When 10 mL of 0.1 M acetic acid (pK_a = 5.0 | vedclass

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(C) The titration involves a weak acid (acetic acid) and a weak base (ammonia). For the titration of a weak acid and a weak base,the $pH$ at the equiv

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$7$

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(C) The titration involves a weak acid (acetic acid) and a weak base (ammonia). For the titration of a weak acid and a weak base,the $pH$ at the equivalence point is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$ Given: $pK_a = 5.0$ $pK_b = 5.0$ Substituting these values into the formula: $pH = 7 + \frac{1}{2}(5.0 - 5.0)$ $pH = 7 + \frac{1}{2}(0)$ $pH = 7$ Therefore,the $pH$ at the equivalence point is $7$.

When $10 \ mL$ of $0.1 \ M$ acetic acid $(pK_a = 5.0)$ is titrated with $10 \ mL$ of $0.1 \ M$ ammonia $(pK_b = 5.0)$,at what $pH$ will the equivalence point be reached?

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