A 0.10 M solution of acetic acid (Ka = 1.8×10^-5) has what approximate pH?

• A. 2.60
• B. 3.00
• C. 2.50
• D. 2.87

Answer: (D)

Weak acids only partially dissociate in water. For acetic acid HA, the equilibrium is HA ⇌ H+ + A− with Ka = [H+][A−]/[HA]. If we call x = [H+] = [A−], then Ka = x^2/(C − x) with C = 0.10 M.

Because Ka is small, x is much smaller than C, so [HA] ≈ C and x ≈ sqrt(Ka C). Compute Ka C = 1.8×10^−5 × 0.10 = 1.8×10^−6. The square root is x ≈ 1.34×10^−3 M. The pH is −log10(x) ≈ −log10(1.34×10^−3) ≈ 2.87.

So the approximate pH is 2.87. This matches the close result from solving the exact expression, since x stays far less than C in this case.