How does pH relate to pKa in a titration?

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How does pH relate to pKa in a titration?

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Answer 1 · 2015-08-07T07:43:28

The $p H$ $pH$ at half-equivalence when a weak acid is titrated with a strong base is precisely the $p K_{a}$ $pK_a$ of the acid.

Explanation:

Consider the dissociation of a weak acid, $H A$ $HA$ :
$H A ⇌ H^{+} + A^{-}$ $HA rightleftharpoons H^+ + A^-$ . As we know,

$K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]},$ $K_a = {[H^+][A^-]}/[[HA]],$

$and {log}_{10} {K_{a}} = {log}_{10} [H^{+}] + {log}_{10} {\frac{[A^{-}]}{[H A]}}$ $and log_10{K_a} = log_10[H^+] + log_10{{[A^-]]/[[HA]]}$ ;

Equivalently (multiplying each by $- 1$ $-1$ ,
$- {log}_{10} {K_{a}} = - {log}_{10} [H^{+}] - {log}_{10} {\frac{[A^{-}]}{[H A]}}$ $-log_10{ K_a} = -log_10[H^+] - log_10{{[A^-]]/[[HA]]}$ .

But, by definition, $- {log}_{10} {K_{a}} = p K_{a}$ $-log_10{ K_a} = pK_a$ , and $- {log}_{10} [H^{+}] = p H$ $-log_10[H^+] = pH$ .

Therefore, $p H$ $pH$ = $p K_{a}$ $pK_a$ $+$ $+$ ${log}_{10} {\frac{[A^{-}]}{[H A]}}$ $log_10{{[A^-]]/[[HA]]}$ . This is a form of the buffer equation, with which you are going to get very familiar.

Now at half-equivalence, by definition, $[H A] = [A^{-}]$ $[HA] = [A^-]$ , and since ${log}_{10} 1$ $log_10 1$ = 0, when plugged back into the equation, $p H = p K_{a}$ $pH = pK_a$ . So, in order to measure $p K_{a}$ $pK_a$ values of weak acids we plot a titration curve with a $p H$ $pH$ meter, and note value at half-equivalence.

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How does pH relate to pKa in a titration?

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