Question

A solution contain $10ml$ of $0.1{NH}_{4}OH$ and $10ml$ $0.5N$ ${\left({NH}_4\right)}_2{SO}_4$. $pH$ of this solution is:

Answer 1

According to Henderson-Hasselbach equation
$P_{OH}=P_{K_b}+\log \left(\left[salt\right]/\left[base\right]\right)$
$P_{K_b}=-\log K_b=-\log 0.85\times {10}^{-5}$
$P_{K_b}=4.733$
$\therefore P_{OH}=4.733\log \left(\frac{0.5}{0.1}\right)$
$=4.733\log 5$
$=4.733\times 0.6989$
$=3.308$
$\therefore P_H=14-P_{OH}=10.692$