Calculate the hydronium ion concentration and $p H$ at the equivalence point in the reaction of $22.0 m L$ of $0.10 M$ acetic acid, $C H_{3} C O O H$ , with $22.0 m L$ of $0.10 M N a O H$ .

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Solution

$22 m l$ of $0.1 M$ acetic acid + $22 m l$ of $0.10 M$ $N a O H$
$C H_{3} C O O H + O H_{-} ⟶ C H_{3} C O O^{-} + H_{2} O$
$n_{a c e t i c} = \frac{22}{1000} \times 0.1 = 0.0022$
$n_{O H} = 0.0022$
Added base completely consumes acetic acic.
$n_{a c e t a t e} = 0.0022 m d C l | n_{a c e t i c} = 0$
$p H = p K_{a} + l o g (\frac{[C H_{3} C O O^{-}]}{[C H_{3} C O O]})$
$p H = 4.75 + l o g (\frac{0.0022}{0.0022})$
$p H = 4.75$ $[H_{3} O^{+}] = 10^{- 4.75} = 1.87 \times 10^{- 5} M$

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