Exam 4B, Spring 2002
Exam 4B, Spring 20021. Complete and balance the following equations.
a. manganese (II) hydroxide plus phosphoric acid
b. CaCl2(s) + K2SO4(aq)
c. S2O82–(aq) + H3PO3(aq) → SO42–(aq) + H3PO4(aq)
d. Ni(OH)3(s) + As(s) → Ni(OH)2(s) + AsO2–(aq)
a. 3 Mn(OH)2(s) + 2 H3PO4(aq) → Mn3(PO4)2(s) + 6 H2O(l)
b. CaCl2(s) + K2SO4(aq) → CaSO4(s) + 2 K+(aq) + 2 Cl–(aq)
c. S2O82–(aq) + H3PO3(aq) → SO42–(aq) + H3PO4(aq)
H3PO3 and H3PO4 indicate that the reaction is run in acid.
Oxidation: H3PO3(aq) + H2O(l) → H3PO4(aq) + 2 H+(aq) + 2 e–
Reduction: S2O82–(aq) + 2 e– → 2 SO42–(aq)
Net: S2O82–(aq) + H3PO3(aq) + H2O(l) → 2 SO42–(aq) + H3PO4(aq) + 2 H+(aq)
d. Ni(OH)3(s) + As(s) → Ni(OH)2(s) + AsO2–(aq)
Ni(OH)3(s) and Ni(OH)2(s) indicate that the reaction is run in base.
Oxidation: As(s) + 2 H2O(l) → AsO2–(aq) + 4 H+(aq) + 3 e–
Reduction: Ni(OH)3(s) + H+(aq) + e– → Ni(OH)2(s) + H2O(l)
Net in acid: 3 Ni(OH)3(s) + As(s) → 3 Ni(OH)2(s) + AsO2–(aq) + H2O(l) + H+(aq)
Neutralization: H+(aq) + OH–(aq) → H2O(l)
Net in base: 3 Ni(OH)3(s) + As(s) + OH–(aq) → 3 Ni(OH)2(s) + AsO2–(aq) + 2 H2O(l)
2. Does Zn(OH)2 have a higher solubility in an unbuffered solution with pH =10.00 or a buffered solution with pH = 10.00? Explain your answer using balanced chemical reactions, showing how you interpret the reactions to arrive at your conclusion. Ksp for Zn(OH)2 = 1.2×10–17.
Zn(OH)2(s) → ← Zn2+(aq) + 2 OH–(aq)
Zn2+(aq) + 4 OH–(aq) → ← [Zn(OH)4]2–(aq)
Net:
Zn(OH)2(s) + 2 OH–(aq) → ← [Zn(OH)4]2–(aq)
In a buffered solution the hydroxide used as the reactant would not become depleted so the net equilibrium could proceed further to the right. Thus, the complex is more soluble in the buffered solution.
3. Consider the reaction between solid barium carbonate and aqueous sulfate ion at 25 °C.
a. Write the balanced reaction.
b. Find ΔH° for the reaction in units of kJ.
c. Find ΔS° for the reaction in units of J/K.
d. Without doing any calculations, would you expect the reaction to be spontaneous or nonspontaneous? Why?
a. BaCO3(s) + SO42–(aq) → BaSO4(s) + CO32–(aq)
b. ΔH° = [(–1473) + (–677.1)] – [(–1216) + (–909.3)] = –25. kJ/mol
c. ΔS° = [(132) + (–56.9)] – [(112) + (20.1)] = –57. J/mol·K
d. ΔH° typically dominates the spontaneity at lower temperatures. Since ΔH° is negative the reaction is likely spontaneous.
4. Ammonium ion, NH4+, is a weak acid.
a. Write the reaction in water.
b. The Ka for ammonium ion was measured at several temperatures and the following plot was made:
Find the standard enthalpy change (in units of kJ/mol) and standard entropy change (in units of J/mol·K).
c. For a 0.1 M solution, does the pH increase or decrease as the temperature is raised? Explain.
a. NH4+(aq) + H2O(l) → ← H3O+(aq) + NH3(aq)
b. The slope of the van t'Hoff plot gives the enthalpy change: ΔH° = –(slope×R) = –(–54300) = 54300 J/mol = 54.3 kJ/mol.
The entropy change is found from the intercept: ΔS° = intercept×R = –1.8 = –1.8 J/mol·K
c. The graph shows that the ln Ka, hence Ka, increases with increasing temperature (to the left on the graph since the x-axis is 1/T). A larger Ka implies more product, hydronium ion, which will lower the pH.
5. Consider the cell: Al(s) | Al3+(aq) || O2(g) | H+(aq) | C(s)
a. What is the balanced oxidation half-reaction?
b. What is the balanced reduction half-reaction?
c. What is the balanced net reaction?
d. What is the standard potential for the cell?
e. Is the reaction spontaneous or nonspontaneous? Why?
f. Does the cell potential increase or decrease as the pH increases? Explain.
a. Al(s) → Al3+(aq) + 3 e–
b. O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l)
c. 4 Al(s) + 3 O2(g) + 12 H+(aq) → 4 Al3+(aq) + 6 H2O(l)
d. E° = E°ox + E°red = 1.676 + 1.229 = +2.905 V
e. Spontaneous because E° > 0.
f. As the pH increases the [H+] decreases, which will drive the reaction towards reactants, which is the nonspontaneous direction. This means that the potential change will be negative, decreasing the cell potential.
