Question

pKa of acetic acid is 4.75. Calculate the pH of a vinegar solution that is 1.5 M in acetic acid. SHOW ICE TABLE

Answer 1

Acetic acid is a weak acid, dissociates as AcOH = AcO^- + H⁺

To determine pH of weak acid, concentration of H⁺ , [H⁺] = (K_a * C)^(1/2)

where K_a is dissociation constant, C = concentration = 1.5 M

Given pK_a = 4.75

Or, -logK_a = 4.75

Or, logK_a = -4.75

Or, K_a = 10^-4.75

So, [H⁺] = (10^-4.75 * 1.5)^(1/2)

= (2.67 * 10^-5)^(1/2)

= 5.16*10^-3

So, pH = -log[H⁺] = -log(5.16*10^-3) = 2.29

So, pH of the solution is 2.29

[ AcOH + H₂O = AcO^- + H₃O⁺

Initial: 1.5 0 0

Change: -x +x +x

Equilibrium: 1.5-x +x +x

K_a = [AcO^- ] * [H₃O⁺ ] / [ AcOH]

= x * x / (1.5 - x)

= x² / (1.5 - x)

As, x is very small, so 1.5-x = 1.5 approximately

Or, K_a = x² / 1.5

Or, x² = K_a * 1.5

Or, x = (K_a * 1.5)^(1/2)

x is the concentration of H₃O⁺ or H⁺ ion, which I have shown previously]