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The pH value of decinormal solution of `NH_(4)OH` which is 20% ionised, isA. `13.30`B. `14.70`C. `12.30`D. `12.95` |
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Answer» Correct Answer - C For `NH_(4)OH`. `[OH^(-)] = C.alpha , c = (1)/(10)M, alpha = 0.2` `[OH^(-)] = (1)/(10) xx 0.2 = 2 xx 10^(-2)M` `pOH = - log [OH^(-)] = log[2 xx 10^(-2)], pOH = 1.7` `pH = 14 - pOH = 14-1.7 = 12.30`. |
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