Exam 2A, Spring 2004

1. Complete and balance the following reactions:

a. H₂SO₄(aq) + KOH(aq)

b. C₆H₅NH₂(aq) + HCl(aq)

c. HOCN(aq) + H₂O(l)

d. Ba(OH)₂(s) + HNO₃(aq)

Answer

a. H₂SO₄(aq) + 2 KOH(aq) → 2 H₂O(l) + 2 K⁺(aq) + SO₄^2–(aq)

b. C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl^–(aq)

c. HOCN(aq) + H₂O(l) → ← H₃O⁺(aq) + OCN^–(aq)

d. Ba(OH)₂(s) + 2 HNO₃(aq) → 2 H₂O(l) + Ba²⁺(aq) + 2 NO₃^–(aq)

2. Complete and balance the following reaction and identify the acid, the base, the conjugate acid, and the conjugate base:

CO₃^2–(aq) + CH₃CH₂CO₂H(aq)

Answer

CO₃^2–(aq) + CH₃CH₂CO₂H(aq) → ← HCO₃^–(aq) + CH₃CH₂CO₂^–(aq)

Acid: CH₃CH₂CO₂H

Base: CO₃^2–

Conjugate Acid: HCO₃^–

Conjugate Base: CH₃CH₂CO₂^–

3. Identify the Lewis acid and the Lewis base in the following reaction.

Al(OH)₃(s) + OH^–(aq) → Al(OH)₄^–(aq)

Answer

Al(OH)₃(s) + OH^–(aq) → Al(OH)₄^–(aq)

Lewis Acid: Al(OH)₃

Lewis Base: OH^–

4. Identify the stronger acid in pair given below. Briefly (5 words or less) justify your reasoning.

a. H₂O or H₂S

b. H₂SeO₄ or H₂TeO₄

Answer

a. H₂O or H₂S

H₂S because of the S size

b. H₂SeO₄ vs. H₂TeO₄

H₂SeO₄ because of the Se electronegativity

5. Calculate the pH and pOH of the following solutions:

a. 2.0×10^–3 M perchloric acid.

b. 2.0×10^–3 M potassium hydroxide.

c. 3.0×10^–3 M calcium hydroxide.

Answer

a. 2.0×10^–3 M perchloric acid.

HClO₄(aq) + H₂O(l) → H₃O⁺(aq) + ClO₄^–(aq)

[H₃O⁺] = [HClO₄] = 2.0×10^–3 M

pH = –log([H₃O⁺]) = –log(2.0×10^–3) = 2.30

pOH = 14.00 – pH = 14.00 – 2.30 = 11.70

b. 2.0×10^–3 M potassium hydroxide.

KOH(aq) → K⁺(aq) + OH^–(aq)

[OH^–] = [KOH] = 2.0×10^–3 M

pOH = –log([OH^–]) = –log(2.0×10^–3) = 2.30

pH = 14.00 – pOH = 14.00 – 2.30 = 11.70

c. 3.0×10^–3 M calcium hydroxide.

Ca(OH)₂(aq) → Ca²⁺(aq) + 2 OH^–(aq)

[OH^–] = 2×[Ca(OH)₂] = 2(3.0×10^–3) = 6.0×10^–3 M

pOH = –log([OH^–]) = –log(6.0×10^–3) = 2.22

pH = 14.00 – pOH = 14.00 – 2.22 = 11.78

6. Consider the reaction shown below, which has ΔH = 131 kJ/mol.

C(s) + H₂O(g) → ← CO(g) + H₂(g)

a. Which way will the reaction shift if the pressure is increased by reducing the volume?

b. Which way will the reaction shift if the temperature is raised?

Answer

C(s) + H₂O(g) + heat → ← CO(g) + H₂(g) (For an endothermic reaction, ΔH > 0, heat is a reactant)

a. Which way will the reaction shift if the pressure is increased by reducing the volume?

The reaction will shift towards reactants to reduce the total pressure: the total moles of gases is 2 for products but only 1 for reactants.

b. Which way will the reaction shift if the temperature is raised?

Raising the temperature adds heat so the reaction will shift towards products to remove the added heat.

7. Estimate the pH of a 2.0×10^–3 M solution of HNO₂. K_a = 7.2×10^–4

Answer

HNO₂(aq) + H₂O(l) → ← H₃O⁺(aq) + NO₂^–(aq)

K_a = [H₃O⁺]_e[NO₂^–]_e [HNO₂]_e = 7.2×10^–4

Initial 2.0×10^–3 0 0

Change –x +x +x

Equilibrium 2.0×10^–3 – x x x

Approximate? 2.0×10^–3/7.2×10^–4 = 2.8: No

7.2×10^–4 = x²/(2.0×10^–3 – x)

x² + 7.2×10^–4 – 1.44×10^–6 = 0

[H₃O⁺]_e = 8.9×10^–4 M

pH = –log[H₃O⁺] = –log[8.9×10^–4] = 3.05