Solution of 0.1  $NN{H}_{4}OH$ and $0.1\text{\hspace{0.17em}}MN{H}_{4}Cl$  has $pH9.25$ . Then find out $p{K}_b$  of  $N{H}_{4}OH$.

Solution:  Given that – Normality of $N{H}_{4}OH=0.1N$   Molarity of  $N{H}_{4}Cl=0.1M$
 $pH=9.25$
According to the Henderson’s equation – 
 $$\begin{gathered} pOH=p{H}_b+\log \frac{\left[Salt\right]}{\left[Base\right]} \\ pOH=pOH=14\left(at\text{\hspace{0.17em}}{25}^{0}C\right) \\ pOH=14-9.5=4.75 \\ 4.75=p{K}_b+\log \frac{\left[1\right]}{\left[1\right]} \\ p{K}_b=4.75 \end{gathered}$$