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

(a) CH₃COONa is a salt of weak acid
(CH₃COOH) and a strong base (NaOH).
Hence, the solutions is alkaline due to hydrolysis.
\(CH_{3}COO^{-}_{(aq)}+H_{2}O_{(aq)}\rightleftharpoons CH_{3}COOH_{(aq)}+OH^{-}_{(aq)}\)
(i)\(h=\sqrt{\frac{K_{w}}{K_{a}\times C}}\)
Given that pK_a =4.74
pK_a = -log K_a
ie., K_a = antilog of (-pK_a)
= antilog of (-4.74)
= antilog of (-5 + 0.26)
= 10^-5 \(\times\) 1.8 = 1.8 \(\times\) 10^-5
[antilog of 0.26 = 1.82 \( \simeq\) 1.8]
\(\therefore\) h=\(\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times0.1}}\)
h = 7.5 x 10^-5
(ii) \(K_{h}=\frac{K_{w}}{K_{a}}=\frac{1\times10^{-14}}{1.8\times10^{-5}}\)
\(=5.56\times10^{-10}\)
iii) \(pH=7+\frac{pK_{a}}{2}+\frac{logC}{2}\)
= \(7+\frac{4.74}{2}+\frac{log0.1}{2}\)
= 7 + 2.37 - 0.5
= 8.87

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

\(Al(OH)_{3}\rightleftharpoons Al^{3+}_{(aq)}+3OH^{-}_{(aq)}\)
\(K_{sp}=[Al^{3+}][OH^{-}]^{3}\)
Al(OH)₃ precipitates when ionic product > K_sp
K_s = 1.0 \(\times\) 10^-15m, [Al³⁺] = 1.0 \(\times\) 10^-3m
1.0 \(\times\) 10^-15 = [1.0 \(\times\) 10^-3][OH^-]³
\(\left[\mathrm{OH}^{-}\right]^3=\frac{1.0 \times 10^{-15}}{1.0 \times 10^{-3}}\)
[OH^-]³ = 1.0 \(\times\) 10^-12
(or)
[OH^-]³ = 10^-4M
[H⁺][OH^-] = 10^-4M
[H⁺] = \(\frac{10^{14}}{10^{-4}}\) =10^-10
Here,
pH = 10
ie., At pH = 10, Al(OH)₃ gets precipitated on the addition of NH₄Cl & NH₄OH solution.

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.

K_sp =1.1 \(\times\)10^-2, [K₂, CrO₄]= 0.1M
\(\underset {s}{Ag_{2}CrO_{4}}\rightleftharpoons \underset{2s}{2Ag^{+}}+\underset{s}{CrO^{2-}_{4}}\)
\(\underset{0.1M}{K_{2}CrO_{4}}\rightleftharpoons \underset{0.2M}{2K^{+}}+\underset{2 \times 0.1M}{CrO^{2-}_{4}}\)
\({\left[\mathrm{Ag}^{+}\right] } =2 \mathrm{~s} ;\left[\mathrm{CrO}_{4}^{2-}\right]=(\mathrm{S}+0.1) \simeq 0.1  \)
\(\mathrm{~K}_{\mathrm{sp}} =\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right]  \)
\(1.1 \times 10^{-12} =(2 \mathrm{~s})^{2}(0.1)  \)
\(1.1 \times 10^{-12} =0.4 \mathrm{~s}^{2}  \)
\(0.4 \mathrm{~s}^{2} =1.1 \times 10^{-12}  \)
\(\mathrm{~s}^{2} =\frac{1.1 \times 10^{-12}}{0.4}=2.75 \times 10^{-12}  \)
\(\mathrm{~s} =\sqrt{2.75 \times 10^{-12}}  \)
\(\mathrm{~s} =1.658 \times 10^{-6} \mathrm{M}\)

\(Ag_{2}CrO_{4}(s)\rightleftharpoons 2Ag^{+}_{aq}+CrO^{2-}_{4}(aq)\)
\(K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}]\)
\(=(5\times10^{-5})^2(4.4\times10^{-4})\)
=\((1.1\times10^{-12})\)

\(\mathrm{CaF}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{Ca}_{\mathrm{(aq)}}^{2+}+2 \mathrm{~F}^{-}_  {(\mathrm{aq}) }\)
                            s           2s 
\({\left[\mathrm{Ca}^{2+}\right] } =3.3 \times 10^{-4} \mathrm{M}  \)
\({\left[\mathrm{F}^{-}\right] } =2 \times 3.3 \times 10^{-4}=6.6 \times 10^{-4} \mathrm{M}  \)
\(\mathrm{K}_{\mathrm{sp}} =\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{F}^{-}\right]^{2}  \)
\(=\left(3.3 \times 10^{-4}\right)\left(6.6 \times 10^{-4}\right)^{2}=1.44 \times 10^{-10}\)

\(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \rightleftharpoons 3 \mathrm{Ca}^{2+}_{(aq)}+2 \mathrm{PO}_{4_{(aq)}}^{3-}\\\)
(s)                           (3s)             (2s)
\(K_{sp}=[Ca^{2+}]^{3}[PO_{4}^{3-}]^{2}\)
\(K_{sp}=(3s)^{3}(2s)^{2}\)
\(K_{sp}=27s^{3}.4s^{2}\)
\(K_{sp}=108s^{5}\)
(or)
\(K_{s p} =m^{m} \cdot n^{n} \cdot(s)^{m+n} \)
\(K_{\text {sp }} =3^{3} \cdot 2^{2} \cdot(s)^{3+2} \)
\(K_{s p} =27 \times 4 \times(s)^{5} \)
\(=108(s)^{5}=108 s^{5}\)

\(\mathrm{Ag}_{2} \mathrm{CrO}_{4(\mathrm{~s})} \rightleftharpoons 2 \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{CrO}_{4{(\mathrm{aq})}}^{2-}\\ \quad s \quad \quad \quad \quad \quad 2s \quad \quad \quad \quad s\)
\(\left[\mathrm{Ag}^{+}\right]=2 \mathrm{~s}+0.01 \)
\(\simeq 0.01 \)
\((\because 2 s<<0.01) \)
\(\left[\mathrm{CrO}_{4}^{2-}\right]=\mathrm{S} \)
\(\mathrm{AgNO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{NO}_{3_{(\mathrm{aq})}}^{-}\\ 0.01M \quad \quad 0.01M \quad \quad 0.01M \quad\)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right]  \)
\(1 \times 10^{-12}=(0.01)^{2}(\mathrm{~s})  \)
\(\mathrm{S}=\frac{1 \times 10^{-12}}{10^{-4}}=1 \times 10^{-8} \mathrm{M}\)

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

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