PbI_2(s) ⇌  Pb²⁺_(aq)  + 2I^-_(aq) 
K_sp = (s) (2s)²
3.2 × 10^-8 = 4s³
s = (3.2 × 10^-8/4)^1/3
= ( 8 x  10^-9)^1/3
= 2 x 10^-3 M

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CO → having lone pair of electron →  Lewis base
F → unshared pair of electron →  Lewis base

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×10^-4mol L^-1
\(\mathrm{C_2O}_{4{}}^{2-} = \frac {2.24 \times 10 ^{-4}}{2}\) mol L^-1
= 1.12 ×10^-4mol L^-1
K_sp = [Ag]² [C₂O₄^2-]
= (2.24 ×10^-4mol L^-1) (1.12 ×10^-4mol L^-1)
= 5.619 × 10^-12mol³L^-3

(i) To resist changes in its pH on the addition of an acid (or) a base, the buffer solution should contain both acidic as well as basic components so as to neutralize the effect of added acid (or) base and at the same time, these components should not consume each other.
(ii) Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa.
(iii) The dissociation of the buffer components occurs as below.
\(\mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{aq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Ha}_{(\mathrm{aq})}^{+}\)

Henderson equation is an equation which is used to determine the pH of an acid buffer with the help of the dissociation constant K_a of the weak acid and concentration of the acid and the salt used.
\(\boxed { pH={ pK }_{ a }+\log\frac { [salt] }{ [acid] } }\)
For a basic buffer \(pH={ pK }_{ b }+\log\frac { [salt] }{ [base] } \)

If we do not consider [H₃O]⁺ from the ionisation of H₂O,
then [H₃O⁺] = [HCl] = 10^-7M
i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7M) Hence, the H₃O⁺ (10^-7M) formed due to the auto ionisation of water cannot be neglected.
so, in this case we should consider  [H₃O⁺] from ionisation of H₂O
[H₃O⁺] = 10^-7 (from HCl) + 10^-7 (from water)
= 10^-7 (1+1)
= \(2\times10^{-7}\)
pH = -log₁₀[H₃O⁺]
=\(-\log_{10}(2\times10^{-7})=-[\log2+\log_{10}10^{-7}]\)
=\(-\log2-(-7)\log_{10}^{10}\)
= 7-log²
= 7-0.3010 = 0.6990 = 6.70
= 6.70

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

(i) Let us consider the reactions between a strong acid, HCI, and a weak base, NH₄OH, to produce a salt, NH₄CI, and water.
HCI_(aq) + NH₄OH_(aq) ⇌ NH₄CI_(aq) + H₂O _(I)
NH₄CI_(aq) ⟶ NH₄⁺ +CI^-_(aq)
(ii) NH₄⁺ is a strong conjugate acid of the weak base NH₄OH and it has a tendency to react with OH^- from water to produce unionised NH₄OH
NH₄⁺ _(aq) + H₂O_(I) ⇌ NH₄OH_(aq) + H⁺_(aq)
(iii) There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
(iv) The K_h and K_b are related by
\(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\)
(Or)
\(\mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\)
Degree of hydrolysis (h)
\(\underset{(1-h)}{\mathrm{NH}_{4}^{+}(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(j)} \rightleftharpoons \underset{h} {\mathrm{NH}_{4} \mathrm{OH}_{(\mathrm{aq})}} +\mathrm{H}^{+} \underset{h}{(\mathrm{aq})}\\ \)
\(\mathrm{K}_{\mathrm{h}} =\frac{\left[\mathrm{NH}_{4} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \\ \)
\(=\frac{\mathrm{hc} \times \mathrm{h}}{(1-\mathrm{h}) \mathrm{c}} \)
\(K_{h}=\frac{h^{2} c}{(1-h)} \)
\(\text{If } \mathrm{h}<<1 ; \mathrm{K}_{\mathrm{h}} \simeq \mathrm{h}^{2} \mathrm{c} \)
\(h^{2}=\frac{K_{h}}{c}\)
\(h=\sqrt{\frac{K_{h}}{c}} \ (or) \ h=\sqrt{\frac{K_{w}}{K_{b} \cdot C}} \quad\left(\because K_{h}=\frac{K_{w}}{K_{b}}\right)\)
Also; \(\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}} \) (or) \(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}}\)
pH = -log [H⁺]

HCN is a weak acid
\({\left[\mathrm{H}^{+}\right] } =\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}  \)
\(=\sqrt{10^{-9} \times 0.4}  \)
\(=\sqrt{4 \times 10^{-10}}   \)
\(=2 \times 10^{-5} \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10}\left(2 \times 10^{-5}\right)  \)
\(=-\left[\log _{10} 2-5 \log 10\right]  \)
\(=5-\log 2\)
= 5-0.3010 = 4.6990

\(\text { Normality }=\text { Molarity } \times \text { Basicity }  \)
\(=0.04 \times 1  \)
\({\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}=0.04=4 \times 10^{-2} } \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \)
\(=-\log \left[4 \times 10^{-2}\right]  \)               log₁₀ 10 =1
\(=-\left[\log _{10} 4+\log _{10} 10^{-2}\right]  \)
\(=-\left[\log _{10} 4-2 \log _{10} 10\right]=2-\log _{10} 4  \)
= 2 - 0.6021 
= 1.3979 \(\simeq\) 1.40

Ostwald's dilution law is applicable only for weak electrolyte and does not hold good for concentrated solution.

Common ion effect decreases the solubility of the electrolyte.

Given that \(\alpha=1.20\)%=\(\frac{1.20}{100}\times1.2\times10^{-2}\)
\(K_{a}=\alpha^{2}c\)
\(=(1.2\times10^{-2})^{2}(0.1)=1.44\times10^{-4}\times10^{-1}\)
=\(1.44\times10^{-5}\)

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = -log₁₀ [H₃O⁺]
= -log₁₀^(0.001)
= -log₁₀(10^-3) = 3

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