A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - pKₐ)?

Solution

Chemistry ma’am · 3 steps

Rearrange the Kₐ expression into Henderson form, then write [A⁻] and [HA] in terms of x.

1. Set up equilibrium. For HA ⇌ H⁺ + A⁻ with starting concentration c, at equilibrium [HA] = c(1-x), [H⁺] = cx and [A⁻] = cx.
2. Take logs. Start from Kₐ = [H⁺][A⁻]/[HA]. Take -log of both sides and rearrange. This gives the Henderson form.

   pH - pKₐ = log[A⁻]/[HA] = log cx/c(1-x)

3. So. The c cancels, leaving pH - pKₐ = log(x/(1-x)), which is option (d).

The idea to keep. pH - pKₐ is always log of (dissociated form over undissociated form). For a weak acid that ratio is x/(1-x). Keep A⁻ on top.

Answer from the NTA final answer key. Solution by PrepWiser.