The $\mathrm{OH}$ concentration in a mixture of $5.0 \mathrm{~mL}$ of $0.0504 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}$ and $2 \mathrm{~mL}$ of $0.0210 \mathrm{M} \mathrm{NH}_{3}$ solution is $x \times 10^{-6} \mathrm{M}$. The value of $x$ is (Nearest integer)
[Given $\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}$ and $\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5}$ ]
$\left[\mathrm{NH}_{4}^{+}\right]=0.0504 \&\left[\mathrm{NH}_{3}\right]=0.0210$
So $\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{HO}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}$
$[\mathrm{HO}]=\frac{\mathrm{K}_{\mathrm{b}} \times\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^{+}\right]}=1.8 \times 10^{-5} \times \frac{2}{5} \times \frac{210}{504}$
$=3 \times 10^{-6}$