25 ml 0.4M a weak base and 75 ml of 0.2M of its salt forms a buffer solution. $K_b$ of weak base is $2\times {10}^{–5}$. The pH of the buffer solution is [log 1.5 = 0.176]

 $p{K}_b=-log\left[K_b\right]$ = -log [$2\times {10}^{–5}$] = 4.70

pOH = $p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}$

= 4.70 + $log\frac{[M_1\times v_1]}{M{[}_2\times v_2]}$

= 4.70 + +$log\frac{0.2\times 75}{0.4\times 25}$

= 4.70 + 0.1760 = 4.876

pH + pOH = 14

pH = 14 – 4.876 = 9.124.

Hence, the correct option is option (A) 9.124.