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pH

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Question

Calculate the hydronium ion concentration and the $p H$ at the equivalence point in a titration of $50.0 m L$ of $0.40 M N H_{3}$ with $0.40 M H C l$ .

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Solution

$50 m l$ of $0.040 M N H_{3} + 0.040 M H C l$
$N H_{3} + H C l ⟶ N H_{4} C l$
$n_{N H_{3}} = \frac{50}{1000} \times 0.40 = n_{H C l} ≅ n_{N H_{4} C l}$
$p H = p K_{a} + l o g \frac{S a l t}{A c i d}$
$p H = p K_{a}$
$p H = - 6.3$

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