Calculate $K_h$ and $pH$ of $0.1 \ M$ $NH_4Cl$ solution. [ $K_w = 1 \times 10^{-14}$,$K_{NH_4OH} = 1.75 \times 10^{-5}$ ]

(A) For a salt of a weak base and a strong acid $(NH_4Cl)$,the hydrolysis constant $K_h$ is given by $K_h = \frac{K_w}{K_b}$.
Substituting the values: $K_h = \frac{1 \times 10^{-14}}{1.75 \times 10^{-5}} = 5.71 \times 10^{-10}$.
The $pH$ of the solution is given by the formula $pH = \frac{1}{2} [pK_w - pK_b - \log C]$.
$pK_w = 14$,$pK_b = -\log(1.75 \times 10^{-5}) \approx 4.76$,and $\log C = \log(0.1) = -1$.
$pH = \frac{1}{2} [14 - 4.76 - (-1)] = \frac{1}{2} [10.24] = 5.12$.