Question

Calculate the $K_h$ and pH of 0.05N $N{H}_{4}Cl$ solution where $K_b=1.8\times {10}^{-5}$

Answer 1

For salt of weak base, strong acid like $N{H}_{3}Cl$,

$K_h=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.56\times {10}^{-10}$

$pH=7-\frac{1}{2}[p{K}_b+\log C]=7-\frac{1}{2}\left\{\right[-\log (1.8\times {10}^{-5})]+\log 0.05\}=pH=5.28$