QUIZ CHAPTER #14
ANSWER KEY
Student’s Name
1. What would be the products of the following acid-base reaction:
−
H2 PO−
4 (aq) + HCO3 (aq) ⇄
Assume dihydrogen phosphate ion, H2 PO−
4 , acts as a Bronsted acid and hydrogen carbonate ion,
HCO−
3 , acts as a Bronsted base.
a) H3PO4 and CO2−
3
b) 𝐇𝐏𝐎𝟐−
𝟒 and H2CO3
−
c) PO3−
4 and HCO3
2−
d) H3O+, PO3−
4 and CO3
2. The Kb value for methylamine, CH3NH2, is 4.2×10-4. What is the pH of a 0.050 M CH3NH2
solution?
CH3NH2(aq) + H2O(aq) ⇄ CH3 NH3+ (aq) + OH − (aq)
a) 1.30
b) 2.34
initial
equilibrium
c) 11.66
Kb =
d) 12.70
0.050 M
0.050−𝑥
−
[CH3 NH+
3 ][OH ]
[CH3 NH2 ]
=
𝑥2
0.050−𝑥
pOH = −𝑙𝑜𝑔[OH − ] = 2.34
= 4.2×10-4,
0
0
𝑥
𝑥
𝑥 = [OH − ] = 4.4×10-3 M
pH = 14.00 − pOH = 14.00 − 2.34 = 11.66
3.Which of the following salt solutions will ALL have an acidic pH?
a) Al(NO3)3, (NH4)2SO4, NaHSO4
b) KI, NaCN, NH4Cl
c) NaF, KNO3, NaHCO3
d) AlCl3, LiCl, KBr
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4. Sulfurous acid, H2SO3, is a diprotic acid with Ka1 = 1.3×10-2, Ka2 = 6.3×10-8. What is a pH of
a 0.25 M aqueous solution of sulfurous acid? What is the concentration of sulfite ion, SO2−
3 , in the
solution?
-8
a) pH = 1.24, [SO2−
3 ] = 6.3×10
1st ionization H2SO3(aq) ⇄ H+(aq) + HSO−
3 (aq)
-2
b) pH = 1.29, [SO2−
3 ] = 5.1×10
initial 0.25 M
equilibrium 0.25−𝑥
-2
c) pH = 1.24, [SO2−
3 ] = 5.1×10
Ka1 =
-8
d) pH = 1.29, [𝐒𝐎𝟐−
𝟑 ] = 6.3×10
[H+ ][HSO−
3]
[H2 SO3 ]
𝑥 = [H + ] = 5.1×10-2 M,
2nd ionization
0.051 M
equilibrium
0.051 −𝑦
[H+ ][SO2−
3 ]
[HSO−
3]
0
𝑥
𝑥
𝑥2
0.25−𝑥
= 1.3×10-2
pH = −𝑙𝑜𝑔[H + ] = 1.29
+
2−
HSO−
3 (aq) ⇄ H (aq) + SO3 (aq)
initial
Ka2 =
=
0
=
0.051
0
0.051 +𝑦
𝑦
(0.051+𝑦)𝑦
0.051−𝑦
Since 𝑦 ≪ 0.051, then 0.051 + 𝑦 ≈ 0.051 and 0.051 − 𝑦 ≈ 0.051. Therefore:
Ka2 =
(0.051+𝑦)𝑦
0.051−𝑦
=
0.051×𝑦
0.051
≈𝑦
-8
𝑦 = [SO2−
3 ] = 6.3×10 M
5. In which of the following chemical reactions aluminum hydroxide, Al(OH)3, acts as a Lewis
acid?
a) Al(OH)3(s) + 𝐎𝐇 − (aq) → 𝐀𝐥(𝐎𝐇)−
𝟒 (aq)
b) Al(OH)3(s) + 3 HCl(aq) → AlCl3(aq) + 3 H2O(l)
c) 2 Al(OH)3(s) → Al2O3(s) + 3 H2O(l)
d) 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
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