The pH of a solution obtained by mixing 5 mL of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ solution with 250 mL of $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ solution is $\_\_\_\_$ $\times 10^{-2}$.(Nearest integer) Given: $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.74$

$$ \begin{aligned} & \log 2=0.30 \\ & \log 3=0.48 \\ & \log 5=0.70 \end{aligned} $$

Consider the dissociation equilibrium of the following weak acid

$$ \mathrm{HA} \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq}) $$

If the pKa of the acid is 4 , then the pH of 10 mM HA solution is $\_\_\_\_$ .(Nearest integer)

[Given: The degree of dissociation can be neglected with respect to unity]

Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$.

A solution containing $0.01 \mathrm{M} \mathrm{X}^{2+}$ and $0.01 \mathrm{M} \mathrm{Y}^{2+}$ is saturated with $\mathrm{H}_2 \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is $\_\_\_\_$ . (Nearest integer)

(Given: $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$, $\left[\mathrm{H}_2 \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{a 1} \times \mathrm{K}_{a 2}\left(\mathrm{H}_2 \mathrm{~S}\right)=1.0 \times 10^{-21}, \log 2=0.30$, $\log 3=0.48, \log 5=0.70)$

The first and second ionization constants of H₂X are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively.

The concentration of ${X^{2-}}$ in $0.1\ \mathrm{M}$ H₂X solution is ______ $\times 10^{-15}\ \mathrm{M}$. (Nearest Integer)