$0.05\, M$ अमोनिया विलयन की आयनन मात्रा तथा $pH$ ज्ञात कीजिए। अमोनिया के आयनन-स्थिरांक का मान तालिका $7.7$ में दिया गया है। अमोनिया के संयुग्मी अम्ल का आयनन स्थिरांक भी ज्ञात कीजिए।

$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$

We use equation $(7.33)$ to calculate hydroxyl ion concentration,

$\left[ OH ^{-}\right]= c \alpha=0.05 \alpha$

$K_{ b }=0.05 \alpha^{2} /(1-\alpha)$

The value of $\alpha$ is small, therefore the quadratic equation can be simplified by neglecting $\alpha$ in comparison to $1$ in the denominator on right hand side of the equation,

Thus,

$K_{ b }= c \alpha^{2}$ or $\alpha=\sqrt{\left(1.77 \times 10^{-5} / 0.05\right)}$ $=0.018$

$\left[ OH ^{-}\right]= c \alpha=0.05 \times 0.018=9.4 \times 10^{-4} \,M$

$\left[ H ^{+}\right]=K_{ w } /\left[ OH ^{-}\right]=10^{-14} /\left(9.4 \times 10^{-4}\right)$

$=1.06 \times 10^{-11}$

$pH =-\log \left(1.06 \times 10^{-11}\right)=10.97$

Now, using the relation for conjugate acid-base pair,

$K_{ a } \times K_{ b }=K_{ w }$

using the value of $K_{ b }$ of $NH _{3}$ from Table

Base	$K _{ b }$
Dimethylamine, $\left( CH _{3}\right)_{2} NH$	$5.4 \times 10^{-4}$
Triethylamine, $\left( C _{2} H _{5}\right)_{3} N$	$6.45 \times 10^{-5}$
Ammonia, $NH _{3}$ or $NH _{4} OH$	$1.77 \times 10^{-5}$
Quinine, ( $A$ plant product)	$1.10 \times 10^{-6}$
Pyridine, $C _{5} H _{5} N$	$1.77 \times 10^{-9}$
Aniline, $C _{6} H _{5} NH _{2}$	$4.27 \times 10^{-10}$
Urea, $CO \left( NH _{2}\right)_{2}$	$1.3 \times 10^{-14}$

We can determine the concentration of conjugate acid $NH _{4}^{+}$

$K_{ a }=K_{ w } / K_{ b }=10^{-14} / 1.77 \times 10^{-5}$

$=5.64 \times 10^{-10}$