Question

Ammonium hydroxide $N{H}_{4}OH$ (a weak base) solution has a concentration of 0.02 M. Then the $pH$ value of this solution is :
The ionization constant $\left(K_b\right)$ for $N{H}_{4}OH$ is $1.8\times {10}^{-5}M$.

• A. 3.22
• B. 10.78
• C. 2.15
• D. 11.85

Answer 1

The correct option is B 10.78

Given, $\left[N{H}_{4}OH\right]=0.02M,K_b=1.8\times {10}^{-5}$
The dissociation is given by :

$N{H}_{4}OH(aq.)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)Initial:C00Equilibrium:C(1-\alpha )C\alpha C\alpha$

$\therefore$ At equilibrium $\left[O{H}^{-}\right]=C\alpha$

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{N{H}_{4}OH}{K}_b=\frac{(C\alpha {)}^2}{C(1-\alpha )}$
Since, $N{H}_{4}OH$ is a weak base.
$\therefore \alpha <<11-\alpha \approx 1$

$K_b\times C=(C\alpha {)}^2{K}_b\times C=[O{H}^{-}{]}^2\left[O{H}^{-}\right]=\sqrt{K_b\times C}=\sqrt{1.8\times {10}^{-5}\times 0.02}\left[O{H}^{-}\right]=6\times {10}^{-4}pOH=-log\left[O{H}^{-}\right]pOH=-log(6\times {10}^{-4})=(4-log(6\left)\right)=(4-0.78)pOH=3.22pH+pOH=14pH=14-3.22pH=10.78$