$50\mathrm{ml}$ of $0.2 {\mathrm{NH}}_2{\mathrm{SO}}_4$ is mixed with $100\mathrm{ml}$ of $0.4 \mathrm{NKOH}$ solution and $1.85$ lit of distilled water is added. The $\mathrm{pH}$ of resulting solution is $(\log 1.5=0.176)$

 $NV of H_2{\mathrm{SO}}_4=50 x 0.2 =10$

$NV of \mathrm{KOH}= 0.4 x100= 40$

NV of [OH^-] = 30

 [OH^-] =$\frac{30}{18500}\approx 1.5 x{10}^{-2}$

 $p^{OH}$ = $-{\log }_{10}\left[O{H}^{-}\right]$  = $-\log 1.5\times {10}^{-2}$ 

${\mathrm{p}}^{\mathrm{OH}} =2-{\log }_{10}1.5=2-0.17 =1.823$ 

$p^H=14-1.823=12.176$ 

Hence the correct option is (D) $12.176$.