Exam 3B, Spring 2002
Exam 3B, Spring 20021. Complete and balance the following reactions.
a. aqueous sulfuric acid plus solid barium hydroxide
b. C6H5NH2(aq) + H2O(l)
c. C6H5NH2(aq) + HClO4(aq)
d. Mg(C2H3O2)2(aq) + NaOH(aq)
e. KF(aq) + Sc(NO3)3(aq)
f. H2PO4–(aq) + HAsO42–(aq)
g. Fe2+(aq) + H2O(l)
a. H2SO4(aq) + Ba(OH)2(s) → BaSO4(s) + 2 H2O(l)
b. C6H5NH2(aq) + H2O(l) → ← C6H5NH3+(aq) + OH–(aq)
c. C6H5NH2(aq) + HClO4(aq) → C6H5NH3+(aq) + ClO4–(aq)
d. Mg(C2H3O2)2(aq) + 2 NaOH(aq) → Mg(OH)2(s) + 2 Na+(aq) + 2 C2H3O2–(aq)
e. 3 KF(aq) + Sc(NO3)3(aq) → ScF3(s) + 3 K+(aq) + 3 NO3–(aq)
f. H2PO4–(aq) + HAsO42–(aq) → ← HPO42–(aq) + H2AsO4–(aq) (Kc = 6.3×10–8/1.0×10–7 = 63 < 100). If dihydrogenphosphate is the base, then Kc ~ 10–9, which is much smaller.)
g. Fe2+(aq) + 2 H2O(l) → ← FeOH+(aq) + H3O+(aq)
2. Find the pH of a 0.0010 M solution of Na2Se. Write all of the appropriate chemical reactions. Ka for HSe– = 1.0×10–11
Na2Se(aq) → 2 Na+(aq)+ Se2–(aq)
Se2–(aq) +H2O(l) → ← HSe–(aq) +OH–(aq)
Kb = [HSe–]e[OH–]e /[Se2–]e = 1.0×10–14/1.0×10–11 = 1.0×10–3
Initial0.001000
Change–x+x+x
Equilibrium0.0010 – xx x
Approximate? 0.0010/1.0×10–3 = 1 < 100 No
1.0×10–3 = x2/(0.0010 – x)
x2 + 1.0×10–3x – 1.0×10–6 = 0
x = 6.2×10–4 or –1.6×10–3
x = [OH–]e = 6.2×10–4
pOH = –log([OH–]) = –log(6.2×10–4) = 3.31
pH = 14.00 – pOH = 14.00 – 3.21 = 10.79
3. A buffer is prepared to be 0.10 M NaH2AsO4 and 0.10 M Na2HAsO4.
a. Find the pH of the buffer at 25 °C. Write all of the appropriate chemical reactions.
b. 0.025 moles of HCl(g) is added to 1.0 L of the buffer. Write any new chemical reactions that occur and find the new pH in the buffer solution.
a. Find the pH of the buffer at 25 °C. Write all of the appropriate chemical reactions.
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
Initial0.1000.10
Change–x+x+x
Equilibrium0.10 – xx 0.10 + x
Approximate? 0.10/1.0×10–7 = 1×106 > 100 Yes
1.0×10–7 = x(0.10)/(0.10)
x = [H3O+]e = 1.0×10–7
pH = –log([H3O+]) = –log(1.0×10–7) = 7.00
b. 0.025 moles of HCl(g) is added to 1.0 L of the buffer. Write any new chemical reactions that occur and find the new pH in the buffer 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
Initial0.1000.10
A/B+0.0250–0.025
Change–x+x+x
Equilibrium0.125 – xx 0.075 + x
Approximate? 0.075/1.0×10–7 = 7.5×105 > 100 Yes
1.0×10–7 = x(0.075)/(0.125)
x = [H3O+]e = 1.7×10–7
pH = –log([H3O+]) = –log(1.7×10–7) = 6.78
4. Consider the solubility of lead carbonate.
a. Find the molar solubility in water. Write the appropriate chemical reactions.
b. Find the molar solubility of lead carbonate in a 0.10 M solution of lead nitrate. Write the appropriate chemical reactions.
a. Find the molar solubility in water. Write the appropriate chemical reactions.
PbCO3(s) → ← Pb2+(aq) +CO32–(aq)
Ksp = [Pb2+]e[CO32–]e = 7.4×10–14
Initial00
Change+x+x
Equilibriumxx
7.4×10–14 = (x)(x) = x2
x = molar solubility = 2.7×10–7 M
b. Find the molar solubility of lead carbonate in a 0.10 M solution of lead nitrate. Write the appropriate chemical reactions.
Pb(NO3)2(aq) → Pb2+(aq) +2 NO3–(aq)
PbCO3(s) → ← Pb2+(aq) +CO32–(aq)
Ksp = [Pb2+]e[CO32–]e = 7.4×10–14
Initial0.100
Change+x+x
Equilibrium0.10 + xx
7.4×10–14 = (0.10)(x) = 0.10x (x << 0.10)
x = molar solubility = 7.4×10–13 M