Calculate the pH of a buffer solution containing 0.45 moles of  

${\mathrm{NH}}_4\mathrm{OH}$ and $\mathrm{x}0.75\text{ mole of }{\mathrm{NH}}_4\mathrm{Cl}.{\mathrm{K}}_b$ for 

${\mathrm{NH}}_4\mathrm{OH}\text{ is }1.8\times {10}^{-2}$

Let volume of buffer solution= one litre 

Then $\left[{\mathrm{NH}}_4\mathrm{OH}\right]=0.45\mathrm{M};\left[{\mathrm{NH}}_4\mathrm{Cl}\right]=0.75\mathrm{M}.$ 

But for a basic  buffer solution: $\begin{array}{l}p\mathrm{OH}=p{K}_b+{\log }_{10}\frac{\left[\text{ Salt }\right]}{\left[\text{ Base }\right]}\\ =-{\log }_{10}\left(1.8\times {10}^{-5}\right)+{\log }_{10}\frac{0.75}{0.45}\end{array}$

or         $p\mathrm{OH}=4.7447+[-0.1249+0.3467]$

           $=4.7447+0.2218=4.9665$

But $\begin{array}{r}\mathrm{pH}=14-p\mathrm{OH}=14-4.9665\\ =9.0335\end{array}$