Which of the following describes the correct relation for $p O H$ for an aqueous salt solution of weak acid and weak base at $25^{\circ} C ?$

p O H = 14 - \frac{1}{2} (p K_{a} - p K_{b})

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p O H = 7 + \frac{1}{2} (p K_{a} - p K_{b})

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p O H = 7 - \frac{1}{2} (p K_{a} - p K_{b})

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p O H = 14 + \frac{1}{2} (p K_{a} - p K_{b})

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Solution

The correct option is C $p O H = 7 - \frac{1}{2} (p K_{a} - p K_{b})$
Theory:
Salt of weak acid and weak base (WA) and (WB) :
$C H_{3} C O O H + N H_{4} O H ⟶ C H_{3} C O O N H_{4}$
Partial dissociation (Close to 100% but not exact)
$C H_{3} C O O N H_{4} (a q) ⟶ N H_{4}^{+} (a q) + C H_{3} C O O^{-} (a q)$
Hydrolysis :
$N H_{4}^{+} (a q) + C H_{3} C O O^{-} (a q) + H_{2} O (l) ⇌ C H_{3} C O O H (a q) + N H_{4} O H (a q)$
$C C 00$
$C - C h C - C h C h C h$
Since both the acid and the base are weak, they will exist in undissociated form.
$K_{h} = \frac{[N H_{4} O H] [C H_{3} C O O H]}{[N H_{4}^{+}] [C H_{3} C O O^{-}]}$
$\frac{K_{w}}{K_{a} K_{b}} = K_{h} = \frac{[H_{3} O^{+}] [O H^{-}]}{\frac{[C H_{3} C O O^{-}] [H_{3} O^{+}]}{[C H_{3} C O O H]} \times \frac{[N H_{4}^{+}] [O H^{-}]}{[N H_{4} O H]}}$

$K_{h} = \frac{(C h) (C h)}{C^{2} (1 - h)^{2}} = \frac{h^{2}}{(1 - h)^{2}}$
$\frac{h}{1 - h} = \sqrt{K_{h}}$
Finding pH:
consider weak acid dissociation,
$K_{a} = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} ⟶ \frac{C (1 - h) [H^{+}]}{C h}$
$[H^{+}] = K_{a} \frac{h}{1 - h} = K_{a} \sqrt{K_{h}}$
$[H^{+}] = K_{a} \sqrt{\frac{K_{w}}{K_{a} K_{b}}}$

$[H^{+}] = \sqrt{\frac{K_{w} K_{a}}{K_{b}}}$
$l o g [H^{+}] = \frac{1}{2} l o g K_{w} + \frac{1}{2} l o g K_{a} - \frac{1}{2} l o g K_{b}$
$- l o g [H^{+}] = - \frac{1}{2} l o g K_{w} - \frac{1}{2} l o g K_{a} + \frac{1}{2} l o g K_{b}$
$p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$
valid only if h<0.1 or 10% or when $C / K_{h} > 100$
so, $p H = 7 + \frac{1}{2} (p K_{a} - p K_{b}) p O H = 14 - p H p O H = 14 - (7 + \frac{1}{2} (p K_{a} - p K_{b}) p O H = 7 - \frac{1}{2} (p K_{a} - p K_{b}) a t 25^{\circ} C$

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