H2O (l) ↔ H3O+ (aq) + OH- (aq)
b) [OH-] = 3.2 x 10-7 M
c) pH = 6.49
d) Since raising the temperature from 25OC to 60OC causes the Kw to increase,
more product must be forming. If raising the temperature causes the reaction to
shift to the right (toward products), then the reaction must be endothermic.
19) a) H3PO4 (aq) ↔ H+ (aq) + H2PO4- (aq)
H2PO4- (aq) ↔ H+ (aq) + HPO42- (aq)
HPO42- (aq) ↔ H+ (aq) + PO42- (aq)
b) Reactions above add to give: H3PO4 (aq) ↔ 3H+ (aq) + PO43- (aq)
c) pH = 0.64
d) [H+] = [H2PO4-] = 0.23 M
[HPO42-] = 6.2 x 10-8 M
[PO43-] = 1.3 x 10-19 M
e) pOH = 13.36