The pH value of decinormal solution of N{H_4} | vedclass

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Question

(c) For $N{H_4}OH$.

$[O{H^ - }] = C\,.\,\alpha $ ; $C = \frac{1}{{10}}M$, $\alpha = 0.2$

$[O{H^ - }] = \frac{1}{{10}} 	imes 0.2 = 2 	imes {10^

Vedclass · Accepted Answer

$12.3$

Vedclass · Answer

(c) For $N{H_4}OH$.

$[O{H^ - }] = C\,.\,\alpha $ ; $C = \frac{1}{{10}}M$, $\alpha = 0.2$

$[O{H^ - }] = \frac{1}{{10}} 	imes 0.2 = 2 	imes {10^{ - 2}}\,M$

$pOH = - \log \,\,[O{H^ - }]$$ = \log \,\,[2 	imes {10^{ - 2}}]$; $pOH = 1.7$

$pH = 14 - pOH$ $ = 14 - 1.7 = 12.30$.

The $pH$ value of decinormal solution of $N{H_4}OH$ which is $20\%$ ionised, is

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