Exam 3B, Spring 2004
Exam 3B, Spring 20041. Complete and balance the following reactions:
a. H2SO4(aq) + CsOH(aq)
b. CH3CH2NH2(aq) + H2O(l)
c. H3AsO4(aq) + H2O(l)
d. Cu(NO3)2(aq) + Li2S(aq)
e. HClO(aq) + CN–(aq)
f. Mn2+(aq) + H2O(l)
g. Ni(ClO4)2(aq) + NaOH(aq)
a. H2SO4(aq) + 2 CsOH(aq) → 2 Cs+(aq) + SO42–(aq) + 2 H2O(l)
b. CH3CH2NH2(aq) + H2O(l) → ← CH3CH2NH3+(aq) + OH–(aq)
c. H3AsO4(aq) + H2O(l) → ← H3O+(aq) + H2AsO4–(aq)
d. Cu(NO3)2(aq) + Li2S(aq) → CuS(s) + 2 Li+(aq) + 2 NO3–(aq)
e. HClO(aq) + CN–(aq) → ← ClO–(aq) + HCN(aq) (Kc = 2.9×10–8/6.2×10–10 = 47 < 100)
f. Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)
g. Ni(ClO4)2(aq) + 2 NaOH(aq) → Ni(OH)2(s) + 2 Na+(aq) + 2 ClO4–(aq)
2. For each of the following salts, indicate if an aqueous solution will be acidic, basic, or neutral. If the salt is not neutral, write the reaction that determines the acidity or basicity.
a. KBr
b. Fe(ClO4)3
c. Na2CrO4
a. KBr(aq) → K+(aq) + Br–(aq)
K+(aq) + 2 H2O(l) → NR
Br–(aq) + H2O(l) → NR
Neutral
b. Fe(ClO4)3(aq) → Fe3+(aq) + 3 ClO4–(aq)
Fe3+(aq) + 2 H2O(l) → ← FeOH2+(aq) + H3O+(aq)
ClO4–(aq) + H2O(l) → NR
Acidic
c. Na2CrO4(aq) → 2 Na+(aq) + CrO42–(aq)
Na+(aq) + 2 H2O(l) → NR
CrO42–(aq) + H2O(l) → ← HCrO4–(aq) + OH–(aq)
Basic
3. A solution is made that is 1.0 mM NaH2AsO4 and 1.0 mM Na2HAsO4 at 25 °C.
a. Calculate the pH of the solution.
b. Calculate the pH of the solution after 0.010 mmol of NaOH is added to 1.0 L of the solution.
c. Calculate the pH of the solution after 0.10 mol of NaOH is added to 1.0 L of the solution.
a. Calculate the pH of the solution.
NaH2AsO4(aq) → Na+(aq)+ H2AsO4–(aq)
Na2HAsO4(aq) → 2 Na+(aq)+ HAsO42–(aq)
H2AsO4–(aq) +H2O(l) → ← H3O+(aq) +HAsO42–(aq)
Ka = [H3O+]e[HAsO42–]e /[H2AsO4–]e = 1.0×10–7
Initial0.001000.0010
Change–x+x+x
Equilibrium0.0010 – xx 0.0010 + x
Approximate? 0.0010/1.0×10–7 = 1×104 > 100 Yes
1.0×10–7 = x(0.0010)/(0.0010)
x = [H3O+]e = 1.0×10–7
pH = –log([H3O+]) = –log(1.0×10–7) = 7.00
b. Calculate the pH of the solution after 0.010 mmol of HCl is added to 1.0 L of the solution.
NaH2AsO4(aq) → Na+(aq)+ H2AsO4–(aq)
Na2HAsO4(aq) → 2 Na+(aq)+ HAsO42–(aq)
H2AsO4–(aq) +H2O(l) → ← H3O+(aq) +HAsO42–(aq)
A/BH2AsO4–(aq) +Cl–(aq) ← HCl(g) +HAsO42–(aq)
Ka = [H3O+]e[HAsO42–]e /[H2AsO4–]e = 1.0×10–7
Initial0.001000.0010
A/B+0.0000100–0.00010
Change–x+x+x
Equilibrium0.0011 – xx 0.0009 + x
Approximate? 0.0009/1.0×10–7 = 9×103 > 100 Yes
1.0×10–7 = x(0.0009)/(0.0011)
x = [H3O+]e = 1.×10–6
pH = –log([H3O+]) = –log(1.×10–6) = 6.0
c. Calculate the pH of the solution after 0.10 mol of HCl is added to 1.0 L of the solution.
In this case the concentration of the strong acid far exceeds the concentration of the base in the buffer. Thus, the concentration of HCl in the solution = 0.10 – 0.0010 = 0.10 M. Thus, pH = –log(0.10) = 1.00.
4. Calculate the molar solubility of lead hydroxide in pure water at 25 °C.
Pb(OH)2(s) → ← Pb2+(aq) +2 OH–(aq)
Ksp = [Pb2+]e[OH–]e2 = 1.2×10–15
Initial00
Change+x+2x
Equilibriumx2x
1.2×10–15 = (x)(2x)2 = 4x3
x = molar solubility = 6.7×10–6 M