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# $pH$ of a mixture of 1 M benzoic acid ($pK_a$ = 4.20) and 1 M solution benzoate is 4.5. In 300 mL of buffer solution, volume of benzoic acid solution is

(a) 200 mL

(b) 150 mL

(c) 100 mL

(d) 50 mL

$pH = pK_a + log \frac{[Salt]}{[Acid]}$

Putting the given values, $4.5 = 4.2 + log \frac{[Salt]}{[Acid]}$
$0.3 = log \frac{[Salt]}{[Acid]}$
$\frac{[Salt]}{[Acid]}$ = antilog 0.3 = 2

Let v mL of benzoic acid is mixed with (300-v) mL of 1M sodium benzoate.
$\therefore \frac{300-v}{v} = 2$
$\Rightarrow 300 - v = 2 v$
v = 100 mL