Ionization Of Acids And Bases

THEORY OF ACID AND BASE

Classical concept of Acids and Bases: According to this concept.

Acid: Acid is a substance which in aqueous solution have the following property.

(i) has a sour test.

(ii) react with active metal to give H₂.

(iii) conduct electricity.

(iv) turns blue litmus red.

(v) react with base and loose its property.

Base: Base is a substance which in aqueous solution gives the following property.

(i) has a bitter test.

(ii) have soapy touch.

(iii) conduct electricity.

(iv) turns red litmus blue.

(v) react with acids and loose its property.

The above definition is based upon certain operations (tests). This definition is replaced by conceptual definition.

Arrhenious concept: According to Arrhenious

Acid: An acid is a substance which gives hydrogen ion (H⁺) in water.

Example HCl is an acid as it gives H⁺ in water.

$HCl+aq.\to H^{+}(aq)+C{l}^{-}(aq)$

Example: Substance like H₂SO₄, HCl, HNO₃ etc. when dissolved in water dissociate completely are called strong acids.

Substance like CH₃COOH, HCN, H₃PO₄, H₂CO₃ etc. when dissolved in water dissociate to a lesser extent called weak acids.

$C{H}_{3}COOH+aq\rightleftharpoons C{H}_{3}CO{O}^{-}(aq)+H^{+}(aq)$

Base: A base is a substance which gives hydroxyl ion (OH^–) in water.

Example: Substance like NaOH, KOH etc. when dissolved in water dissociate completely are called strong bases.

$NaOH+aq\to N{a}^{+}(aq)+O{H}^{-}(aq)$

Neutralisation reaction involves the combination of H⁺ ion from acid and OH^– ion from base.

$H^{+}(aq)+C{l}^{-}(aq)+N{a}^{+}(aq)+O{H}^{-}(aq)\to N{a}^{+}(aq)+C{l}^{-}(aq)+H_{2}O(\ell )$

i.e. $H^{+}(aq)+O{H}^{-}(aq)\to H_{2}O(\ell )$

Limitations of Arrhenious concept

(i) Nature of H⁺ and OH^– ion

In aqueous solutions H⁺ and OH^– ions can not exist as such but exist as hydrated ions as H⁺(aq) and OH^–(aq).

(ii) It is failed to explain the acidic and basic character of certain substances which do not contain H⁺ and OH^– ions.

e.g. CO₂, SO₂, SO₃ etc. are acids and NH₃, Na₂CO₃, CaO etc. are bases.

(iii) It failed to explain the reaction between an acid and base in absence of water.

e.g. $CaO(s)+S{O}_3(g)\to CaS{O}_4(s)$

$N{H}_3(g)+HCl(g)\to N{H}_{4}Cl(s)$

(iv) This theory is limited to aqueous solution only.

Bronsted Lowry concept (proton donor acceptor concept):

According to Bronsted Lowry,

Acid: Acid is a substance that donates proton (H⁺).

Base: Base is a substance that accepts proton (H⁺).

An acid after loosing H⁺, the residual part becomes a conjugated base. Similarly a base after accepting H⁺ becomes a conjugate acid.

$Acid\rightleftharpoons H^{+}+\text{conjugated}\text{base}$

$Base+H^{+}\rightleftharpoons conjugated\text{acid}$

If an acid is strong then its conjugated base will be weak and vice versa.

Water can act both as acid as well as base. Hence they are amphiprotic. Similarly $HC{O}_3^{-},H_{2}P{O}_4^{-},$ $HP{O}_4^{2-},H_{2}P{O}_3^{-},HS{O}_3^{-}$ are amphiprotic or amphoteric.

e.g. $HC{O}_3^{-}+H_{2}O\rightleftharpoons C{O}_3^{2-}+H_3{O}^{+}$

$HC{O}_3^{-}+H_{2}O\rightleftharpoons H_{2}C{O}_3+O{H}^{-}$

Advantages of Bronsted Lowry concept

(i) This theory includes the ionic species which act as cation or anion.

(ii) This theory explains the acid-base reaction involving in non aqueous solution.

(iii) It can explain the nature of basic oxide like Na₂O, CaO which do not contain H⁺ and OH^– ions.

Limitations

(i) It can not explain the acidic character of substance like BF₃, AlCl₃ etc. which do not have any H⁺ but known to be acids.

(ii) It can not explain the reaction between the acidic oxides like CO₂, SO₂, SO₃ etc. and basic oxides like CaO, BaO, MgO etc. taking place in absence of solvent.

$CaO+S{O}_3\to CaS{O}_4$

Illustration 1. Which is the stronger base towards a proton $N{H}_2^{-}$ or $P{H}_2^{-}$ and why?

Solution: Bond energy (N—H > P—H) ionisation suggests that $N{H}_2^{-}$ will be stronger base. This is constituent with the relative strengths of the respective conjugate acids: NH₃ < PH₃.

Illustration 2. H₃BO₃ is ……………….acid

(A) Monobasic (B) Dibasic

(C) Tribasic (D) None

Solution: H₃BO₃ is monobasic acid

H₃BO₃ + H₂O$\to$ B(OH)₄^– + H⁺

Hence, (A) is correct.

Lewis concept of acids and base: According to this concept

Acid is a substance which accept a pair of electrons.

Base is a substance which donate a pair of electrons.

In simple way; Acid – electron pair acceptor and Base – electron pair donor

All electrophiles are Lewis acids e.g. AlCl₃, BF₃, H⁺ etc.

All nucleophiles are Lewis base e.g.$\stackrel{\bullet \bullet }{N}{H}_3,$ H₂O, R–NH₂, R–OH, F^–, OH^–, Br^– etc.

Example: BF₃ is an acid as it can accept an electron pair and NH₃ is a base as it donate an electron pair.

The neutralisation reaction involves the donation of electron pair from base to acid resulting the formation of coordinate bond.

$B{F}_3+\stackrel{\bullet \bullet }{N}{H}_3\to F_{3}B\leftarrow N{H}_3$

Type of Lewis acids

(i) Molecules having incomplete octet of central atom e.g. BF₃, AlCl₃, MgCl₂, BCl₃ etc.

(ii) Simple cations Ca²⁺, Ag⁺, H⁺, Fe³⁺ etc.

(iii) Molecules having vacant d-orbitals of central atom e.g. SnCl₄, PF₅, SiF₄, RX₃, TiCl₄ etc.

(iv) Molecules containing multiple bond between the two atoms of different electronegativity.

$O=C=O\rightleftharpoons O=\stackrel{+\delta }{C}=\stackrel{-\delta }{O}$

Type of Lewis bases

(i) Neutral molecules like H₂O, NH₃, R–OH, R–NH₂ etc. having atleast one lone pair of electrons.

(ii) All negative ions like OH^–, Cl^–, Br^–,$N{H}_2^{-}$etc.

Advantages

Lewis concept is most general idea out of all the concept which explain acidic, basic nature of oxides, and other substances which do not contain H⁺ and OH^– ions etc. But it has following limitations.

Limitations of Lewis concept

(i) Lewis concept can not explain the strength of acids and bases.

(ii) In acid base reaction it involves the formation of coordinate bond which is not always true e.g. HCl and H₂SO₄ do not contain coordinate bond.

(iii) Acid-base reactions are very fast but formation of coordinate bond is slow.

Illustration 3. Which salt undergoes hydrolysis?

(A) CH₃COONa (B) KNO₃

(C) NaCl (D) K₂SO₄

Solution: Salt of strong acid and strong base does not undergo hydrolysis.

Hence, (A) is correct.

RELATIVE STRENGTH OF WEAK ACIDS AND BASES

The relative strength of weak acids and bases are generally determined by their dissociation constant K_a and K_b respectively.

For weak acid CH₃COOH:

$\begin{gathered} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}COOH + aq \stackrel{k_a}\rightleftharpoons C{H_3}CO{O^ - }(aq) + {H^ + }(aq) \hfill \\ {\text{Initially}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \hfill \\ {\text{At}}\,\,{\text{equilibrium}}\,\,\,\,\,C - C\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C\alpha \hfill \\ \end{gathered} $

$K_a=\frac{C\alpha \times C\alpha }{C-C\alpha }$ $=\frac{C{\alpha }^2}{1-\alpha }$

($1-\alpha \approx 1$assuming $\alpha$ is very small).

$\therefore K_a=C{\alpha }^2$

Where K_a is equilibrium constant of weak acid which depends on temperature. Greater the value of K_a, more is the strength of an acid.

Similarly for base NH₄OH

$K_b=C{\alpha }^2$

where K_b is the equilibrium constant for weak base.

For two acids having equi-molar concentrations

$\frac{\text{Strength}\text{of}\text{acid}\text{I}}{\text{Strength}\text{of}\text{acid}\text{II}}=\sqrt{\frac{K_{a_1}}{K_{a_2}}}$

Similarly for two base

Similarly for two base

$\frac{\text{Strength}\text{of}base\text{I}}{\text{Strength}\text{of}base\text{II}}=\sqrt{\frac{K_{b_1}}{K_{b_2}}}$

SELF IONIZATION OF WATER (IONIC PRODUCT OF WATER)

Pure water is a weak electrolyte and itself ionises as

$H_{2}O+H_{2}O\rightleftharpoons H_3{O}^{+}+O{H}^{-}$

or simply

$H_{2}O{H}^{+}+O{H}^{-}$

Applying law of mass action

$K=\frac{[H^{+}][O{H}^{-}]}{[H_{2}O]}$

or $K=\frac{[H_3{O}^{+}][O{H}^{-}]}{{[H_{2}O]}^2}$

where K is called dissociation constant.

Since the ionisation of water is very small it means very few H₂O molecule are dissociated into H and OH^– ions i.e.

[H₂O] = constant $=\frac{1000}{18}=55.55\text{mole/litre}$

At 25^oC, the value of K is $\frac{{10}^{-14}}{55.55}$

$K\times [H_{2}O]=[H^{+}][O{H}^{-}]=\frac{{10}^{-14}}{55.55}\times 55.55$

or $K_w=[H^{+}][O{H}^{-}]={10}^{-14}$

K_w is called ionic product of water.

pH SCALE

Sorenson gave the best way of expressing the acidic or basic nature of solution. This is a logarithmic scale in which hydrogen ion concentration ranges from 10^–14 to 1 mole/litre.

pH may be defined as logarithm of reciprocal of [H₃O⁺].

or negative logarithm of hydronium ion concentration.

pH = –log [H₃O⁺]

pH = – log [H⁺]

The pH range is taken from 0 to 14. The acidity or alkalinity of a solution can be represented on pH scale as

In similar manner, we can define pOH scale as

pOH = – log [OH^–]

Both pH and pOH scale are related as

[H⁺] [OH^–] = 10^–14 at 25^oC

–log [H⁺] – log [OH^–] = 14

pH + pOH = 14

Limitations of pH scale

(i) pH value of solutions do not give immediate idea of their relative strength.

(ii) pH value can be negative too.

(iii) A 10^–8 M solution of acid can not have pH = 8.

Since the value of pH of acid may close to 7 but <7.

Illustration 4. Which of the following concentration has the largest degree of dissociation for a weak acid?

(A) 1.0 M (B) 0.5 M

(C) 0.10 M (D) 0.01 M

Solution: (D)

Illustration 5. Which of the following solutions can be titrated with HCl as well as NaOH?

(A) Glycine (B) Pyruvic acid

(C) Triethyl amine (D) Aniline

Solution: (A)

Determination of pH of acids and bases

(A) Strong Acid

A strong acid is defined as a substance which completely dissociates to give all the maximum possible H⁺ ions that it is capable of giving.

For example

HCl $\to$ H⁺ + Cl^–

Therefore [HCl] = [H⁺]

If we have to find the pH of 10^–4 M HCl,

[HCl] = [H⁺]

$\therefore$ pH = –log 10^–4 = 4.

Now to find the pH of 10^–7 M HCl, by applying the above logic, the pH would come out to be 7. This is not possible since the solution of HCl is acidic and it should have pH less than 7. The error that is made here is that, [H⁺] which comes from acid is being considered, whereas the H⁺ from acid and water should be taken into account.

Assuming that a is the amount of H⁺ (or OH^–) coming from water. In the case of pure water

a x a = 10^–14 ---------------------- (1)

In the case of 10^–7 M HCl, the total [H⁺] will be (10^–7+a) if a is the amount of H⁺ coming from water.

Since [OH^–] = a'.

[H⁺]_T x [OH^–]_T = (10^–7 + a') x a' = 10^–14 ---------------------- (2)

Where [H⁺]_T is the total H⁺ coming from both acid and water.

Therefore a' (the [H⁺] coming from water in the presence of 10^–7 M HCl) can be calculated from equation (2) and therefore the total [H⁺] would be equal to [H⁺]_T = a'+10^–7 and pH = –log [H⁺]_T

Note: In order to figure out when to take the contribution of water, it should be noted that when [H⁺]_A $⩾$ 10^–6 M, water contribution need not be taken and when [H⁺]_A < 10^–6 M, water contribution should be taken.

(B) Weak Acids

By definition weak acids are acids which dissociate weakly or feebly in water. For example CH₃COOH $\rightleftharpoons$ CH₃COO^– + H⁺. The equilibrium constant is represented as K_a.

$K_a=\frac{[C{H}_{3}CO{O}^{-}][H^{+}]}{C{H}_{3}COOH}$

If we start with `C' moles of acetic acid and $\alpha$ is the degree of dissociation which is defined as the number of moles of acetic acid which dissociates from one mole of acetic acid, then the no. of moles dissociated would be `C$\alpha$'

CH₃COOH $\rightleftharpoons$ CH₃COO^– + H⁺

Initial C 0 0

At eqb. C–$\alpha$ C$\alpha$ C$\alpha$

$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$=

$\frac{C\alpha \times C\alpha }{C(1-\alpha )}=\frac{C{\alpha }^2}{1-\alpha }$

Generally, for weak acids `$\alpha$' is very small and to a reasonable approximation we can neglect $\alpha$ in comparison to 1.

K_a = C$\alpha$²;

$\Rightarrow$

Where [H⁺]_A is the [H⁺] coming from the weak acid. Again if [H⁺]_A <10^–6 M, we must take into account H⁺ coming from water.

Let us follow the above discussion and calculate the pH of 10^–6 M CH₃COOH with

K_a = 1.8 x 10^–5. Following the above procedure

[H⁺]_A = $\sqrt{1.8\times {10}^{-5}\times {10}^{-6}}$ = 4.24 x 10^–6

$\Rightarrow$pH = –log 4.24 x 10^–6

= 6–log 4.24 = 6–0.6274 = 5.37

(C) Strong Acid + Weak Acid

Let us assume that we have a strong acid HA and a weak acid HB. Let the concentration of HA and HB be C₁ and C₂ respectively and let the dissociation constant of HB be K_a.

HA $\to$ H⁺ + A^–

HB $\rightleftharpoons$ H⁺ + B^–

C₂(1–$\alpha$) C_2$\alpha$+C₁ C_2$\alpha$

[$\alpha$ is the degree of dissociation of HB in the presence of HA

[HB]=C₂ (1–$\alpha$); [B^–] = C_2$\alpha$; [H⁺]_T = C_2$\alpha$ + C₁

Therefore K_a = $\frac{{[H^{+}]}_T[B^{-}]}{[HB]}$= $\frac{\left(C_2\alpha +C_1\right)C_2\alpha }{C_2\left(1-\alpha \right)}$

Solving this equation we can get the value of $\alpha$ and then determine the [H⁺] and find the pH.

(D) Two Weak Acids

If we have two weak acids, HA and HB with concentrations C₁ & C₂ and dissociation constants Ka₁ and Ka₂, then

HA $\rightleftharpoons$ H⁺ + A^–,

C₁(1–$\alpha$₁) C_1$\alpha$1+C_2$\alpha$2 C_1$\alpha$1

HB $\rightleftharpoons$ H⁺ + B^–

C₂(1–$\alpha$₂) C_1$\alpha$1+C_2$\alpha$2 C_2$\alpha$2

[$\alpha$₁ and $\alpha$₂ are the degree of dissociation of HA & HB in presence of each other]

[HA] = C₁ (1–$\alpha$₁); [A] = C_1$\alpha$1; [HB] = C₂ (1–$\alpha$₂);

[B^–] = C₂$\alpha$₂ and [H⁺] = C₁$\alpha$₁ + C₂$\alpha$₂

$K_{a_1}=\frac{\left[H^{+}\right]\left[A^{-}\right]}{\left[HA\right]}=\frac{\left(C_1{\alpha }_1+C_2{\alpha }_2\right)C_1{\alpha }_1}{C_1\left(1-{\alpha }_1\right)}$

$K_{a_2}=\frac{\left[H^{+}\right]\left[B^{-}\right]}{\left[HB\right]}=\frac{\left(C_1{\alpha }_1+C_2{\alpha }_2\right)C_2{\alpha }_2}{C_2\left(1-{\alpha }_2\right)}$

The above two equations can be solved for $\alpha$₁ and $\alpha$₂ from which the H⁺ concentration can be calculated and the pH can be found out.

Note: The pH of bases will be calculated just like we do for acids, with the difference that instead of H⁺ ions we would have OH^– ions, instead of Ka, we would have Kb and instead of pH, we would first calculate pOH. Then pH can be calculated as pH = 14–pOH.

Illustration 6. 0.2 (M) solution of monobasic acid is dissociated to 0.95% Calculate its dissociation constant. Given a = 0.0095; C = 0.2 mole lit^–1

Solution: K_a = C$\alpha$² = 0.2 x (0.0095)² = 1.8 x 10^–5

Illustration 7. Calculate pH for (i) 0.01 (N) Ca(OH)₂, (ii) 10²(M) HCl

Solution: (i) $\underset{\underset{0}{{10}^{-2}N}}{Ca{\left(OH\right)}_2}\to \underset{\underset{{10}^{-2}}{0}}{C{a}^{2+}}+\underset{\underset{2\times {10}^{-2}}{0}}{2O{H}^{-}}$

$\therefore$ [OH] = 10^–2 x 2 gm equivalent / lit = 10^–2 gm mole / lit

pOH = 2, pH = 12

(ii) HCl $\to$ H⁺ + Cl^–

[H⁺] = 10² (M) or pH = –2.

This is not possible because pH range is 0–14.

pH = – log_{10$a_{H^{+}}$}

Where $a_{H^{+}}$ = activity = molar concentration x activity coefficient

_$\therefore$ Unless and unit activity coefficient is given it is not possible to calculate pH of solution.

DETERMINATION OF pH DUE TO HYDROLYSIS

When a salt is dissolved in a solvent, it first dissociates into its constituent ions. This process is called dissolution. Now, if these ions chemically react with water, the process is called hydrolysis. The salts that undergo hydrolysis after dissolution are:

(a) Salts of weak acids + strong bases

(b) Salts of weak bases + strong acids

(c) Salts of weak acids + weak bases

(a) Salt of a Weak Acid and Strong Base

Let us take a certain amount of weak acid (CH₃COOH) and add to it the same amount (equivalents) of a strong base (NaOH). They will react to produce CH₃COONa.

CH_3­COOH + NaOH $\to$ CH₃COONa + H₂O

CH₃COONa being a strong electrolyte, completely dissociates into its constituent ions.

CH₃COONa $\to$ CH₃COO^– + Na⁺

Now, the ions produced would react with H₂O. This process is called hydrolysis

Na⁺ + CH₃COO^– + H₂O $\rightleftharpoons$ CH₃COOH + NaOH

We know that NaOH is a strong base and therefore it would be completely dissociated to give Na⁺ and OH^– ions.

$\therefore$ Na⁺ + CH₃COO^– + H₂O $\rightleftharpoons$ CH₃COOH + OH^– + Na⁺

Canceling Na⁺ on both the sides,

$\Rightarrow$ CH₃COO^– + H₂O v CH₃COOH + OH^–

We can note here that ions coming from strong bases do not get hydrolysed. We should note here that the solution will be basic. This is because the amount of CH₃COOH produced and OH^– produced are equal. But CH₃COOH will not completely dissociate to give H⁺ ions. Therefore [OH^–] ions will be greater than [H⁺] ions.

Since the reaction is at equilibrium,

K_C = $\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$

This equilibrium constant K_c is given a new symbol, K_h.

$\therefore$$K_h=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{[C{H}_{3}CO{O}^{-}]}$If we multiply and divide the above equation by [H⁺] of the solution, then K_h = $\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]\left[H^{+}\right]}{[C{H}_{3}CO{O}^{-}][H^{+}]}$ = $\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{[C{H}_{3}CO{O}^{-}][H^{+}]}{[C{H}_{3}COOH]}}$K_h = $\frac{K_w}{K_a}$ $\Rightarrow$ K_h = $\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{[C{H}_{3}CO{O}^{-}]}$ = $\frac{K_w}{K_a}$\begin{array}{*{20}{c}}{}{C{H_3}CO{O^ - }} + {{H_2}O} \rightleftharpoons {C{H_3}COOH} + {O{H^ - }} \\{Initial:}C{}{}{}0{}0 \\{At\,\,eq:}{C\left( {1 - \alpha } \right)}{}{}{}{C\alpha }{}{Ca}\end{array}Where $\alpha$ is the degree of hydrolysis of CH₃COO^– ion. $\therefore$ $\frac{C\alpha \times C\alpha }{C\left(1-\alpha ;\right)}$= $\frac{K_w}{K_a}$If $\alpha$ is very much less than 1, C$\alpha$² =$\frac{K_w}{K_a}$, $\alpha$ = $\sqrt{\frac{K_w}{K_{a}C}}$As [OH^–] = C$\alpha$, [OH^–] = C x$\sqrt{\frac{K_w}{K_{a}C}}=\sqrt{\frac{K_{w}C}{K_a}}$[H⁺] = $\frac{K_w}{[O{H}^{-}]}=\sqrt{\frac{K_w{K}_a}{C}}$or pH = – log [H⁺] = –log $\left(\sqrt{\frac{K_W\cdot K_a}{C}}\right)$ = $\frac{1}{2}$$p{K}_W+\frac{1}{2}p{K}_a+\frac{1}{2}\log C$
(b)Salt of a Weak Base and a Strong AcidLet the acid be HCl and the base be NH₄OH.Therefore the salt would be NH₄Cl.NH₄Cl completely dissociates into$N{H}_4^{+}$ and Cl^– ions.$\therefore$NH₄Cl $\to$$N{H}_4^{+}$ + Cl^–Cl^– + $N{H}_4^{+}$ +H₂O $\rightleftharpoons$NH₄OH + HClHCl being a strong acid dissociates completely to give H⁺ ions and Cl^– ions.$\therefore$Cl^– + $N{H}_4^{+}$ + H₂O $\rightleftharpoons$NH₄OH + H⁺ + Cl^–$N{H}_4^{+}$+ H₂O $\rightleftharpoons$NH₄OH + H⁺In this hydrolysis, NH₄OH and H⁺ are being produced. This implies that the solution is acidic. To calculate pH,\begin{array}{*{20}{c}}{}{NH_4^ + } + {{H_2}O} \rightleftharpoons {N{H_4}OH} + {{H^ + }} \\{Initial:}C{}{}{}0{}0 \\{At\,\,eq:}{C\left( {1 - \alpha } \right)}{}{}{}{C\alpha }{}{C\alpha }\end{array}Where $\alpha$ is the degree of hydrolysis of$N{H}_4^{+}$.K_h = $\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{[N{H}_4^{+}]}$Multiplying and dividing by OH^– and rearrangingK_h = $\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]\left[O{H}^{-}\right]}{[N{H}_4^{+}][O{H}^{-}]}$=$\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{[N{H}_4^{+}][O{H}^{-}]}{[N{H}_{4}OH]}}$ = $\frac{K_w}{K_b}$$\therefore$ K_h­ = $\frac{K_w}{K_b}$Now, substituting the concentrations, K_h = $\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{[N{H}_4^{+}]}=\frac{K_w}{K_b}=\frac{C\alpha \times C\alpha }{C\left(1-\alpha \right)}=\frac{C{\alpha }^2}{1-\alpha }$If $\alpha$ $⩽$ 0.1, then, C$\alpha$² =$\frac{K_w}{K_b}$,$\alpha$ = $\sqrt{\frac{K_w}{K_{b}C}}$Since [H⁺] = C$\alpha$, [H⁺] = C $\sqrt{\frac{K_w}{K_{b}C}}$ = $\sqrt{\frac{K_{w}C}{K_b}}$or pH = $\frac{1}{2}$$p{K}_W-\frac{1}{2}p{K}_b-\frac{1}{2}\log C$

(c) Salt of a Weak Acid and Weak Base

Let the weak acid be CH₃COOH and the weak base be NH₄OH. Therefore, the salt of these is CH₃COONH₄.

The salt completely dissociates.

CH₃COONH₄ $\to$ CH₃COO^– + $N{H}_4^{+}$

The ions get hydrolysed according to the reaction.

CH₃COO^– + $N{H}_4^{+}$ + H₂O $\rightleftharpoons$ NH₄OH + CH₃COOH

Initial: C C 0 0

At equilibrium: C(1–$\alpha$) C(1–$\alpha$) C$\alpha$ C$\alpha$

K_h = $\frac{\left[N{H}_{4}OH\right]\left[C{H}_{3}COOH\right]}{[N{H}_4^{+}][C{H}_{3}CO{O}^{-}]}$

Multiplying and dividing by H⁺ & OH^– and rearranging,

K_h = $\frac{\left[N{H}_{4}OH\right]\left[C{H}_{3}COOH\right]\left[H^{+}\right]\left[O{H}^{-}\right]}{[N{H}_4^{+}][C{H}_{3}CO{O}^{-}][H^{+}][O{H}^{-}]}$

= $\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{[C{H}_{3}CO{O}^{-}][H^{+}]}{[C{H}_{3}COOH]}\times \frac{[N{H}_4^{+}][O{H}^{-}]}{[N{H}_{4}OH]}}$ = $\frac{K_w}{K_a{K}_b}$

$\therefore$ K_h = $\frac{K_w}{K_a{K}_b}$

Substituting the concentration terms,

K_h = $\frac{K_w}{K_a{K}_b}=\frac{C\alpha \times C\alpha }{C\left(1-\alpha \right)C\left(1-\alpha \right)}=\frac{{\left(C\alpha \right)}^2}{\left[C\left(1-\alpha \right)\right]{}^2}=\frac{{\alpha }^2}{{\left(1-\alpha \right)}^2}$

$\therefore$ $\frac{\alpha }{1-\alpha }=\sqrt{\frac{K_w}{K_a{K}_b}}$

There is an important issue that needs clarification before we move on further. In this case, we can see that both the ions (i.e., cation and anion) get hydrolysed to produce a weak acid and a weak base (hence, we can't predict whether the solution is acidic, basic or neutral). We have considered the degree of hydrolysis of both the ions to be the same. Now we present an explanation as to why this is incorrect and then state reasons for the validity of this assumption.

Actually the hydrolysis reaction given earlier,

CH₃COO^– + $N{H}_4^{+}$+ H₂O $\rightleftharpoons$ CH₃COOH + NH₄OH

is made up of the following three reactions,

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH+O{H}^{-}$

$N{H}_4^{+}+H_{2}O\rightleftharpoons N{H}_{4}OH+H^{+}$

$H^{+}+O{H}^{-}\rightleftharpoons H_{2}O$

In fact the equilibrium between NH₄OH, $N{H}_4^{+}$and OH^– also exists.

Now, we calculate the pH of the solution as,

CH₃COOH $\rightleftharpoons$ CH₃COO^– + H⁺

C$\alpha$ C(1–$\alpha$)

K_a = $\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$ = $\frac{C\left(1-\alpha \right)[H^{+}]}{C\alpha }$

$\therefore$ [H⁺] = K_a x $\left(\frac{\alpha }{1-\alpha }\right)$

Substituting $\frac{\alpha }{1-\alpha }$ as $\sqrt{\frac{K_W}{K_a{K}_b}}$

[H⁺] = K_ax $\left(\frac{\alpha }{1-\alpha }\right)$ = K_a x $\sqrt{\frac{K_w}{K_a{K}_b}}$ = $\sqrt{\frac{K_w{K}_a}{K_b}}$

or pH = $\frac{1}{2}$$p{K}_W+\frac{1}{2}p{K}_a-\frac{1}{2}p{K}_b$

Illustration 7. Calculate the pH at the equivalence point between the titration of 0.1 M, 25 ml CH₃COOH with 0.05 M NaOH solution. K_a (CH₃COOH) = 1.8 x 10^–5

Solution: Volume of NaOH required to reach equivalence point

$=\frac{0.1\times 25}{0.05}=50mL.$

$\therefore$ Concentration of salt formed =$\frac{mil\lim olesofacid}{totalvolume}$

= $\frac{25\times 0.1}{75}=\frac{0.1}{3}$

Since [H⁺] = $\sqrt{\frac{K_w\times K_a}{C}}=\sqrt{\frac{{10}^{-14}\times 1.8\times {10}^{-5}}{\frac{0.1}{3}}}$

$\therefore$ pH = 8.63

Illustration 8. When 0.2 M CH₃COOH is neutralised with 0.2 M NaOH in 0.5 litre of water the resulting solution is slightly alkaline. Calculate pH of resulting solution K_a(CH₃COOH) = 1.8 x 10^–5.

Solution: $\underset{C\left(1-x\right)}{C{H}_{3}CO{O}^{-1}}+H_{2}O\rightleftharpoons \underset{Cx}{C{H}_{3}COOH}+\underset{Cx}{O{H}^{-}}$

$K_h=\frac{C{x}^2}{1-x}=C{x}^2,x\gg 1$

$K_h=\frac{K_w}{K_a}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$

So, $K_h=C{x}^2=5.5\times {10}^{-10}$

x² = 55 x 10^–10

$\therefore$ $\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{Cx}=\frac{{10}^{-14}}{7.42\times {10}^{-6}}=1.3477\times {10}^{-9}M$

pH = -log[H⁺] = -log(1.3477 x 10^-9) = 8.87