Question

What is the $pH$ of millimolar solution of ammonium hydroxide which is $20\%$ dissociated ?

• A. $3.699$
• B. $10.301$
• C. $4.691$
• D. $9.301$

Answer 1

Answer: (B)

Concentration of $N{H}_{4}OH$ solution $=0.001M$

Degree of dissociation$=20\%=\frac{20}{100}=0.2$

Concentration of $O{H}^{-}$ i.e., $\left[O{H}^{-}\right]=$

Concentration of solution $\times$ degree of dissociation

$=0.001\times 0.02=2\times 10^{-4}$

$\because \left[H^{+}\right]\left[O{H}^{-}\right]=1\times 10^{-14}$

$\therefore \left[H^{+}\right]=\frac{1\times 10^{-14}}{2\times 10^{-4}}$

$=\frac{1}{2}\times 10^{-10}$

Again, $\because pH=-\log \left[H^{+}\right]$

$\therefore pH=-\log \left(\frac{1}{2}\times 10^{-10}\right)$

$=-\log (-10-0.3010)=10.3010$