Exam 3A, Spring 2005
Exam 3A, Spring 20051. Complete and balance the following reactions:
a. H3PO4(aq) + Ba(OH)2(s)
b. CH3CH2NH2(aq) + H2O(l)
c. Ag+(aq) + EDTA4–(aq)
d. Mn(NO3)2(aq) + Na2S(aq)
e. Hydrogen phosphate ion plus chlorite ion
f. Ti2+(aq) + H2O(l)
g. Cu(CH3CO2)(aq) + HCl(aq)
a. 2 H3PO4(aq) + 3 Ba(OH)2(s) → Ba3(PO4)2(s) + 6 H2O(l)
b. CH3CH2NH2(aq) + H2O(l) → ← CH3CH2NH3+(aq) + OH–(aq)
c. Ag+(aq) + EDTA4–(aq) → ← [Ag(EDTA)]3–(aq)
d. Mn(NO3)2(aq) + Na2S(aq) → MnS(s) + 2 Na+(aq) + 2 NO3–(aq)
e. HPO42–(aq) + ClO2–(aq) → ← PO43–(aq) + HClO2(aq) (Kc = 4.2×10–13/1.1×10–2 = 3.8×10–11 < 100)
f. Ti2+(aq) + 2 H2O(l) → ← TiOH+(aq) + H3O+(aq)
g. Cu(CH3CO2)(aq) + HCl(aq) → CuCl(s) + CH3CO2H(aq)
2. Identify if the following ionic salts are acidic or basic in aqueous solution. Show the reaction that defines the acidity or basicity.
a. NaHSO4
b. Fe(NO3)2
c. Mg(IO3)2
a. NaHSO4(aq) → Na+(aq) + HSO4–(aq)
Na+(aq) + 2 H2O(l) → NR
HSO4–(aq) + H2O(l) → ← H3O+(aq) + SO42–(aq)
Acidic
b. Fe(NO3)2(aq) → Fe2+(aq) + 2 NO3–(aq)
Fe2+(aq) + 2 H2O(l) → ← FeOH+(aq) + H3O+(aq)
NO3–(aq) + H2O(l) → NR
Acidic
c. Mg(IO3)2(s) → ← Mg2+(aq) + 2 IO3–(aq)
Mg2+(aq) + 2 H2O(l) → NR
IO3–(aq) + H2O(l) → ← HIO3(aq) + OH–(aq)
Basic
3. A buffer is prepared to be 0.10 M in H2S and 0.10 M in NaHS.
a. Calculate the pH of the solution, showing all of your work.
b. 1 mL of HCl(g) is added to 1 L of the buffer. Calculate the new pH.
a. Calculate the pH of the solution, showing all of your work.
NaHS(aq) → Na+(aq)+ HS–(aq)
H2S(aq) +H2O(l) → ← H3O+(aq) +HS–(aq)
Ka = [H3O+]e[HS–]e /[H2S]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. 1 mL of HCl(g) is added to 1 L of the buffer. Calculate the new pH.
Assume 25 °C = 298 K and the Ideal Gas Law to find the molarity of HCl: (1 atm)(0.001 L) = n(0.0821 L·atm mol–1K–1)(298 K)
n = (1)(0.001)/(0.0821)(298) = 4×10–5 mol = 4×10–5 M
NaHS(aq) → Na+(aq)+ HS–(aq)
H2S(aq) +H2O(l) → ← H3O+(aq) +HS–(aq)
A/BH2S(aq) +Cl–(aq) ← HCl(aq) +HS–(aq)
Ka = [H3O+]e[HS–]e /[H2S]e = 1.0×10–7
Initial0.1000.10
A/B+ 4×10–50 – 4×10–5
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
4. Will the solubility in water increase, decrease, or stay the same for the following salts if they are added to a pH = 14 solution? Briefly explain your answer.
a. CaC2O4
b. Cr(OH)3
a. CaC2O4
CaC2O4(s) + 2 OH–(aq) → Ca(OH)2(s) + C2O42–(aq)
Solubility decreases because of precipitation of calcium hydroxide.
b. Cr(OH)3
Cr(OH)3(s) + OH–(aq) → ← [Cr(OH)4]–(aq
Solubility increases because of the complex ion formation.
5. Determine the solubility of lead(II) bromide in a buffer with pH = 11.0.
PbBr2(s) → ← Pb2+(aq) +2 Br–(aq)
Pb2+(aq) +3 OH–(aq) → ← [Pb(OH)3]–(aq)
NetPbBr2(s) +3 OH–(aq) → ← [Pb(OH)3]–(aq) +2 Br–(aq)
Kc = [Pb(OH)3–)]e[Br–]e2/ [OH–]e3 = Ksp×Kf = 4.0×10–5×2.8×1014 = 1.5×1010
Initial1×10–30 0
Change– 0+x+2x
Equilibrium1×10–3x 2x
1.5×1010 = (x)(2x)2/(1×10–3)3 = 4×109x3
x = molar solubility = 2 M
6. Calculate ΔH° and ΔS° for the reaction of solid silver nitrate with aqueous hydrochloric acid.
AgNO3(s) + HCl(aq) + H2O(l) → AgCl(s) + H3O+(aq) + NO3–(aq)
ΔH° = [–127.1 + –285.8 + –205.0] – [–124.4 + –167.2 + –285.8] = +531.1 kJ/mol
ΔS° = [96.2 + 69.9 + 146.4] – [140.9 + 56.5 + 69.9] = +45.2 J/mol·K