\(K_h= { \frac { { K }_{ w } }{ K_b } } = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}}\)
= 0.55 x 10^-9 = 5.5 x 10^-10

pH = -log₁₀ [H⁺]
\(\therefore\) [H⁺] = 10^-pH
10⁰ = 1
[H⁺] = 1 M
The solution is strongly acidic

KOH → K⁺ + OH^-
10^-5M    10^-5M 10^-5M
[OH^-] = 10^-5M
pH = 14 - pOH
pH= 14-(-log [OH^-])
= 14 + log [OH^-]
= 14 + log 10^-5
= 14 - 5 = 9

x ml of 0.1 M NaOH + x mL of 0.01 M HCI
No. of moles of NaOH = 0.1 x X x 10^-3
= 0.1 X x 10^-3
No. of moles of HCI = 0.01 x X x 10^-3
= 0.01 X x 10^-3
No. of moles of NaOH after mixing
= 0.1 X x 10^-3 - 0.01 X x 10^-3
= 0.09 X x 10^-3
Concentration of NaOH\(= (\frac{0.09x \times 10^{-3}}{2x \times 10^{-3}}) = 0 .045\)
[OH^-] = 0.045
pOH =-log (4.5 x 10^-2)
= 2 -log 4.5
= 2 - 0.65 = 1.35
pH = 14 - 1.35 = 12.65

pH = -log₁₀ [H⁺]
∴  [H⁺] = 10^-pH
Let the volume be x ml.
V₁M₁ +V₂M₂ +V₃M₃ = VM
∴   x ml of 10^-1 M+ x ml of 10^-2 M + x ml of 10^-3 M
= 3 x ml of [H⁺]
\(∴  [H^+] = \frac{ x[0.1 + 0.01 + 0.001]}{3x}\)
\(= \frac{ [0.1 + 0.01 + 0.001]}{3}\)
\(= \frac{ [0.111]}{3}\)
= 0.037
= 3.7 x 10^-2

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%

HCOONa + HOH  ⇌ NaOH + H-COOH
                               strong base weak acid
Basic in nature.
C₆H₅NH₃Cl + HOH  ⇌ H₃O⁺ + C₆H₅ - NH₂ + Cl^-
                                      acidic
KCN + H - OH ⇌ KOH + HCN
Base                  strong base weak acid
basic, acidic, basic is correct.

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

Basic buffer is the solution which has weak base and its salt.
NH₄OH + HCI → NH₄CI +H₂O + NH₄OH
200 ml   100 ml    salt       100 ml  weak base

Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
Given that pH = 9
pOH = 14 - 9 = 5
[pOH = - log₁₀ [OH]]
[OH^-] = 10 ^[pOH]
[OH] = 10^-5 M
Ksp = [Ca²⁺] [OH^-]
= 10^-5/2 x (10^-5)²  = 0.5 x 10^-15

\(\mathrm{Ag}_{2} \mathrm{C_2O_4} \rightleftharpoons 2 \mathrm{Ag}_{}^{+}+\mathrm{C_2O}_{4{}}^{2-}\)
\(\left[\mathrm{Ag}^{+}\right]=2 .24 \) ×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

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

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

(i) \(\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}\left(25^{\circ} \mathrm{C}\right)\)
(ii) Ionic product of water is defined as the product of the concentration of hydronium and hydroxide ions of pure water. Its value at 25^oC is \(1 \times 10^{-14} \mathrm{~mol}^{2} \mathrm{dm}^{-2}\)

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

1. If \(\left[\mathrm{OH}^{-}\right]>1 \times 10^{-7} \mathrm{M}\), the solution is basic. \(2.5 \times 10^{-6} \mathrm{M}>1 \times 10^{-7} \mathrm{M}\)
2. \(\therefore\) The solution is basic.

(i) Lewis acid: It is a species that accepts an electron pair. Eg: \(\mathrm{Ag}^{+} ; \mathrm{BF}_{3} ; \mathrm{A} / \mathrm{Cl}_{3}\)
(ii) Lewis base: It is a species that donates an electron pair. Eg: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

(i) Ostwald's dilution law relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (α) and the concentration (c).
where  \(\alpha=\frac{\text { Number of moles dissociated }}{\text { Total number of moles }}\)
(ii) The dissociation of acetic acid can be represented as
\(\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{H}^{+}+\mathrm{CH}_{3} \mathrm{COO}^{-}\)
The dissociation constant of acetic acid is,
\({ K }_{ a }=\frac { \left[ { H }^{ + } \right] \left[ { CH }_{ 3 }COO^{ - } \right] }{ \left[ { CH }_{ 3 }COOH \right] } \)  ........(1) 

	CH3COOH	H+	CH3COO-
Initial number of moles	1	-	-
Degree of dissociation of CH3COOH	α	-	-
Number of moles at equilibrium	1-α	α	α
Equilibrium concentration	(1-α)C	αC	αC

Substituting the equilibrium concentration in equation (1)
\({ K }_{ a }=\cfrac { \left( \alpha C \right) \left( \alpha C \right) }{ \left( 1-\alpha \right) C } \) 
\({ K }_{ a }=\cfrac { { \alpha }^{ 2 }C }{ 1-\alpha } \) .......(2) 
(iii) We know that weak acid dissociates only to a very small extent compared to one, a is so small and hence in the denominator (1 - α) ⋍1. The above expression (2) now becomes,
k_a =a²C \(\Rightarrow { \alpha }^{ 2 }=\cfrac { { k }_{ a } }{ C } \) ; \(\alpha =\sqrt { \cfrac { { K }_{ a } }{ C } } \) 
(iv) When dilution increases, the degree of dissociation of weak electrolyte also increases. This is called Ostwald's dilution law
Also \(;\left[\mathrm{H}^{+}\right]=\alpha \mathrm{C}\) and \(\left[\mathrm{H}^{+}\right]=\left(\sqrt{\frac{\mathrm{K}_{\mathrm{a}}}{\mathrm{C}}}\right) \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \mathrm{C}^{2}}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{C}}\)
Similarly for a weak base
\(\begin{aligned} & \mathrm{K}_{\mathrm{b}}=\alpha^2 \mathrm{C} ; \quad \therefore \alpha=\sqrt{\frac{\mathrm{k}_{\mathrm{b}}}{\mathrm{C}}}, \\ \end{aligned}\)
\(\begin{aligned} & {\left[\mathrm{OH}^{-}\right] \alpha \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}}} \times \mathrm{C}=\sqrt{\frac{\mathrm{K}_{\mathrm{b}} \mathrm{C}^2}{\mathrm{C}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}}} \end{aligned}\)

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

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

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

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

\(h=\sqrt{K_{h}}=\sqrt{\frac{K_{w}}{K_{a}K_{b}}}=\sqrt{\frac{1\times10^{-14}}{1.8\times10^{-5}\times{1.8\times10^{-5}}}}\)
\(= \frac{1 \times10^{-7}}{1.8\times10^{-5}}\)
=\(0.7453\times10^{-2}\)
\(pH=\frac{1}{2}pK_{w}+\frac{1}{2}pK_{a}-\frac{1}{2}pK_{b}\)
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