Question

a) A solution was prepared by dissolving 0.02 moles ethanoic acid in

water to give 1 L of solution. What is the pH?

b) to this solution was added 0.008 moles of concentrated sodium

hydroxide (NaOH). What is the new pH? (In this question you may

ignore changes in volume due to addition of NaOH).

Answer 1

a)

An ethanoic acid is an acetic acid (CH₃COOH);

n(CH₃COOH) = 0.02 mol;

V(H₂O) = 1 L;

C(CH₃COOH) = n(CH₃COOH)/V(H₂O) = 0.02/1 = 0.02 M;

pK_a(CH₃COOH) = 4.76;

K_a(CH₃COOH) = 1.74 * 10^-5;

CH₃COOH = H⁺ + CH3COO^-;

0.02 0 0

-x x x

0.02-x x x

K_a(CH₃COOH) = [H⁺] * [CH₃COO^-]/[CH₃COOH];

1.74 * 10^-5 = x * x/(0.02-x) = x²/(0.02-x);

x² = 1.74 * 10^-5 * (0.02-x);

x² = 3.48 * 10^-7 - 1.74 * 10^-5 * x;

x² + 1.74 * 10^-5 * x - 3.48 * 10^-7 = 0

x₁ = -0.00060;

x₂ = 0.00058;

So, [H⁺] = [CH₃COO^-] = 0.00058 M.

pH = -log[H⁺] = -log(0.00058) = 3.24.

b)

n(NaOH) = 0.008 mol

CH₃COOH + NaOH = CH₃COONa + H₂O;

If n(NaOH) = 0.008 mol, then also n(CH₃COOH) = 0.008 mol and n(CH₃COONa) = 0.008 mol and n(H₂O) = 0.008 mol;

n(CH₃COOH)_remaining = 0.02 - 0.008 = 0.012 mol;

pH = pKa + log([CH₃COONa]/[CH₃COOH]) = 4.76 + log(0.008/0.012) = 4.76 - 0.18 = 4.58.

Answer: a) 3.24

b) 4.58.