Question

Calculate pH of $0.002NN{H}_{4}OH$ having 2 % dissociation.

• A. 7.6
• B. 8.6
• C. 9.6
• D. 10.6

Answer 1

The correct option is C 9.6

$N{H}_{4}OH$ is a weak base and partially dissociated
Concentration
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
Before:$100$

after $1-\alpha \alpha \alpha$
$SinceN{H}_{4}OHispartiallydissociated\left[O{H}^{-}\right]=Concentration\left(C\right)\times Dissociation\left(\alpha \right)Here\alpha =2\%=\frac{2}{100}C=0.002N=2\times {10}^{-3}$
$\therefore \left[O{H}^{-}\right]=C\alpha =2\times {10}^{-3}\times \frac{2}{100}=4\times {10}^{-5}MpOH=-log\left[O{H}^{-}\right]=-\log (4\times {10}^{-5})=4.4pH=14-4.4=9.6$