A given weak acid $(0.01 M)$ has $p K_{a} = 6$ . The pH of this solution is:

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Solution

$p K_{a} = 6$ means $K_{a} = 10^{- 6}$
$H A ⇌ H^{+} + A^{-}$
$C$
$C - C α$ $C α$ $C α$
$K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]} = \frac{C α C α}{C (1 - α)}$
Let $α << 1$
$K_{a} = C α^{2} = 10^{- 6} = 0.01 (α^{2})$
$α = 0.01$
$[H^{+}] = C α = 10^{- 4}$ means $p H = - log ([H^{+}]) = 4$

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p K_{a} = 6

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