According to Henderson equation
pH = pK_a + log [Acid]/[Salt]
i.e., -log [H⁺] = - log K_a + log [Acid]/[Salt]
-log [H⁺] =  log [Acid]/[Salt] x 1/ K_a
log 1/[H⁺] = log [Acid]/[Salt] x 1/ K_a
[H⁺] = K_a  [Acid]/[Salt]

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

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

H₂O + H₂O ⇌  H₃O⁺ + OH^-
acid 1  base 1  acid 2   base 2
HF + H₂O ⇌  H₃O⁺ + F^-
acid 1  base 1  acid 2   base 2
∴ Conjugate bases are OH^- and F^- respectively.

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

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.

(i) Acidic buffer solution: a solution containing a weak acid and its salt.
Example: Solution containing acetic acid and sodium acetate
(ii) Basic buffer solution: a solution containing a weak base and its salt.
Example: Solution containing NH₄OH and NH₄Cl.

(i) Arrhenius theory does not explain the behaviour of acids and bases in non aqueous solvents such as acetone, Tetrahydrofuran etc ..
(ii) This theory does not account for the basicity of the substances like ammonia (NH₃) which do not possess hydroxyl group.

(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.

Conjugate Pairs:
\((a) \mathrm{HS}_{\text {(aq) }}^{-} \& \mathrm{H}_{2} \mathrm{~S}_{\text {(aq) }} \) \((b) \mathrm{HF}_{\text {(aq) }} \& \mathrm{~F}_{\text {(aq) }}^{-} \)
\((a) \mathrm{HPO}_{4}^{2-} \& \mathrm{PO}_{4}^{3-} \) \((b) \mathrm{SO}_{3}^{2-} \& \mathrm{HSO}_{3}^{-} \)
\((a) \mathrm{NH}_{4}^{+} \& \mathrm{NH}_{3} \) \((b) \mathrm{CO}_{3}^{2-} \& \mathrm{HCO}_{3}^{-}\)

(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) The concentration of hydronium ion in an acidic buffer solution depends on the ratio of the concentration of the weak acid to the concentration of its conjugate base present in the solution i.e.,
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] }_{ aq } }{ [{ base] }_{ aq } } \)
(ii) The weak acid is dissociated only to a small extent. Moreover, due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly, the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.
\(\left[ { H }_{ 3 }{ O }^{ + } \right] ={ K }_{ a }\frac { [{ acid] } }{ [{ salt] } } \)
(iii) Here [acid] and [salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution
Taking logarithm on both sides of the equation
\(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={ \log K }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
reverse the sign on both sides
- \(\log\left[ { H }_{ 3 }{ O }^{ + } \right] ={- \log K }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
We know that
pH = -log [H₃O⁺] and pK_a = -log K_a
\(\Rightarrow pH={ pK }_{ a }-\log\frac { [{ acid] } }{ [{ salt] } } \)
\(​​\Rightarrow pH={ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)
Similarly for a basic buffer,
pOH = \({ pK }_{ a }+\log\frac { [{ acid] } }{ [{ salt] } } \)

Let us explain the buffer action in a solution containing CH₃COOH and CH₃COONa.
The dissociation of the buffer components occurs as below.
\( \mathrm{CH}_{3} \mathrm{COOH}_{(\mathrm{sq})} \rightleftharpoons \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{H}_{3} \mathrm{O}_{(\mathrm{aq})}^{+} \)
\(\mathrm{CH}_{3} \mathrm{COONa}_{(\mathrm{s})} \stackrel{\mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{COO}_{(\mathrm{aq})}^{-}+\mathrm{Na}_{(\mathrm{aq})}^{+}\)
If an acid is added to this mixture, it will be consumed by the conjugate baseCH₃COO^- to form the undissociated weak acid i.e, the increase in the concentration of H⁺ does not reduce the pH significantly.
\({ { CH }_{ 3 }COO }_{ (aq) }^{ - }+{ H }_{ (aq) }^{ + }\rightarrow { CH }_{ 3 }{ COOH }_{ (aq) }\)
If a base is added, it will be neutralized by H₃O⁺, and the acetic acid is dissociated to maintain the equilibrium. Hence the pH is not significantly altered.

\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
The concentration of the salt and the acid should be in moles/lit.
Number of moles of acetic acid
= \(\frac{weight \ of \ acetic \ acid}{molecular \ weight \ of \ acetic \ acid}=\frac{7.5}{60}\)
Number of moles of sodium acetate
= \(\frac{weight \ of \ sodium \ acetate}{molecular \ weight \ of \ sodium acetate}=\frac{10.25}{82}\)
pK_a = -log K_a
= -Iog 1.8 x 10^-5
\(=log\frac { 1 }{ 1.8\times { 10 }^{ -5 } } \)
= log 1 - log 1.8 - log 10^-5
= 5 - 0.2553 = 4.7447
\(\therefore pH=4.7447+\log\frac { 0.12 }{ 0.08 } \)
pH = 4.7447.

(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) 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.

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

Solubility can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution.
For a solute X_mY_n,
\({ X }_{ m }{ Y }_{ n(s) }\rightleftharpoons m{ X }_{ (aq) }^{ n+ }+{ nY }_{ (aq) }^{ m- }\)
From the above stoichiometrically balanced equation we have come to know that 1 mole of X_m Y_n(s) dissociated to furnish 'm' moles of Xⁿ⁺ and 'n' moles of Y^m- if 's' is molar solubility of X_mY_n then
[Xⁿ⁺] = ms and [Y^m-] = ns
ஃK_sp = [Xⁿ⁺]^m [Y^m-]ⁿ
K_sp = (ms)^m (ns)ⁿ
K_sp = (m)^m (n)ⁿ (s)^m+n

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.
\({ Ca(OH) }_{ 2 }\overset { { H }_{ 2 }O }{ \rightleftharpoons } { Ca }_{ (aq) }^{ 2+ }+{ 2OH^- }_{ (aq) }\)

This is a buffer mixture containing a weak base and its salt. Hence the equation to be used is
\(pH={ pK }_{ a }+{ \log }\frac { [salt] }{ [acid] } \)
pK_b = logK_b
= -log 1.8 - log 10^-5 = 4.7447
∴ \(pOH=4.7447+\log\frac { 0.15 }{ 0.10 } \)
pOH = 4.7447 + log 1.5
= 4.7447 + 0.1761 = 4.9208
pH + POH = 14
pH + 4.9208 = 14
pH = 9.08

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

\(Al(OH)_{3}\rightleftharpoons Al^{3+}_{(aq)}+3OH^{-}_{(aq)}\)
\(K_{sp}=[Al^{3+}][OH^{-}]^{3}\)
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.

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

(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

Considering Ba(OH)₂ to be a strong base:
\(\text { Normality } =\text { Molarity } \times \text { Acidity }  \)
\(=1.5 \times 10^{-3} \times 2\)
\({\left[\mathrm{OH}^{-}\right] } =3 \times 10^{-3}  \)
\(\mathrm{pOH} =-\log _{10}[\mathrm{OH}^-]  \)
\(=-\log _{10}\left(3 \times 10^{-3}\right)  \)
\(=-\left[\log _{10} 3+3 \log 10\right]  \)
= 3-log 3 
= 3-0.4771 
= 2.5229 
\(\mathrm{pH} =14-\mathrm{pOH} \)
= 14-2.5229 
pH = 11.4771 = 11.48