\(\mathrm{pH} +\mathrm{pOH}=14  \)
\(\mathrm{pH} =14-\mathrm{pOH}  \)
\(=14-\left(-\log _{10}\left[\mathrm{OH}^{-}\right]\right)  \)
\(=14+\log _{10}\left[\mathrm{OH}^{-}\right] \)
\(\mathrm{pH}=14+\log \left(\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}\right)^{1 / 2}\left[\because\left(\mathrm{OH}^{-}\right)=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}\right]  \)
\(\mathrm{pH}=14+\log \left(\frac{\mathrm{K}_{w}}{\mathrm{~K}_{\mathrm{a}}} \cdot \mathrm{C}\right)^{1 / 2}\left[\because \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}}}\right] \)
\(\mathrm{pH}=14+1 / 2 \log \mathrm{K}_{\mathrm{w}}+1 / 2 \log \mathrm{C}-1 / 2 \log \mathrm{K}_{\mathrm{a}}\)
As \(1 / 2 \log \mathrm{K}_{\mathrm{w}}=1 / 2 \log 10^{-14}=-14 / 2 \log 10=-7 \quad\left[\because \mathrm{K}_{\mathrm{w}}=10^{-14}\right]\)
So eqn (4) becomes
\(\mathrm{pH}=14-7+1 / 2 \log \mathrm{C}-1 / 2 \log \mathrm{K}_{\mathrm{a}}  \)
\(\text { As }-\log \mathrm{K}_{\mathrm{a}}=\mathrm{pK}_{\mathrm{a}}  \)
\(\mathrm{pH}=7+1 / 2 \log \mathrm{C}+1 / 2 \mathrm{pK}_{\mathrm{a}}  \)
\(\mathrm{pH}=7+1 / 2 \mathrm{pK}_{\mathrm{a}}+1 / 2 \log \mathrm{C}\)
b) A strong acid and a weak base
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]  \)
\({\left[\mathrm{H}^{+}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{C}}=\sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}} \cdot \mathrm{C}}=\left(\frac{\mathrm{K}_{w} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}\right)^{1 / 2}}\)
So. Eqn (1) becomes
\(\mathrm{pH}=-\log \left(\frac{\mathrm{K}_{\mathrm{w}} \cdot \mathrm{C}}{\mathrm{K}_{\mathrm{b}}}\right)^{1 / 2} \)
\(\mathrm{pH}=-1 / 2 \log \mathrm{K}_{\mathrm{w}}-1 / 2 \log \mathrm{C}+1 / 2 \log \mathrm{K}_{\mathrm{b}}  \)
\(\text { As }-1 / 2 \log \mathrm{K}_{\mathrm{w}}=-1 / 2 \log 10^{-14}=14 / 2 \log 10=7  \)
\(1 / 2 \log \mathrm{K}_{\mathrm{b}}=-1 / 2 \mathrm{pK} \quad\left(\because-\log \mathrm{K}_{\mathrm{b}}=\mathrm{pK}_{\mathrm{b}}\right)\)
So. Eqn (3) becomes
\(\mathrm{pH}=7-1 / 2 \log \mathrm{C}-1 / 2 \mathrm{pK}_{\mathrm{b}}  \)
\(\mathrm{pH}=7-1 / 2 \mathrm{pK}_{\mathrm{b}}-1 / 2 \log \mathrm{C}\)
c) A weak acid and a weak base
\(\mathrm{pH}=-\log _{10}^{\cdot}\left[\mathrm{H}^{+}\right] \)
\(As \left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \mathrm{h} \)
\(\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \sqrt{\mathrm{K}_{\mathrm{h}}} \)
\(\left(\because h=\sqrt{\mathrm{K}_{\mathrm{h}}}\right) \)
\(\left[\mathrm{H}^{+}\right]=\mathrm{K}_{\mathrm{a}} \cdot \sqrt{\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}}} \)
\((2) \left(\because \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}}\right) \)
\(\left[\mathrm{H}^{+}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}}=\left(\frac{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\right)^{1 / 2}\)
Sub. (3) in (1)
\(\mathrm{pH}=-\log \left(\frac{\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{b}}}\right)^{1 / 2} \)
\(\mathrm{pH}=-1 / 2 \log \mathrm{K}_{\mathrm{a}}-1 / 2 \log \mathrm{K}_{\mathrm{w}}+1 / 2 \log \mathrm{K}_{\mathrm{b}}\)
As \(a) 1 / 2 \log \mathrm{K}_{\mathrm{w}}=1 / 2 \log 10^{-14}=-14 / 2 \log 10=-7\)
\(b) -1 / 2 \log \mathrm{K}_{\mathrm{a}}=1 / 2 \mathrm{pK} \)
\(c) -1 / 2 \log \mathrm{K}_{\mathrm{b}}=1 / 2 \mathrm{pK}_{\mathrm{b}} \Rightarrow 1 / 2 \log \mathrm{K}_{\mathrm{b}}=-1 / 2 \mathrm{pK}_{\mathrm{b}}\)
Eqn. (4) becomes
\(\mathrm{pH}=1 / 2 \mathrm{pK}_{\mathrm{a}}-(-7)-1 / 2 \mathrm{pK}_{\mathrm{b}}\)
(Or)
\(\mathrm{pH}=7+1 / 2 \mathrm{pK}_{\mathrm{a}}-1 / 2 \mathrm{pK}_{\mathrm{b}}\)

(i) According to Lewis (1923), an acid is a species that accepts an electron pair while base is a species that donates an electron pair. We call such species as Lewis acids and bases.
(ii) A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a anion (or) neutral molecule with at least one lone pair of electrons.
(iii) Let us consider the reaction between Boron tri fluoride and ammonia

(iv) Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond.
(v) In coordination compounds, the Ligands act as a lewis base and the central metal atom or ion that accepts a pair of electrons from the ligand behaves as a Lewis acid.

(i) According to Lewis (1923), an acid is a species that accepts an electron pair while base is a species that donates an electron pair. We call such species as Lewis acids and bases.
(ii) A Lewis acid is a positive ion (or) an electron deficient molecule and a Lewis base is a anion (or) neutral molecule with at least one lone pair of electrons.
(iii) Let us consider the reaction between Boron tri fluoride and ammonia

(iv) Here, boron has a vacant 2p orbital to accept the lone pair of electrons donated by ammonia to form a new coordinate covalent bond.
(v) In coordination compounds, the Ligands act as a lewis base and the central metal atom or ion that accepts a pair of electrons from the ligand behaves as a Lewis acid.

A relation between pH and pOH can be established using their following definitions
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \) ..........(1)
\(\mathrm{pOH} =-\log _{10}\left[\mathrm{OH}^{-}\right]\) ..........(2)
Adding equation (1) and (2)
\(\mathrm{pH}+\mathrm{pOH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\log _{10}\left[\mathrm{OH}^{-}\right] \)
\(=-\left(\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\log _{10}\left[\mathrm{OH}^{-}\right]\right)\)
\(\mathrm{pH}+\mathrm{pOH}=-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)
\([\because \log a+\log b=\log a b]\)
We know that \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\)
\(\Rightarrow \mathrm{pH}+\mathrm{pOH}=-\log _{10} \mathrm{~K}_{\mathrm{w}}\)
\(\Rightarrow \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}}\)
(3) \(\left[\because \mathrm{pK}_{\mathrm{w}}=-\operatorname{Iog}_{10} \mathrm{~K}_{\mathrm{w}}\right]\)
At 25^oC, the ionic product of water, \(\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}\)
\(\mathrm{pK}_{\mathrm{w}}=-\log _{10} 10^{-14}=14 \log _{10} 10=14 \)
\(\therefore(3) \Rightarrow \mathrm{At} 25^{\circ} \mathrm{C}, \mathrm{pH}+\mathrm{pOH}=14\)

A relation between pH and pOH can be established using their following definitions
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \) .............(1)
\(\mathrm{pOH} =-\log _{10}\left[\mathrm{OH}^{-}\right]\) ............(2)
Adding equation (1) and (2)
\(\mathrm{pH}+\mathrm{pOH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\log _{10}\left[\mathrm{OH}^{-}\right] \)
\(=-\left(\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\log _{10}\left[\mathrm{OH}^{-}\right]\right)\)
\(\mathrm{pH}+\mathrm{pOH}=-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]\)
\([\because \log a+\log b=\log a b]\)
We know that \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{\mathrm{w}}\)
\(\Rightarrow \mathrm{pH}+\mathrm{pOH}=-\log _{10} \mathrm{~K}_{\mathrm{w}}\)
\(\Rightarrow \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}}\)
(3) \(\left[\because \mathrm{pK}_{\mathrm{w}}=-\operatorname{Iog}_{10} \mathrm{~K}_{\mathrm{w}}\right]\)
At 25^oC, the ionic product of water, \(\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}\)
\(\mathrm{pK}_{\mathrm{w}}=-\log _{10} 10^{-14}=14 \log _{10} 10=14 \)
\(\therefore(3) \Rightarrow \mathrm{At} 25^{\circ} \mathrm{C}, \mathrm{pH}+\mathrm{pOH}=14\)

There are two types of buffer solutions
1) Acidic buffer solution
(i) A solution containing a weak acid and its salt.
(ii) Example: Solution containing acetic acid and sodium acetate.
2) Basic buffer solution
(i) A solution containing a weak base and its salt.
(ii) Example: Solution containing NH₄OH and NH₄Cl

(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⁺]
\(={ \left( \frac { { K }_{ w }.C }{ { K }_{ b } } \right) }^{ \frac { 1 }{ 2 } }\)
= \(-\frac { 1 }{ 2 } \log { K }_{ w }-\frac { 1 }{ 2 } \log C+\frac { 1 }{ 2 } \log{ K }_{ b }\)
\(pH=7-\frac { 1 }{ 2 } p{ K }_{ b }-\frac { 1 }{ 2 } \log C\)

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

Buffer index β, as a quantitative measure of the buffer capacity. It is defined as the number of gram equivalents of acid or base added to 1litre of the buffer solution to change its pH by unity.
\(\beta =\frac { dB }{ d(pH) } \)
Here,
dB = number of gram equivalents of acid / base added to one litre of buffer solution
d(pH) = The change in the pH after the addition of acid / base.

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