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Factors Affecting Alpha

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Question

Calculate the accurate pH of $5 \times 10^{- 3} M$ formic acid solution.
$K_{a} (H C O O H) = 2 \times 10^{- 4}$

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Solution

$K_{a} = 2 \times 10^{- 4} = 0.0002$
$C = 5 \times 10^{- 3} = 0.005$
Formic acid solution equilibrium state will be
${[H^{+}]}^{2} + K_{a} [H^{+}] - C K_{a} = 0$
let $[H^{+}] = x$
$x^{2} + 0.0002 x - 0.005 \times 0.0002 = 0$
$x = [H^{+}] = 0.0009$
$p H = - {log}_{10} [H^{+}] = - l o g (0.0004) = 3.039$

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