CHEM 2 problems?

1.
Moles of HC₇H₅O₂ = (0.10 mol/L) × (300.0/1000 L) = 0.03 mol
Moles of C₇H₅O₂⁻ = (0.30 mol/L) × (200.0/1000 L) = 0.06 mol
[C₇H₅O₂⁻]/[HC₇H₅O₂] = (Moles of C₇H₅O₂⁻)/(Moles of HC₇H₅O₂)] = 0.06/0.03 = 2

Henderson-Hasselbalch equation:
pH = pKa + log([C₇H₅O₂⁻]/[HC₇H₅O₂])
pH = -log(6.5 × 10⁻⁵) + log(2) = 4.49

The answer: b. 4.49

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2.
Moles of NH₄⁺ = (0.15 mol/L) × (500.0/1000 L) = 0.075 mol
Moles of NH₃ = (0.12 mol/L) × (400.0/1000 L) = 0.048 mol
[NH₄⁺]/[ NH₃] = (Moles of NH₄⁺)/(Moles of NH₃)] = 0.075/0.048

Henderson-Hasselbalch equation:
pOH = pKb + log([NH₄⁺]/[ NH₃])
pOH = -log(1.8 × 10⁻⁵) + log(0.075/0.048) = 4.94
pH = pKw - pOH = 14.00 - 4.94 = 9.06

The answer: c. 906

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c.
Henderson-Hasselbalch equation:
pH = pKa + log([C₂H₃O₂⁻]/[HC₂H₃O₂])
5.14 = 4.74 + log([C₂H₃O₂⁻]/[HC₂H₃O₂])
log([C₂H₃O₂⁻]/[HC₂H₃O₂]) = 0.4
[C₂H₃O₂⁻]/[HC₂H₃O₂] = 10⁰˙⁴ = 2.51

The answer: c. 2.51