Question

Degree of dissociation of ${CH}_{3}COOH$ and ${NH}_{4}OH$ are the same, If $0.01M$ solution of ${CH}_{3}COOH$ has $pH=3.0$; then pH of $0.01M$ ${NH}_{4}OH$ will be:

• A. $4$
• B. $7$
• C. $10$
• D. $11$

Answer 1

The correct option is D $11$

Solution -

$\to$ Degree of dissociation of $C{H}_{3}COOH$

= Degree of dissociation of $N{H}_{4}OH$

$=\alpha$

$\to$ concentration given $=0.01M$

For $C{H}_{3}COOH$

$conc=0.01m$

$pH=3$

i.e, $\left[H^{+}\right]={10}^{-3}$

Relation of $\left[H^{+}\right]$ & $\alpha$

$\left[H^{+}\right]=c\alpha$

${10}^{-3}=0.01\alpha$

$0.1=\alpha$

For $N{H}_{4}OH$

$\alpha =0.1$

$conc=0.01M$

Hence, $\left[O{H}^{-1}\right]=c\alpha$

$\left[O{H}^{-}\right]=0.01\left[0.1\right]$

$\left[O{H}^{-}\right]={10}^{-3}$

$K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

${10}^{-14}/{10}^{-3}={10}^{-11}$ $pH=11$ Ans