\(\mathrm{Hg}_{2} \mathrm{Cl}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Hg}_{2}^{2+} \text { (aq) }+2 \mathrm{Cl^-}_{(\mathrm{aq})}\\ s \quad \quad \quad \quad \quad \quad s \quad \quad \quad \quad \quad 2s\)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Hg}_{2}^{2+}\right]{\left[\mathrm{Cl}^{-}\right]^{2}}  \)
\(=(\mathrm{s})(2 \mathrm{~s})^{2} \)
\(\mathrm{~K}_{\mathrm{sp}} =4 \mathrm{~s}^{3}\)

K_sp = 1.8 \(\times\)10^-10, [AgNO₃]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
                    s            s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)

a) \(BaSO_{4}(s)\overset{H_{2}O}{\rightleftharpoons }Ba^{2+}(aq)+SO^{2+}_{4}(aq)\)
\(K_{sp}=[Ba^{2+}][SO^{2-}_{4}]\) =   (s)   (s)
K_sp = s²
b) \(Ag_{2}CrO_{4}(s)\overset{H_{2}O}{\rightleftarrows }2Ag^{+}(aq)+CrO_{4}^{2-}(aq)\)
\(K_{sp}=[Ag^{+}]^{2}[CrO)^{2-}_{4}]\)
=  (2s)² (s)
K_sp = 4S³

According to Henderson – Hasselbalch equation,
\(pH=pK_{a}+\log\frac{[salt]}{[acid]}\)
\(p{K_{a}}=-\log K_{a}=-\log(1.8\times10^{-5})=4.74\)
[Salt]=\(\frac{\text {Number of moles of sodium acetate}}{\text {Volume of the solution (litre)}}\)
Number of moles of sodium acetate =\(\frac{\text {mass of sodium acetate}}{\text {molar mass of sodium acetate}}\)
\(=\frac{8.2}{82}=0.1\)
\(\therefore [Salt]=\frac{0.1\ mole}{1/2 \ Litre}=0.2M\)
\([acid]=\frac{(\frac{mass \ of \ CH_{3}COOH}{molar \ mass \ of \ CH_{3}COOH})}{\text{Volume of solution in litre}}\)
=\(\frac{(\frac{6}{60})}{\frac{1}{2}}\)=0.2 M
\(\therefore pH=4.74+log\frac{(0.2)}{(0.2)}\)
pH = 4.74 + log1
pH = 4.74 + 0 = 4.74

When two are more solution are mixed, the resulting concentrations are different from the original.
\(\text { Molarity }=\frac{n}{\mathrm{~V}} \text { (or) } \mathrm{n}=\text { Molarity } \times \mathrm{v}  \)
Total Volume of the mixture = 0.15 + 0.1
= 0.25 L
\(\underset{0.1M}{Pb(NO_{3})_{2}}\rightleftharpoons \underset{0.1M}{Pb^{2+}}+2\underset{0.2M}{2NO^{-}_{3}}\)
n_Pb²⁺ \(=0.1\times0.15=0.015 \ mol\)
\([Pb^{2+}]_{mix}= \frac{n}{v} = \frac{0.1\times0.15}{0.25}=0.06M\)
\(\underset{0.2M}{NaCl}\rightleftharpoons \underset{0.2M}{Na^{+}}+\underset{0.2M}{Cl^{-}}\)
\(\mathrm{n}_{\mathrm{Cl^-}}=0.2 \times 0.1=0.02 \mathrm{~mol} \)
\(\left[\mathrm{Cl}^{-}\right]_{\text {mix }}=\frac{0.02}{0.25}=0.08 \mathrm{M} \)
\(\therefore Ionic \ Product =\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2} \)
\(=0.06 \times(0.08)^{2} \)
\(IP =3.84 \times 10^{-4}\)
\(\therefore 3.84 \times 10^{-4}>1.2 \times 10^{-5}\)
(or) \(\mathrm{IP}>\mathrm{K}_{\mathrm{sp}}\)
\(\therefore\) PbCl₂ will be precipitated.