Question

For a ammonium chloride solution in water, the given equilibrium reaction occur:
$N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\stackrel{hydrolysis}{\rightleftharpoons }N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)$
Which of the following describes correct relation ?

• A. $K_h=\frac{K_w}{K_b}$
• B. $K_h=K_w\times K_b$
• C. $K_h=\frac{K_b}{K_w}$
• D. None of the above

Answer 1

The correct option is A $K_h=\frac{K_w}{K_b}$

$N{H}_4^{+}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons N{H}_{4}OH\left(aq\right)+H^{+}\left(aq\right)$

$C00$ $C-ChChCh$

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 $K_h$

$\therefore$ $\frac{K_w}{K_b}=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{\left[N{H}_4\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}}$

$\frac{K_w}{K_b}=\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{N{H}_4^{+}}$

$K_h=\frac{K_w}{K_b}$