Question

The degree of dissociation of $0.1$ M $N{H}_3$ at ${25}^{\circ }C$ is $1\times {10}^{-5}$. The pH of solution is:

• A. $3$
• B. $5$
• C. $7$
• D. $8$

Answer 1

The correct option is D $8$

$\alpha =1\times {10}^{-5}$

${NH}_{4}OH\rightleftharpoons {NH}_4^{+}+{OH}^{-}$

$\left[{OH}^{-}\right]=C\alpha$

$=0.1\times 1\times {10}^{-5}$

$=1\times {10}^{-6}$

$pOH=log1+6=6$

$pH=14-6=8$