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 $0.10 M$ $N a O H$ .
$K_{a} = 2 \times 10^{- 5}$

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Solution

$22 m l$ of $0.1 N$ acetic acid will just neutralise $22 m l$ of $0.1 N$ $N a O H$ . The resulting solution will be neutral.
$p H = 7$
$[H^{+}] = 1 \times 10^{- 7} m o l / L$

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Calculate the hydronium ion concentration and pH at the equivalence point in the reaction of 22.0 ml of 0.10 M acetic acid, CH3COOH, with 22.0 mL of 0.10 M NaOH. Ka =2×10−5

22ml of 0.1N acetic acid will just neutralise 22ml of 0.1N NaOH. The resulting solution will be neutral.pH=7[H+]=1×10−7mol/L

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 $0.10 M$ $N a O H$ .
$K_{a} = 2 \times 10^{- 5}$

$22 m l$ of $0.1 N$ acetic acid will just neutralise $22 m l$ of $0.1 N$ $N a O H$ . The resulting solution will be neutral.
$p H = 7$
$[H^{+}] = 1 \times 10^{- 7} m o l / L$