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# The $pK_a$ of weak acid (HA) is 4.5 . The $pOH$ of an aqueous buffered solution of HA in which 50% of acid is ionized is

(a) 7.0

(b) 4.5

(c) 2.5

(d) 9.5

When HA is 50% ionised, [HA] = [$A^-$]
$pH = PK_a + log \frac{[A^-]}{[HA]} = pK_a = 4.5 (Given)$
$pOH$ = 14 - $pH$ = 14 - 4.5 = 9.5