$\begin{array}{r}\mathrm{BaSO}_4(\mathrm{~s}) \rightleftharpoons \mathrm{Ba}^{2+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq}) \\ K_{\text {sp }} \text { for } \mathrm{BaSO}_4=\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=\mathrm{s} \times \mathrm{s}=\mathrm{s}^2\end{array}$

$\begin{array}{lr}\text { But } & s=8 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \\ \therefore & K_{\mathrm{sp}}=\left(8 \times 10^{-4}\right)^2=64 \times 10^{-8}\end{array}$

In the presence of $0.01 \mathrm{MH}_2 \mathrm{SO}_4$, the expression for $K_{\text {sp }}$ will be

$$\begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Ba}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right] \\ K_{\mathrm{sp}} & =(s) \cdot(s+0.01) \quad\left(0.01 \mathrm{M} \mathrm{SO}_4^{2-} \text { ions from } 0.01 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\right) \\ 64 \times 10^{-8} & =s \cdot(s+0.01) \\ s^2+0.01 s-64 \times 10^{-8} & =0 \end{aligned}$$

$\begin{aligned} S & =\frac{-0.01 \pm \sqrt{(0.01)^2+\left(4 \times 64 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}+\left(256 \times 10^{-8}\right)}}{2} \\ & =\frac{-0.01 \pm \sqrt{10^{-4}\left(1+256 \times 10^{-4}\right)}}{2} \\ & =\frac{-0.01 \pm 10^{-2} \sqrt{1+0.0256}}{2}=\frac{10^{-2}(-1 \pm 1.012719)}{2} \\ & =5 \times 10^{-3}(-1+1.012719)=6.4 \times 10^{-5} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$

$\begin{aligned} \mathrm{pH} \text { of } \mathrm{HOCl} & =2.85 \\ \text{But,}\quad -\mathrm{pH} & =\log \left[\mathrm{H}^{+}\right] \\ \therefore\quad -2.85 & =\log \left[\mathrm{H}^{+}\right] \\ \Rightarrow \quad \overline{3} .15 & =\log \left[\mathrm{H}^{+}\right] \\ \Rightarrow \quad {\left[\mathrm{H}^{+}\right] } & =1.413 \times 10^{-3}\end{aligned}$

For weak monobasic acid $\left[\mathrm{H}^{+}\right]=\sqrt{K_a \times C}$

$\begin{aligned} \Rightarrow \quad K_a & =\frac{\left[\mathrm{H}^{+}\right]^2}{C}=\frac{\left(1.413 \times 10^{-3}\right)^2}{0.08} \\ & =24.957 \times 10^{-6}=2.4957 \times 10^{-5}\end{aligned}$

pH of solution $A=6$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{~mol} \mathrm{~L}$

pH of solution $B=4$. Hence, $\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$

On mixing 1 L of each solution, molar concentration of total $\mathrm{H}^{+}$is halved.

$\begin{aligned} & \text{Total,}\quad {\left[\mathrm{H}^{+}\right] }=\frac{10^{-6}+10^{-4}}{2} \mathrm{~mol} \mathrm{~L}^{-1} \\ & {\left[\mathrm{H}^{+}\right] }=\frac{1.01 \times 10^{-4}}{2}=5.05 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} \\ & {\left[\mathrm{H}^{+}\right] }=5.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-[\log 5+(-5 \log 10)] \Rightarrow \mathrm{pH}=-\log \left(5.0 \times 10^{-5}\right) \\ & \mathrm{pH}=5-\log 5=5-0.6990 \Rightarrow \mathrm{pH}=-\log 5+5 \\ & \mathrm{pH}=4.3010 \approx 4.3\end{aligned}$

Thus, the pH of resulting solution is 4.3.

Let S be the solubility of Al(OH)$$_3$$.

Concentration of species at $t=0$

Concentration of various species at equilibrium

$\begin{aligned} K_{\mathrm{sp}} & =\left[\mathrm{Al}^{3+}\right]\left[\mathrm{OH}^{-}\right]^3=(\mathrm{S})(3 \mathrm{~S})^3=27 \mathrm{~S}^4 \\ S^4 & =\frac{K_{\mathrm{sp}}}{27}=\frac{2.7 \times 10^{-11}}{27}=1 \times 10^{-12} \\ S & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\end{aligned}$

(i) Solubility of $\mathrm{Al}(\mathrm{OH})_3$

Molar mass of $\mathrm{Al}(\mathrm{OH})_3$ is 78 g . Therefore,

Solubility of $\mathrm{Al}(\mathrm{OH})_3$ in $\mathrm{g}^{-1}=1 \times 10^{-3} \times 78 \mathrm{~g} \mathrm{~L}^{-1}=78 \times 10^{-3} \mathrm{~g} \mathrm{~L}^{-1}$

$=7.8 \times 10^{-2} \mathrm{~g} \mathrm{~L}^{-1}$

$\begin{aligned} \text{(ii) pH of the solution}\quad \mathrm{S} & =1 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ {\left[\mathrm{OH}^{-}\right] } & =3 \mathrm{~S}=3 \times 1 \times 10^{-3}=3 \times 10^{-3} \\ \mathrm{pOH} & =3-\log 3 \\ \mathrm{pH} & =14-\mathrm{pOH}=11+\log 3=11.4771\end{aligned}$

Suppose, solubility of PbCl$$_2$$ in water is s mol L$$^{-1}$$

$\begin{aligned} \mathrm{PbCl}_2(\mathrm{~s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}(\mathrm{aq}) \\ \mathrm{s}_{\mathrm{sp}} & =\left[\mathrm{Pb}^{2 \uparrow}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\ K_{\mathrm{sp}} & =[\mathrm{s}][2 \mathrm{~s}]^2=4 s^3 \\ 3.2 \times 10^{-8} & =4 \mathrm{~s}^3 \\ s^3 & =\frac{3.2 \times 10^{-8}}{4}=0.8 \times 10^{-8} \\ s^3 & =8.0 \times 10^{-9}\end{aligned}$

$$\begin{aligned} \text{Solubility of}\quad \mathrm{PbCl}_2, \mathrm{~s} & =2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \\ \text{Solubility of}\quad \mathrm{PbCl}_2 \text { in } \mathrm{gL}^{-1} & =278 \times 2 \times 10^{-3}=0.556 \mathrm{~g} \mathrm{~L}^{-1} \end{aligned}$$

$\left(\because\right.$ Molar mass of $\left.\mathrm{PbCl}_2=207+(2 \times 35.5)=278\right)$

0.556 g of $\mathrm{PbCl}_2$ dissolve in 1 L of water.

$\therefore \quad 0.1 \mathrm{~g}$ of $\mathrm{PbCl}_2$ will dissolve in $=\frac{1 \times 0.1}{0.556} \mathrm{~L}$ of water $$=0.1798 \mathrm{~L}$$

To make a saturated solution, dissolution of $0.1 \mathrm{g~PbCl}_2$ in $0.1798 \mathrm{~L} \approx 0.2 \mathrm{~L}$ of water will be required.