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

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

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

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

FeCl₃ when dissolved in water gives strong acid HCl and a weak base Fe(OH)₃ A strong acid ionises completely so concentration of H⁺ ion becomes much greater than OH^- ion. So, aqueous solution of FeCl₃ is acidic.

According to Lewis concept of Acid and bases, any species capable of accepting an electron pair is an acid. BF₃ is electron deficient so accepts a pair of electron, Hence termed as acid

Magnesium has to precipitate as Mg(OH)₂ when treated with NH₄OH, but addition of NH₄CI will suppress the ionisation of NH₄OH due to common ion effect. So ionic product of Mg²⁺ and OH^- ions will be less than the solubility product (K_sp) of Mg(OH)₂. Hence will not precipitate.

When an acid reacts with a base, a salt and water are formed and the reaction is called neutralization.

Buffer is a solution which consists of a mixture of a weak acid and its conjugate base (or) a weak base and its conjugate acid.

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

Given that H₃O⁺ = \(2\times10^{-3}M\)
\(K_{w}=[H_{3}O^{+}][OH^{-}]\)
\(\therefore [OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1\times10^{-14}}{2\times10^{-3}}=0.5\times10^{-11}M\)
\(2\times10^{-3} >>0.5\times10^{-11}\)
i.e., [H₃O⁺]>>[OH^-], hence the juice is acidic in nature

[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.

(i)  According to Lewis theory a species that donates a pair of electron is called Lewis base.
(ii) Nitrogen more in NH₃ is less electro negative than oxygen in water. So the non - bonded electron pair on nitrogen is more available for sharing than a non - bonded electron pair on oxygen atom. So NH₃ is a stronger lewis base than H₂O.

(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: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

Let us find a relation between the equilibrium constant for the hydrolysis reaction (hydrolysis constant) and the dissociation constant of the acid.
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }][{ H }_{ 2 }O] } \)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }COOH][{ OH }^{ - }] }{ [{ CH }_{ 3 }{ COO }^{ - }] } \)  ...(1)
\({ CH }_{ 3 }{ COONH }_{ (aq) }\rightleftharpoons { C }{ H }_{ 3 }COO_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\)
\({ K }_{ h }=\frac { [{ CH }_{ 3 }CO{ O }^{ - }][{ H }^{ + }] }{ [{ CH }_{ 3 }{ COO }H] } \)  ...(2)
(1) x (2)
⇒ K_b . K_a = [H⁺][OH^-]
we know that [H⁺] [OH^-] = K_w
K_h· K_a = K_w
K_h value in terms of degree of hydrolysis (h) and the concentration of salt (C) for the equilibrium can be obtained as in the case of Ostwald's dilution law. K_h = h²C. and i.e [OH^-] = \(\sqrt { { K }_{ h }.C } \)
pH of salt solution in terms of K_a and the concentration of the electrolyte
pH + pOH = 14
pH = 14 - pOH = 14 - {-log [OH^-]}
= 14 + log [OH^-]
∴ pH = 14 + log (K_hC)^\(\frac12\)
pH =14 + log \({ \left( \frac { { K }_{ w }C }{ { K }_{ a } } \right) }^{ \frac { 1 }{ 2 } }\)
pH = 14 +  (\(\frac12\) log K_w + \(\frac12\) log C - \(\frac12\) log K_a)
[∴ K_w = 10^-14]
\(pH=14-7+\frac { 1 }{ 2 } \log \ C+\frac { 1 }{ 2 } p{ K }_{ a }\frac { 1 }{ 2 } \log{ K }_{ w }=\frac { 1 }{ 2 } \times { \log10 }^{ -14 }=\frac { -14 }{ 2 } (1)=-7\)
\(pH=7+\frac { 1 }{ 2 } { pK }_{ a }+\frac { 1 }{ 2 } \log \ C\)   [-log K_a = pK_a]

(NaOH strong base fully ionised)
\(pOH=\log\frac { 1 }{ { OH }^{ - } } =\log\frac { 1 }{ .01 } \)
\(=\log\frac { 100 }{ 1 } =2.0\)
pOH = 14 - pOH = 14 - 2 = 12.

\(N{ H }_{ 4 }\rightleftharpoons { N{ H }_{ 4 } }^{ + }+{ OH }^{ - }\)
If α is the degree of dissociation, and C is the concentration in moles/lit. then
\({ K }_{ b }=\frac { { \alpha }^{ 2 }{ C } }{ 1-\alpha } =\frac { 0.04\times 0.042\times 0.01 }{ -0.042 } \)
K_b = 1.84 x 10^-5.

pH = - log [H₃O⁺]
(i) pH - log [10⁴]
pH = log 1 - log 10⁴
pH =  - 4
(ii) pH = - log [10^-7]
pH = log 1 - log 10^-7.
pH = 7
(iii) pH = - log [H₃O⁺]
pH = - log [6.8 x 10^-3]
pH = log 1 - log 6.8 - log 10^-3
= 3 - 0.8325
pH = 2.17 (or) 2.2
(iv) pH = - log [3.2 x 10^-5]
pH = log 1 - log 3.2 - log 10^-5
= 5 - 0.5051
pH = 4.49 (or) 4.5
(v) pH = - log [0.035]
pH = log 1 - log 0.035
= 2 - 0.5441
pH = 1.46 (or) 1.5
(vi) pH = - log [0.25]
pH = log 1 - log 0.25
= 1 - 0.3979
pH = 0.602 (or) 0.60
(vii) pH = -log [H₃O⁺]
pH = log 1 - log 5.4 - log 10^-9
= 7 - 0.7324
pH = 8.267 (or) 8.3
(viii) pH = - log [7.1 x 10^-7]
pH = log 1 - log 7.1 - log 10^-7
= 7 - 0.8513
pH = 6.2.

(i) The dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.
(ii) Consider the dissociation of a weak acid, acetic acid whose ionisation is incomplete.
CH₃COOH_(aq) ⇌ H⁺_(aq) + CH₃COO^-_(aq)
(iii) If the salt sodium acetate with common ion CH₃COO^- is added to the above equilibrium, it dissociates completely increasing.
CH₃COONa_(aq)⟶Na⁺_(aq) + CH₃COO^-_(aq)
(iv) Hence, the overall concentration of CH₃COO^- is increased, and the acid dissociation equilibrium is disturbed.
(v) So, in order to maintain the equilibrium, the excess CH₃COO^- ions combines with H⁺ ions to produce much more unionized CH₃COOH i.e, the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed.

(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 \(pH={ pK }_{ b }+\log\frac { [salt] }{ [base] } \)

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.

According to Arrhenius, an acid is a substance that dissociates to give hydrogen ions in water. For example, HCl, H₂SO₄ etc., are acids. Their dissociation in aqueous solution is expressed as
\({ HCl }_{ (g) }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { H }_{ (aq) }^{ + }+{ Cl }_{ (aq) }^{ - }\)
Similarly a base is a substance that dissociates to give hydroxyl ions in water. For example, substances like NaOH, Ca(OH)² etc., are bases.
\({ Ca(OH) }_{ 2 }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { Ca }_{ (aq) }^{ 2+ }+{ 2OH^- }_{ (aq) }\)

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

\(pH=pK_{a}+\log[\frac{salt}{acid}]\)
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

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

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

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

(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⁺]

\(\mathrm{M} =\frac{\mathrm{V}_{1} \mathrm{M}_{1}-\mathrm{V}_{2} \mathrm{M}_{2}}{\mathrm{~V}_{1}+\mathrm{V}_{2}}  \)
\(=\frac{(50 \times 0.05)-(50 \times 0.025)}{50+50}  \)
\(\text { Molarity }=\frac{\text { Number of millimoles }}{\mathrm{V}_{\mathrm{m} l}}\)
\(=\frac{2.5-1.25}{100}=\frac{1.25}{100}=0.0125 \mathrm{M}  \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}^{+}\right]  \)
\(=-\log _{10} 0.0125  \)
Normality = Molarity \(\times\) basicity
\(=-\log _{10}\left(1.25 \times 10^{-2}\right)  \)
\(=-\left[\log _{10} 1.25-2 \log _{10} 10\right]  \)
\(=2-\log _{10} 1.25=2-0.0969 \)
pH = 1.9031