What is the $pH$ of a solution made by mixing $25.0 mL$ of $1.00\times {10}^{-3} \mathrm{M} {\mathrm{HNO}}_3\text{ and }25.0 \mathrm{mL}\text{ of }1.00\times$${10}^{-3} \mathrm{M} {\mathrm{NH}}_3?\left[K_b\text{ for }{\mathrm{NH}}_3=1.8\times {10}^{-5}\right]$

$\frac{{\mathrm{HNO}}_3}{S\mathrm{A}}+\frac{{\mathrm{NH}}_3}{\omega B}⟶{\mathrm{NH}}_4{\mathrm{NO}}_3$

$p^H=7-\frac{1}{2}\left(p^{K_b}+\log C\right)$

$C=\frac{n}{v}$

$\left[{\mathrm{HNO}}_3\right]=\left[{\mathrm{NH}}_3\right]=25\times {10}^{-3}\mathrm{m}\text{ moles }$

$\left[salt\right]=\left[N{H}_{4}N{O}_3\right]$ $=\frac{25\times {10}^{-3}}{50}$

$\therefore C=\frac{{10}^{-3}}{2}$

$\log C=\log \left(\frac{{10}^{-}}{2}\right)=-13.30$

$p{K}_b=-\log \left(K_b\right)=4.75$

$p^H=7-\frac{1}{2}(4.75-3.30)$

$=7-\frac{1}{2}\left(1.45\right)$

$=7-0.72$

$p{K}_b$$=6.28$