Addition of salt KY (having a common ion Y^-) decreases the solubility of MY and NY₃ due to common ion effect.
Option (a) and (b) are wrong
For salt MY, MY ⇌ M⁺ + Y^-
K_sp = (s) (s)
6.2 × 10^-13 = s²
\(\therefore\) s = \(\sqrt{6.2 \times 10^{-13}} = 10^{-7}\)
For salt NY₃,
NY₃ ⇌ N³⁺ + 3Y^-
K_sp = (s) (3s)³
K_sp = 27s⁴
\(s = (\frac{6.2 \times 10^{-13}}{27})^{1/4}\)
s = 10^-4
The molar solubility of MY in water is less than of NY₃

C₅H₅N + H-OH  ⇌  C₅H₅ ⁺NH + OH^-
\(\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =K_b\)
\(\alpha\)²C \(\approx\) K_b
\(\alpha = \sqrt {\frac{K_b} C} = \sqrt {\frac{1.7 \times 10^{-9}} {0.1}}\)
\(= \sqrt{1.7} \times 10^{-4}\)
Percentage of dissociation =\(= \sqrt{1.7} \times 10^{-4}\) x 100
= 1.3 x 10^-2 = 0.013%

BF₃ → electron deficient →  Lewis acid
PF₃ → electron rich →  Lewis base
CF₄ → neutral →  neither Lewis acid nor base
SiF₄ → neutral →  neither Lewis acid nor 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

Common ion effect decreases the solubility of the electrolyte.

pK_w = pH + pOH = 10^-12
For a neutral solution, pH = pOH
So pH = pOH = 6.

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
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

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co - efficient in a balanced equilibrium equation.
X_mY_n (s) \(\rightleftharpoons\) mXⁿ⁺ _(aq) + nY^m-_(aq)
K_sp=[Xⁿ⁺]^m[Y^m-]ⁿ

[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

NH₄CI is a salt of a strong acid HCI and weak base NH₄OH.
\({ HCl }_{ (aq) }+{ NH }_{ 4 }OH_{(aq)}\rightleftharpoons { { NH }_{ 4 }Cl }_{ (aq) }+{ H }_{ 2 }O(I)\)
\({ NH }_{ 4 }^{ + }\) 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 shown below.
\({ NH }_{ 4 }^{ + }+{ H }_{ 2 }O(1)\rightleftharpoons { NH }_{ 4 }{ { { OH }_{ (aq) }+ }H }_{ (aq) }^{ + }\)
There is no such tendency shown by Cl^- and therefore [H⁺] > [OH^-]; the solution is acidic and the pH is less than 7.
As discussed in the salt hydrolysis of strong base and weak acid. In this case also, we can establish a relationship between the K_h and K_b as 
K_h·K_b = K_w
Let us calculate the K_h value in terms of degree of hydrolysis (h) and the concentration of salt
K_b = h²C and \([{ H }^{ + }]=\sqrt { { K }_{ h }.C } \)
= \({ [H }^{ + }]=\sqrt { \frac { { K }_{ w } }{ { K }_{ b } } .C } \)
pH = - log [H⁺]
\(={ \left( \frac { { K }_{ w }.C }{ { K }_{ b } } \right) }^{ \frac { 1 }{ 2 } }\)
= \(-\frac { 1 }{ 2 } \log { K }_{ w }-\frac { 1 }{ 2 } \log C+\frac { 1 }{ 2 } \log{ K }_{ b }\)
\(pH=7-\frac { 1 }{ 2 } p{ K }_{ b }-\frac { 1 }{ 2 } \log C\)

Pure water itself has a little tendency to dissociate. i.e, one water molecule donates a proton to an another water molecule. This is known as auto ionisation of water and it is represented as below.

(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

\(pH=pK_{a}+\log[\frac{salt}{acid}]\)
Given that K_a = \(1.8\times10^{-5}\)
\(\therefore pK_{a}=-\log(1.8\times10^{-5})\)
= 5 - log 1.8
= 5 - 0.26
= 4.74
\(\therefore pH=4.74+\log\frac{0.20}{0.18}\)
= 4.74 + log(10/9)
= 4.74 + log10 - log9
= 4.74 + 1 - 0.95
= 5.74 - 0.95
= 4.79

\(h=\sqrt{K_{h}}=\sqrt{\frac{K_{w}}{K_{a}K_{b}}}=\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times{1.8\times10^{-5}}}}\)
\(= \frac{1 \times10^{-7}}{1.8\times10^{-5}}\)
=\(0.7453\times10^{-2}\)
\(pH=\frac{1}{2}pK_{w}+\frac{1}{2}pK_{a}-\frac{1}{2}pK_{b}\)
Given that \(K_{a}=K_{b}=1.8\times10^{-5}\)
if K_a = K_b, then, pK_a = pK_b 
\(\therefore pH= \frac{1}{2}pK_{w}=\frac{1}{2}(14)=7\)
pH = 7

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

pH + pOH = 14.0
∴ pOH = 14.0 - pH
= 14.0 - 5.5 = 8.50
pOH = 8.5 ∴ antilog [-pOH]|
∴  [OH^-] = antilog [-8.5] = 3.2 x 10^-9 M.

Degree if dissociation \(\alpha =\sqrt { \frac { { K }_{ B } }{ C } , } C\alpha =\sqrt { { K }_{ a }.C } \)
∴ \(pH=\log\frac { 1 }{ \sqrt { { K }_{ a }.C } } =\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -5 }\times .1 } } \)
= \(\log\frac { 1 }{ \sqrt { 1.8\times { 10 }^{ -3 } } } =\log\frac { { 10 }^{ 3 } }{ \sqrt { 1.8 } } \)
\(=3-\frac { 1 }{ 2 } \log1.8\)
\(=3-\frac { 1 }{ 2 } \times 0.2553=3-0.1276=2.8724\)
Alternating,
\(\left[ { H }^{ + } \right] =\sqrt { { K }_{ a }\times C } \)
\(=\sqrt { 1.8\times { 10 }^{ -5 }\times 0.1 } =\sqrt { 1.8\times { 10 }^{ -6 } } \)
= 1.341 x 10^-3
∴ -log [H⁺] = -log (1.341 x 10^-3)
= 3 - 0.1277 = 2.8723.