A solution contains $0.1 M$ acetic acid ( $K_{a} = 1.8 \times 1 0^{- 5}$ ) and $2.0 \times 1 0^{- 3} M$ $HCl$ . Taking $[HA] \approx 0.1 M$ but solving exactly for the acetate formed, i.e. $K_{a} = x (2 \times 1 0^{- 3} + x) /0.1$ , find the pH to two decimals.

Numerical answer type, tolerance 0.01.

Answer: 2.57

Let

x = [A X^{-}]

. Then

x^{2} + (2 \times 1 0^{- 3} + K_{a}) x - K_{a} c = 0

gives

x = 6.70 \times 1 0^{- 4} M

[H X^{+}] = 2 \times 1 0^{- 3} + x = 2.67 \times 1 0^{- 3} M

, so

pH = 2.574 \approx 2.57