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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.

Verified Answer :

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\(K_{a}\) = 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



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