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Q.

If for an aqueous solution of a weak acid, $pH = {pK}_{a} + 2$ at $25^{°} C$ , the approximate percentage fraction of the acid in the dissociated form is

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answer is 99.

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Detailed Solution

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$\begin{array}{l} H A ⇌ H^{+} + A^{-} \\ C (1 - α) C α C α \end{array}$

$p H = p K a + \log_{10} \frac{(A^{-})}{(H A)}$
$p H = p K a + 2$ gives
Then $\frac{(A^{-})}{(H A)} = 100$

$\frac{α}{1 - α} = 100$

$α = 100 - 100 α$
$α = \frac{100}{101} = 0.99$

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