(^26) › STEP 2. Determine Your Test Readiness
- FeS(s) + 2 H+(aq) Fe^2 +(aq) + H 2 S(aq)
What is the equilibrium constant for the above
reaction? The successive acid dissociation con-
stants for H 2 S are 9.5 × 10 -^8 (Ka1) and 1 × 10 -^19
(Ka2). The Ksp, the solubility product constant,
for FeS equals 5.0 × 10 -^18.
(A) 9.5 × 10 -^27 /5.0 × 10 -^18
(B) 5.0 × 10 -^18 /9.5 × 10 -^27
(C) 5.0 × 10 -^18 /9.5 × 10 -^8
(D) 9.5 × 10 -^8 /5.0 × 10 -^18
- The Ksp for Cr(OH) 3 is 1.6 × 10 -^30. What is the
molar solubility of this compound in water?
(A) ×(^4) 1.6 10 −^30
(B)^4 1.6×10 /27−^30
(C) 1.6 × 10 -^30
(D) 1.6 × 10 -^30 /27
Chapter 16
- I-(aq) + H+(aq) + MnO 4 - (aq) → Mn^2 +(aq) +
H 2 O(l) + I 2 (s)
What is the coefficient of H+ when the above
reaction is balanced?
(A) 12
(B) 32
(C) 16
(D) 8
- How many moles of Au will deposit on the cath-
ode when 0.60 Faradays of electricity is passed
through a 1.0 M solution of Au^3 +?
(A) 0.60 mol
(B) 0.30 mol
(C) 0.40 mol
(D) 0.20 mol
80. Sn^2 +(aq) + 2 Fe^3 +(aq) → Sn^4 +(aq) + 2 Fe^2 +(aq)
The reaction shown above was used in an electro-
lytic cell. The voltage measured for the cell was
not equal to the calculated E° for the cell. Which
of the following could explain this discrepancy?
(A) Both of the solutions were at 25°C instead
of 0°C.
(B) The anode and cathode were different sizes.
(C) The anion in the anode compartment was
chloride instead of nitrate, as in the cathode
compartment.
(D) One or more of the ion concentrations was
not 1 M.
Questions 81–82 refer to the following half-reaction
in an electrolytic cell:
2 SO 42 - (aq) + 10 H+(aq) + 8 e- →
S 2 O 32 - (aq) + 5 H 2 O(l)
81. Choose the correct statement from the following
list.
(A) The sulfur is oxidized.
(B) This is the cathode reaction.
(C) The oxidation state of sulfur does not change.
(D) The H+ serves as a catalyst.
82. If a current of 0.60 amperes is passed through
the electrolytic cell for 0.75 h, how should you
calculate the grams of S 2 O 32 - (aq) formed?
(A) (0.60) (0.75) (3,600) (112)/(96,500) (8)
(B) (0.60) (0.75) (3,600) (112)/(96,500) (10)
(C) (0.60) (0.75) (60) (32)/(96,500) (8)
(D) (0.60) (0.75) (3,600) (112)/(10)
83. 2 BrO 3 - (aq) + 12 H+(aq) + 10 e- → Br 2 (aq) +
6 H 2 O(l)
If a current of 5.0 A is passed through the elec-
trolytic cell for 0.50 h, how should you calculate
the number of grams of Br 2 that will form?
(A) (5.0) (0.50) (3,600) (159.8)/(10)
(B) (5.0) (0.50) (3,600) (159.8)/(96,500) (10)
(C) (5.0) (0.50) (60) (159.8)/(96,500) (10)
(D) (5.0) (0.50) (3,600) (79.9)/(96,500) (10)