Determine the degree of ionization and $pH$ of a $0.05 \,M$ of ammonia solution. The ionization constant of ammonia can be taken from Table $7.7 .$ Also, calculate the ionization constant of the conjugate acid of ammonia.
$NH _{3}+ H _{2} O \rightleftharpoons NH _{4}^{+}+ OH ^{-}$
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 $NH _{4}^{+}$
$K_{ a }=K_{ w } / K_{ b }=10^{-14} / 1.77 \times 10^{-5}$
$=5.64 \times 10^{-10}$
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