Question

Calculate the pH of 0.1 M $N{H}_{4}Cl$ solution in water :
$K_b\left(N{H}_{4}OH\right)=1.8\times {10}^{-5}$

• A. 4.5
• B. 7.63
• C. 6.53
• D. 5.13

Answer 1

The correct option is D 5.13

$K_b=1.8\times {10}^{-5}p{K}_b=-log\left(K_b\right)p{K}_b=-log(1.8\times {10}^{-5})=(5-0.25)=4.75$
$N{H}_{4}Cl$ is a salt of strong acid and weak base.
$pH=7-\frac{1}{2}(p{K}_b+logC)pH=7-\frac{1}{2}(4.75+log(0.1\left)\right)pH=7-\frac{3.74}{2}pH=5.125$



Theory :
Salt of strong acid (SA) and weak base(SB):
Consider a salt $\left(MX\right)$ added to water and produce cations $\left(M^{+}\right)$and anions $\left(X^{-}\right)$ in water.
For example :- $N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)$
$C00$
$C-ChChCh$
In this reaction the $\left(K_h\right)$ hydrolysis constant is:
$K_h=\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}=\frac{(Ch{)}^2}{C(1-h)}$
$N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
In this reaction the $K_b$ constant is:
$K_b=\frac{\left[N{H}_4\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$
$H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
In this reaction the $K_w$ constant is:
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
Using he value of $K_{b}and{K}_w$ we can derive the value of $K_h$
$\therefore$ $\frac{K_w}{K_b}=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{\left[N{H}_4\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}}$
$\frac{K_w}{K_b}=\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{N{H}_4^{+}}$
$K_h=\frac{K_w}{K_b}$
Form earlier equation of $K_h$
$K_h=\frac{(Ch{)}^2}{C(1-h)}$ Generally h<<1. So, neglect(1-h)
If h<<1
$C{h}^2=K_h=\frac{K_w}{K_b}$
Then $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_{b}C}}$
$\left[H^{+}\right]=Ch$
$\left[H^{+}\right]=C\times \sqrt{\frac{K_w}{K_{b}C}}=\sqrt{\frac{K_{w}C}{K_b}}$
Taking log on both sides, we get
$log\left[H^{+}\right]=\frac{1}{2}log{K}_w-\frac{1}{2}log{K}_b+\frac{1}{2}logC$
$pH=\frac{1}{2}p{K}_w-\frac{1}{2}p{K}_b-\frac{1}{2}logC$
Since $p{K}_b=-log{K}_b$ and $p{K}_w=-log{K}_w$
Valid if h<0.1 or 10%
$pH=\frac{1}{2}[p{K}_w-p{K}_b-logC]$
At ${25}^{\circ }$ $pH=7-\frac{1}{2}p{K}_b-\frac{1}{2}logC$
If h obtained from $\frac{K_h}{C}$ > 0.1, then solve the quadratic equation and get h. Then, $\left[H^{+}\right]=Ch$ and now pH calculation can be done.