$p{K}_b$ for $N{H}_{4}OH$ at certain temperature is $4.74$. The $pH$ of basic buffer containing equimolar concentration of $N{H}_{4}OH$ and $N{H}_{4}Cl$ will be :

The correct option is A $9.26$
The $pH$ of a basic buffer solution can be determined by using Henderson-Hasselbalch equation.
As we know,
$pOH=p{K}_b+\text{log}\frac{\text{[Salt]}}{\text{[Base]}}$
As the $\left[N{H}_{4}OH\right]=\left[N{H}_{4}Cl\right]$
$pOH=p{K}_b=4.74$
So,
$\Rightarrow pH=14-4.74=9.26$