Exam 3A, Spring 2001
Exam 3A, Spring 20011. Complete and balance the following reactions:
(a) C5H5N(aq) + H2O(l)
(b) C5H5N(aq) + HCl(aq)
(c) Co2+(aq) + H2O(l)
(d) Ca(OH)2(aq) + HF(aq)
(e) H3PO4(aq) + CaI2(aq)
(f) NO2–(aq) + H2O(l)
(g) C6H5NH2(aq) + HOCl(aq)
(a) C5H5N(aq) + H2O(l) → ← C5H5NH+(aq) + OH–(aq)
(b) C5H5N(aq) + HCl(aq) → C5H5NH+(aq) + Cl–(aq)
(c) Co2+(aq) + 2 H2O(l) → ← CoOH+(aq) + H3O+(aq)
(d) Ca(OH)2(aq) + 2 HF(aq) → CaF2(s) + 2 H2O(l)
(e) 2 H3PO4(aq) + 3 CaI2(aq) + 6 H2O(l) → Ca3(PO4)2(s) + 6 H3O+(aq) + 6 I–(aq)
(f) NO2–(aq) + H2O(l) → ← HNO2(aq) + OH–(aq)
(g) C6H5NH2(aq) + HOCl(aq) → ← C6H5NH3+(aq) + OCl–(aq) (Kc = 2.9×10–8/1.4×10–5 = 2.1×10–3 < 100)
2. What is the pH of a 0.0025 M solution of methylamine, CH3NH2, at 25 °C?
CH3NH2(aq) +H2O(l) → ← CH3NH3+(aq) +OH–(aq)
Kb = [CH3NH3+]e[OH–]e /[CH3NH2]e = 1.0×10–14/2.4×10–11 = 4.2×10–4
Initial0.002500
Change–x+x+x
Equilibrium0.0025 – xx x
Approximate? 0.0025/4.2×10–4 = 6 < 100 No
4.2×10–4 = x2/(0.0025 – x)
x2 + 4.2×10–4x – 1.05×10–6 = 0
x = 8.4×10–4 or –1.3×10–3
x = [OH–]e = 8.4×10–4
pOH = –log([OH–]) = –log(8.4×10–4) = 3.08
pH = 14.00 – pOH = 14.00 – 3.08 = 10.92
3. Indicate whether the following salts are acidic, basic, or neutral.
(a) NaNO3
(b) Ca(C2H3O2)2
(c) KHSO4
(d) Co(OH)2
(e) CsNO2
(a) NaNO3(aq) → Na+(aq) + NO3–(aq)
Na+(aq) + 2 H2O(l) → NR
NO3–(aq) + H2O(l) → NR
Neutral
(b) Ca(C2H3O2)2(aq) → Ca2+(aq) + 2 C2H3O2–(aq)
Ca2+(aq) + 2 H2O(l) → NR
C2H3O2–(aq) + H2O(l) → ← HC2H3O2(aq) + OH–(aq)
Basic
(c) KHSO4(aq) → K+(aq) + HSO4–(aq)
K+(aq) + 2 H2O(l) → NR
HSO4–(aq) + H2O(l) → ← H3O+(aq) + SO42–(aq)
Acidic
(d) Co(OH)2(s) → ← Co2+(aq) + 2 OH–(aq)
Basic
(e) CsNO2(aq) → Cs+(aq) + NO2–(aq)
Cs+(aq) + 2 H2O(l) → NR
NO2–(aq) + H2O(l) → ← HNO2(aq) + OH–(aq)
Basic
4. A buffer is prepared that is 0.50 M boric acid, H3BO3, and 0.40 M sodium dihydrogen borate.
(a) What is the pH of the buffer at 25 °C?
(b) 0.05 moles of perchloric acid is added to 1.0 L of the buffer prepared in part (a). What is the new pH?
(a) What is the pH of the buffer at 25 °C?
NaH2BO3(aq) → Na+(aq) + H2BO3–(aq)
H3BO3(aq) +H2O(l) → ← H3O+(aq) + H2BO3–(aq)
Ka = [H3O+]e[H2BO3–]e /[H3BO3]e = 5.9×10–10
Initial0.5000.40
Change–x+x+x
Equilibrium0.50 – x0.40 + x x
Approximate? 0.40/5.9×10–10 = 7×108 > 100 Yes
5.9×10–10 = x(0.40)/(0.50)
x = [H3O+]e = 7.4×10–10
pH = –log([H3O+]) = –log(7.4×10–10) = 9.13
(b) 0.05 moles of perchloric acid is added to 1.0 L of the buffer prepared in part (a). What is the new pH?
[HClO4] = 0.05 mol/1.0 L = 0.05 M
NaH2BO3(aq) → Na+(aq) + H2BO3–(aq)
H3BO3(aq) +H2O(l) → ← H3O+(aq) + H2BO3–(aq)
A/BH3BO3(aq) +ClO4–(aq) ← HClO4(aq) + H2BO3–(aq)
Ka = [H3O+]e[H2BO3–]e /[H3BO3]e = 5.9×10–10
Initial0.5000.40
A/B+0.050–0.05
Change–x+x+x
Equilibrium0.55 – x0.35 + x x
Approximate? 0.35/5.9×10–10 = 6×108 > 100 Yes
5.9×10–10 = x(0.35)/(0.55)
x = [H3O+]e = 9.3×10–10
pH = –log([H3O+]) = –log(9.3×10–10) = 9.03
5. Which is more soluble at 25 °C, lead fluoride or lithium phosphate?
Lead fluoride, PbF2
PbF2(s) → ← Pb2+(aq) +2 F–(aq)
Ksp = [Pb2+]e[F–]e2 = 2.7×10–8
Initial00
Change+x+2x
Equilibriumx2x
2.7×10–8 = (x)(2x)2 = 4x3
x = molar solubility = 1.9×10–3 M
Lithium phosphate, Li3PO4
Li3PO4(s) → ← 3 Li+(aq) +PO43–(aq)
Ksp = [Li+]e3[PO4–]e = 3.2×10–9
Initial00
Change+3x+x
Equilibrium3xx
3.2×10–9 = (3x)3(x) = 27x4
x = molar solubility = 3.3×10–3 M
Lithium phosphate is slightly more soluble than lead fluoride.