A $0.1 M$ solution of a weak monobasic acid has $K_{a} = 1.0 \times 1 0^{- 3}$ . Solving the quadratic $x^{2} + K_{a} x - K_{a} c = 0$ exactly (no approximation), find the pH to two decimal places.

Numerical answer type, tolerance 0.01.

Answer: 2.02

x = \frac{- K _{a} + K _{a}^{2} + 4 K _{a} c}{2} = \frac{- 0.001 + 4.01 \times 1 0 ^{- 4}}{2} = 0.00951 M

pH = - lo g x = 2.02