Exam 3B, Spring 2000
Exam 3B, Spring 20001. Complete and balance the following reactions:
a. C5H11N(aq) + H2O(l)
b. (CH3)2NH(aq) + HOBr(aq)
c. Fe2+(aq) + H2O(l)
d. Cr(NO3)3(aq) + K3PO4(aq)
e. Ca(C2H3O2)2(aq) + H2SO4(aq)
f. Zn(NO3)2(aq) + NH3(aq)
a. C5H11N(aq) + H2O(l) → ← C5H11NH+(aq) + OH–(aq)
b. (CH3)2NH(aq) + HOBr(aq) → (CH3)2NH2+(aq) + OBr–(aq) (Kc = 2.5×10–9/1.7×10–11 = 150 > 100)
c. Fe2+(aq) + 2 H2O(l) → ← FeOH+(aq) + H3O+(aq)
d. Cr(NO3)3(aq) + K3PO4(aq) → CrPO4(s) + 3 K+(aq) + 3 NO3–(aq)
e. Ca(C2H3O2)2(aq) + H2SO4(aq) → CaSO4(s) + 2 HC2H3O2(aq)
f. Zn(NO3)2(aq) + 4 NH3(aq) → ← [Zn(NH3)]42+(aq) + 2 NO3–(aq)
2. Consider a 0.65 M solution of triethylamine, (C2H5)3N, at 25 °C.
a. Find the pH of the solution.
b. Enough triethylammonium chloride, (C2H5)3NHCl, is added to the solution to make the concentration of the triethylammonium ion 0.70 M. Find the pH of this buffer.
c. 5.0 g of perchloric acid is added to 1.0 L of the buffer prepared in part b. What is the new pH?
a. Find the pH of the solution.
(CH3)3N(aq) +H2O(l) → ← (CH3)3NH+(aq) +OH–(aq)
Kb = [(CH3)3NH+]e[OH–]e [(CH3)3N]e = 10–(14.00 – pKa) = 10–(14.00 – 10.72) = 10–3.28 = 5.2×10–4
Initial0.6500
Change–x+x+x
Equil.0.65 – xxx
Approximate? 0.65/5.2×10–4 = 1250 > 100 so Yes
5.2×10–4 = x2/0.65
x = [OH–]e = 1.8×10–2
pOH = –log(1.8×10–2) = 1.74
pH = 14.00 – 1.74 = 12.26
b. Enough triethylammonium chloride, (C2H5)3NHCl, is added to the solution to make the concentration of the triethylammonium ion 0.70 M. Find the pH of this buffer.
(CH3)3N(aq) +H2O(l) → ← (CH3)3NH+(aq) +OH–(aq)
Kb = [(CH3)3NH+]e[OH–]e [(CH3)3N]e = 10–(14.00 – pKa) = 10–(14.00 – 10.72) = 10–3.28 = 5.2×10–4
Initial0.650.700
Change–x+x+x
Equil.0.65 – x0.70 + xx
Approximate? 0.65/5.2×10–4 = 1250 > 100 so Yes
5.2×10–4 = x(0.70)/0.65
x = [OH–]e = 4.8×10–4
pOH = –log(4.8×10–4) = 3.32
pH = 14.00 – 3.32 = 10.68
c. 5.0 g of perchloric acid is added to 1.0 L of the buffer prepared in part b. What is the new pH?
[HClO4] = 5.0 g/(1.0 g/mol + 35.5 g/mol + 4(16.0) g/mol)/L = 0.050 M
(CH3)3N(aq) +H2O(l) → ← (CH3)3NH+(aq) +OH–(aq)
(CH3)3N(aq) +HClO4(aq) → (CH3)3NH+(aq) + ClO4–(aq)
Kb = [(CH3)3NH+]e[OH–]e [(CH3)3N]e = 10–(14.00 – pKa) = 10–(14.00 – 10.72) = 10–3.28 = 5.2×10–4
Initial0.650.700
A/B rxn–0.050+0.0500
Change–x+x+x
Equil.0.60 – x0.75 + xx
Approximate? 0.60/5.2×10–4 = 1150 > 100 so Yes
5.2×10–4 = x(0.75)/0.60
x = [OH–]e = 4.2×10–4
pOH = –log(4.2×10–4) = 3.38
pH = 14.00 – 3.38 = 10.62
3. Consider silver bromide at 25 °C.
a. What is the molar solubility of silver bromide in pure water?
b. What is the molar solubility of silver bromide in 0.035 M sodium bromide solution?
c. What is the molar solubility of silver bromide in 1.0 M ammonia solution?
a. What is the molar solubility of silver bromide in pure water?
AgBr(s) → ← Ag+(aq) +Br–(aq)
Ksp = [Ag+]e[Br–]e = 5.0×10–13
Initial00
Change+x+x
Equil.xx
5.0×10–13 = x2
x = molar solubility = 7.1×10–7 M
b. What is the molar solubility of silver bromide in 0.035 M sodium bromide solution?
NaBr(aq)→ Na+(aq) +Br–(aq)
AgBr(s) → ← Ag+(aq) +Br–(aq)
Ksp = [Ag+]e[Br–]e = 5.0×10–13
Initial00.035
Change+x+x
Equil.x0.035 + x
x will be small, so 0.035 + x ~ 0.035
5.0×10–13 = x(0.035)
x = molar solubility = 1.4×10–11 M
c. What is the molar solubility of silver bromide in 1.0 M ammonia solution?
AgBr(s) → ← Ag+(aq) +Br–(aq)
Ag+(aq)+ 2 NH3(aq) → ← [Ag(NH3)2]+(aq)
Net:AgBr(s)+ 2 NH3(aq) → ← [Ag(NH3)2]+(aq) +Br–(aq)
Kc = [[Ag(NH3)2]+]e[Br–]e [NH3]e2 = Ksp×Kf = 5.0×10–13×1.6×107 = 8.0×10–6
Initial1.000
Change–2x+x+x
Equil.1.0 – 2xxx
8.0×10–6 = x2/(1.0 – 2x)2
2.8×10–3 = x/(1.0 – 2x)
x = molar solubility = 2.8×10–3 M
4. Indicate whether the following salts are acidic, basic, or neutral.
a. NH4NO3
b. Mn(OH)2
c. Mn(HSO4)2
d. NaC2H3O2
e. NaClO4
a. NH4NO3(aq) → NH4+(aq) + NO3–(aq)
NH4+(aq) + H2O(l) → ← H3O+(aq) + NH3(aq)
NO3–(aq) + H2O(l) → NR
Acidic
b. Mn(OH)2(s) → ← Mn2+(aq) + 2 OH–(aq)
Basic
c. Mn(HSO4)2(aq) → Mn2+(aq) + 2 HSO4–(aq)
Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)
HSO4–(aq) + H2O(l) → ← H3O+(aq) + SO42–(aq)
Acidic
d. NaC2H3O2(aq) → Na+(aq) + C2H3O2–(aq)
Na+(aq) + H2O(l) → NR
C2H3O2–(aq) + H2O(l) → ← HC2H3O2(aq) + OH–(aq)
Basic
e. NaClO4(aq) → Na+(aq) + ClO4–(aq)
Na+(aq) + H2O(l) → NR
ClO4–(aq) + H2O(l) → NR
Neutral