Solutions-Class-12-Chemistry-Chapter-3-Ionic Equilibria-Maharashtra Board

(c) less than 7

(d) 75 mL of 0.2M HCl + 25mLof 0.2M NaOH

(c) CH₃COOH+CH₃COONa in water

(a) 7.2 × 10^-7

(a) 7.4

(c) [Zn(H₂O)₃OH]⁺

(b) < 10^-7M

Because cations lack electrons, they receive a pair of electrons and are hence Lewis acids.

• Since KCI is a salt of strong base KOH and strong acid HCl, it does not undergo hydrolysis in its aqueous solution.
• Due to strong acid and strong base, concentrations [H₃O⁺] = [OH^—] and the solution is neutral.

• A basic buffer solution is made by combining aqueous solutions of a weak base, such as NH₄OH, with its salt of a strong acid, such as NH₄
• A weak base is chosen based on the needed pH or pOH of the solution and the weak base's dissociation constant.

Given : Ka = 1.8 x 10⁻⁵; C= 0.01 M, Percent dissociation = ?

K_a = \(\frac{Cα^2}{1-α}≈Cα^2\)

∴  α = \(\sqrt{\frac{K_a}{C}}\) = \(\sqrt{\frac{1.8×10^{-5}}{0.01}}\)

= 4.242 x 10^-2

Per cent dissociation = α x 100

= 4.242 x 10^-2 x 102

= 4.242 %

Answer is: Per cent dissociation = 4.242 %.

Properties of a buffer solution :

• The pH of a buffer solution is maintained appreciably constant.
• By addition of a small amount of an acid or a base pH does not change.
• On dilution with water, pH of the solution doesn’t change.

Given : pH = 6.06, [H⁺] = ?

pH = −log₁₀ [H⁺]

log₁₀ [H⁺] = −pH

[H⁺] = Antilog −pH

= Antilog −6.06

= Antilog .94

= 8.714 x 10⁻⁷ M

Answer is : [H⁺] = 8.714 x 10⁻⁷ M

Given : C = 0.01 M H₂SO₄, pH = ?

H₂SO₄(aq) →2H⁺(aq) + SO₄²⁻(aq)

∴ [H₃O⁺] = 2 x 0.01 = 0.02 M

pH = −log₁₀[H₃O⁺]

= − log₁₀ 0.02

= − ( .3010)

= 2 − 0.3010

= 1.6990

Answer is : pH = 1.6990.

The weak dibasic acid H2S is dissociated as follows :

H₂S (aq) + 2H₂O (l) ⇌  2H₃O⁺ (aq) + S^2— (aq)

The dissociation constant K_a is represented as,

K_a = \(\frac{[H_3O^+]^2[S^{2-}]}{[H_2S]}\)

When HCl is added, it increases the concentration of common ion H₃O⁺.

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^— (aq)

Hence by Le Chaterlier’s principle, the equilibrium is shifted from right to left, suppressing the dissociation of weak electrolyte H2S.

• CuSO₄ is a salt of strong acid H₂SO₄ and weak base Cu(OH)₂.
• CuSO₄ in aqueous solution undergoes hydrolysis and forms a precipitate of Cu(OH)₂. And solution becomes turbid.

CuSO₄ + 2H₂O ⇌ Cu(OH)₂↓ + H₂SO₄

OR

CuSO4 + 4H₂O ⇌ Cu(OH)2 + 2H₃O⁺ + SO₄^2-

.                             turbidity

• When H₂SO₄ is added, the hydrolysis equilibrium is shifted to left hand side and Cu(OH)₂ dissolves giving clear solution.

(a) Acidic buffer (CH₃COOH + CH₃COONa)

(b) Basic buffer (NH₄OH + NH₄Cl)

(c) Acidic buffer (Sodium benzoate + benzoic acid)

(d) Basic buffer (Cu(OH)₂ + CuCl₂)

According to Arrhenius theory acids and bases are defined as follows :

Acid : Acid is a substance which contains hydrogen and gives rise to H⁺ ions in aqueous solution.

For example : HCl

HCl (aq) → H⁺(aq) + Cl⁻(aq)

CH₃COOH(aq) \(\underleftrightarrow{water}\) CH₃COO (aq)+ H⁺(aq)

Base : A substance that contains OH group and produces hydroxide ions (OH⁻) in aqueous solution is called a base.

For example : NaOH

NaOH(aq) → Na⁺(aq) + OH⁻(aq) .

NH₄OH(aq) ⇌  NH₄⁺(aq) + OH⁻(aq)

Conjugate acid-base pair : A pair of an acid and a base differing by a proton is called a conjugate acid-base pair.

(a) HCl  +   H₂O    ⇌  H₃O⁺   +   Cl⁻

Acid-1      Base-2         Acid-2       Base-1

(b) CO₃²⁻  +  H₂O  ⇌  OH⁻ +  HCO₃⁻

Base-1        Acid-2      Base-2    Acid-1

H₂O(l)   +   HCl    ⇌   H₃O⁺ +   Cl

Base—1        Acid—2        Acid-1      Base—2

Since water accepts a proton, it acts as a base

• Since ammonia molecule, NH₃ has a lone pair of electrons to donate it acts as a Lewis base.
• AlCl₃ is a molecule with incomplete octet hence it is electron deficient and acts as a Lewis acid.

Given : C = 0.1 M; Dissociation = 5%, K_a = ?

α = \(\frac{Percent\,dissociation}{100}=\frac{5}{100}\) = 0.05

K_a = \(\frac{Cα^2}{1-α}\) = \(\frac{0.1×0.05^2}{1-0.05}\)   = 2.63 x 10⁻⁴

Answers is : Dissociation constant of acid = K_a = 2.63 x 10⁻⁴

Relationship between pH and pOH :

The ionic product of water, Kw is given by,

K_w = [H₃O⁺] x [OH⁻]

At 298 K, Kw = 1 x 10⁻¹⁴.

pK_w = −log₁₀K_w = − log₁₀1 x 10⁻¹⁴.= 14

[H₃O⁺] x [OH⁻]  = 1 x 10⁻¹⁴

Taking logarithm to base 10 of both sides,

log₁₀[H₃O⁺] + log_1o[OH⁻] = log₁₀1 x 10⁻¹⁴

Multiplying both the sides by − 1,

−log₁₀[H₃O⁺] − log_1o[OH⁻] = −log₁₀1 x 10⁻¹⁴

‘.’ pH = —log₁₀[H₃O⁺],  pOH = — log₁₀[OH⁻]

pK_w = −log₁₀K_w

∴ pH + pOH = pKw

Or pH + pOH = 14

(A) (i) Sodium carbonate is a salt of weak acid and strong base.

(ii) In aqueous solution it undergoes hydrolysis.

Na₂CO₃(aq) + 2H₂O(l) ⇌ 2NaOH(aq) + H₂CO₃(aq)

salt                                   strong base         weak acid

2Na⁺(aq) + CO₃²⁻ 2H₂O(aq) ⇌ 2Na⁺(aq) + 2OH⁻(aq) + H₂CO₃(aq)

(iii) Strong base dissociates completely while weak acid dissociates partially hence [OH⁻] > [H₃O⁺] and the solution is basic.

(B) (i) Ammonium chloride is a salt of strong acid and weak base.

(ii) In aqueous solution it undergoes hydrolysis.

NH₄Cl(aq) + H₂O(aq)  ⇌ NH₄OH(aq) + HCl(aq)

salt                                         weak base       strong acid

NH₄⁺ +Cl⁻(aq) + H₂O(aq)  ⇌ NH₄OH(aq) + H⁺ + Cl⁻(aq)

(iii) Since [H⁺] or [H₃O⁺] > [OH⁻] the solution is acidic.

Given : pH = 3.2; C = 0.02 M; K_a = ?

K pH = −1og₁₀ [H⁺]

[H⁺] = Antilog −pH

= Antilog −3.2

= Antilog .8

= 6.31 x 10⁻⁴  M

HA(aq)    ⇌  H⁺ + A^—

c(1 − α)       cα        cα

‘.’ [H⁺] = cα

α = \(\frac{H^+}{c}=\frac{6.31×10^{-4}}{0.02}\)  = 0.0315

K_a = cα²  = 0.02 x (0.0315)² = 1.984 x 10^—5

Answer is: Dissociation constant = K_a = 1.984 x 10^—5

Given: [OH⁻] = 2.87 x 10⁻⁴ M, pH = ?

pOH = −log₁₀[OH⁻]

= − log₁₀ 2.87 x 10⁻⁴

= −( .4579)

= (4 − 0.4579)

= 3.5421

pH + pOH = 14

pH =14 − pOH = 14 − 3.5421 = 10.4579

Answer is : pH = 10.4579.

Degree of dissociation : It is defined as a fraction of total number of moles of an electrolyte that dissociate into its ions at equilibrium. It is denoted by symbol ∝ and given by

α = \(\frac{\text{number of mole dissociated}}{\text{total number of moles}}\)

Percent dissociation = α × 100

If 'c' is the molar concentration of an electrolyte the equilibrium concentration of cation or anion is (α × c) mol dm^—3.

Ostwald's dilution law for the CH₃COOH.

Consider the dissociation of a weak electrolyte HA. Let V dm³ of a solution contain one mole of CH₃COOH.

Then the concentration of a solution is, C = 1/V mol dm⁻³

Let α be the degree of dissociation of CH₃COOH,

Applying the law of mass action to this dissociation equilibrium, we have,

K_a = \(\frac{[CH_3COO^-]×[H^+]}{[CH_3COOH]}\)

∴ K_a = \(\frac{α/V\,×α/V}{(1-α)/V}=\frac{α^2}{(1-α)V}\)

‘.’ C = 1/V mol dm³

∴ K_a = \(\frac{Cα^2}{(1-α)}\)

This is the Ostwald’s dilution law.

• pH : The negative logarithm, to the base 10, of the molar concentration of hydrogen ions, H⁺ is known as the pH of a solution.

pH = − log₁₀[H⁺]

• pOH : The negative logarithm, to the base 10, of the molar concentration of hydroxyl ions, OH⁻ is known as the pOH of a solution.

pOH = −log₁₀[OH⁻]

Relationship between pH and pOH :

The ionic product of water, Kw is given by,

Kw = [H₃O⁺] x [OH⁻]

At 298 K, Kw =1 x 10⁻¹⁴.

pK_w = −log₁₀K_w= − log_1o1 x 10⁻¹⁴.= 14

[H₃O⁺] x [OH⁻]  = 1 x 10⁻¹⁴

Taking logarithm to base 10 of both sides,

log₁₀[H₃O⁺] + log_1o[OH⁻] = log_1o1 x 10⁻¹⁴

−log₁₀[H₃O⁺] − log_1o[OH⁻] = −log_1o1 x 10⁻¹⁴   …(Multiplied both sides by −1)

‘.’ pH = −log₁₀[H₃O⁺],  pOH = − log_1o[OH⁻]

pK_w = −log₁₀K_w

∴ pH + pOH = pKw

Or pH + pOH = 14

Hydrolysis : A reaction in which the cations or anions or both the ions of a salt react with water to produce acidity or basicity or sometimes neutrality is called hydrolysis.

Consider hydrolysis of CH₃COONH₄.

CH₃COO^— + NH₄⁺ H₂O   ⇌  CH₃COOH + NH₄OH

Salt                                           weak acid       weak base

Since K_a = K_b, the weak acid CH₃COOH and weak base NH₄OH dissociate to the same extent, hence, [H₃O⁺] = [OH^—] and the solution reacts neutral after hydrolysis.

The weak acid HCN is dissociated as follows :

HCN (aq) + H₂O (l) ⇌  H₃O (aq) + CN^— (aq)

The dissociation constant K_a is represented as,

K_a = \(\frac{[H_3O^+]^2×[CN^-]}{[HCN]}\)

When HCl is added, it increases the concentration of H₃O⁺, hence in order to keep the ratio constant, then by Le Chatelier’s principle, the equilibrium is shifted from right to left, suppressing the dissociation of HCN.

Consider the dissociation of a weak electrolyte HA. Let V dm³ of a solution contain one mole of the electrolyte.

Then the concentration of a solution is, C = 1/V mol dm⁻³

Let α be the degree of dissociation of the electrolyte,

Applying the law of mass action to this dissociation equilibrium, we have,

K = \(\frac{[H^+]×[A^-]}{[HA]}\)

∴ K = \(\frac{α/V\,×α/V}{(1-α)/V}=\frac{α^2}{(1-α)V}\)

As the electrolyte is weak, ∝ is very small as compared to unity,

(1 — α) ≈ 1.

∴ K = \(\frac{α^2}{V}\)

∴  α = \(\sqrt{KV}\)

∴  α ∝ \(\sqrt{V}\)

1/V = C, where C = concentration in mol dm⁻³

∴ K = ∝²C

∴  α = \(\sqrt{\frac{K}{C}}\)

∴  α = \(\sqrt{KV}\)   (‘ .’ C = 1/V or V = 1/C)

∴  α = \(\sqrt{\frac{K}{C}}\)

This is the expression of Ostwald’s dilution law. Thus, the degree of dissociation of a weak electrolyte is directly proportional to the square root of the volume of the solution containing 1 mole of an electrolyte.

• In qualitative analysis, the cations of group II are precipitated as sulphides, namely HgS, CuS, PbS, etc., while cations of group IIIB are precipitated as sulphides, namely, CoS, NiS, ZnS.
• The sulphides of group II have extremely low solubility product (K_sp) about 10⁻²⁹ to 10⁻⁵³ while the sulphides of group IIIB have slightly higher K_sp values about 10⁻²⁰ to 10⁻³⁰
• In group II, sulphides are precipitated by adding H₂S in acidic solution while in IIIB group they are precipitated in a basic solution like ammonical solution.
• In acidic medium due to common ion H⁺, H₂S is dissociated to very less extent but gives sufficient S²⁻ ion to exceed solubility product of group II sulphides of cations and precipitate them.

HCl (aq) → H⁺ (aq) + Cl^— (aq)

H₂S (aq) ⇌ 2H⁺(aq) + S^2— (aq)

• In basic medium, H⁺ from H₂S are removed by OH^— in solution, or by NH₄OH, increasing the dissociation of H₂S and concentration of S^2—, so that IP > K_sp.

H₂S (aq) ⇌ 2H⁺ (aq) + S^2— (aq)

NH₄OH(aq) + H⁺ (aq) → NH₄⁺(aq) + H2O (l)

• Therefore, group II cations are precipitated in an acidic medium while cations of group IIIB are precipitated in ammonical solution.

Consider the solubility equilibrium of a sparingly soluble salt, AgCl.

AgCl (s) ⇌ Ag (aq) + Cl^— (aq)

The solubility product, K_sp is given by,

K_sp = [Ag⁺] x [Cl^—]

Consider addition of a strong electrolyte AgNO₃ with a common ion Ag⁺.

AgNO3(aq) _’ Aggq) + Noiaq)

The concentration of Ag⁺ in the solution is increased, hence by Le Chatelier’s principle the equilibrium of AgCl is shifted to left hand side since the value of K_sp is constant.

Thus in the presence of a common ion, the solubility of a sparingly soluble salt is suppressed.

Given : pH = 5.1, [H⁺] = ?

pH = — log₁₀ [H⁺]

log₁₀ [H⁺] = −pH

[H⁺] = Antilog −pH

= Antilog −5.1

= Antilog .9

= 7.943 x 10⁻⁶ M

Answer is: [H⁺] = 8.714 x 10⁻⁷ M

• If ionic product and solubility product are indicated by IP and K_(sp) respectively then,
• When IP = K_(sp) the solution is saturated.
• When IP > K_(sp) the solution is supersaturated and hence precipitation will occur, when two solutions are mixed.
• When IP < K_(sp) the solution is unsaturated and precipitation will not occur, when two solutions are mixed.