Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0116 M solution. The pH of the resulting solution is 2.35. Calculate the Ka for the acid.

Answer 1

$K_{a}$ = 1.72 × 10⁻³

HA + H₂O ⇌ H₃O⁺ + A⁻

$K_{a} = \frac{[H_{3} O^{+}][ A^{-}] }{[HA]}$

[HA] = 0.0116 mol/L

pH = 2.35

[H₃O⁺] = $10^{-pH}$ mol/L= $10^{-2.35}$ mol/L = 4.47 × 10⁻³ mol/L

[A⁻] = [H₃O⁺] = 4.47 × 10⁻³ mol/L

$K_{a} = \frac{[H_{3} O^{+}][ A^{-}] }{[HA]}$ = $\frac{4.47 \times 10^{-3} \times 4.47 \times 10^{-3}}{ 0.0116}$ = 1.72 × 10⁻³