What is the pH of a 0.20 M solution of a weak monoprotic acid with Ka = 5.0e-5?

• A. 2.30
• B. 2.50
• C. 3.00
• D. 4.00

Answer: (B)

In a solution of a weak monoprotic acid, the extent of dissociation is set by the acid’s Ka. For HA ⇌ H+ + A−, let x be the amount that dissociates, so [H+] = [A−] = x and [HA] ≈ C − x, where C is the initial concentration.

Ka = x^2 / (C − x). If x is small compared with C, you can approximate by x^2 ≈ Ka·C. Then x ≈ sqrt(Ka·C), and pH = −log10(x).

Here, Ka = 5.0×10^−5 and C = 0.20 M. Ka·C = 5.0×10^−5 × 0.20 = 1.0×10^−5. So x ≈ sqrt(1.0×10^−5) ≈ 3.16×10^−3 M, and pH ≈ −log10(3.16×10^−3) ≈ 2.50.

So the solution is acidic with pH about 2.50.