Explore topic-wise InterviewSolutions in .

(1) Lewis acid :

Cu⁺⁺, BF₃, Ag⁺

(2) Lewis bases : 

CN^-,Cl^-,S^2-, OH^-.

1. Since KCl is a salt of strong base KOH and strong acid HCl, it does not undergo hydrolysis in its aqueous solution. 

2. Due to strong acid and strong base, concentrations [H₃O⁺] = [OH^-] and the solution is neutral.

CH₃COONH₄ is a strong electrolyte since in aqueous solution it dissociates completely. Sparingly soluble salts AgCl, CuSO₄ are also strong electrolytes.

(1) Sodium chloride is a salt of strong acid HCl and strong base NaOH.

(2) In water, it reacts forming HCl and NaOH.

(3) As both are strong, they dissociate almost completely to liberate H⁺ and OH^- ions, respectively.

(4) H⁺ and OH^- ions combine together to form weakly dissociating H₂O. As there are no free H⁺ ions and OH^- ions, the solution is neutral and the salt does not undergo hydrolysis.

NaCl + H₂O ⇌ NaOH + HCl

Ionic equation :

Na⁺ + Cl^- +H₂O ⇌ Na⁺ + OH^- + H⁺ + Cl^-

H₂O ⇌ H⁺ + OH^-

Since the solution contains equal number of H⁺ and OH^- ions, it is neutral.

Hence the salt of strong acid and strong base does not undergo hydrolysis.

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.

(A) \(CH_3 COONa\)

The solution of salts of weak acid and strong base are basic in nature.

Here, \(CH_3 COONa\) is salts of weak acid (\(CH_3COOH\)) and strong base \((NaOH)\).

Therefore, its aqueous solution is basic in nature.

Option : (a) CH₃COONa

(C) Sodium nitrate

Only salts of (weak acid + strong base) and (strong acid + weak base) get hydrolysed and show alkalinity or acidity in water. salt of strong acid and strong base does not get hydrolysed in water.

Here, sodium acetate (\(CH_3COONa\)) (weak acid + strong base)

Sodium carbonate (strong acid + weak base)

Sodium cyanide (\(NaCN\)), (strong base + weak acid)

While - Sodium nitrate is salt of strong base and strong acid.

Therefore, it does not get hydrolyse in water.

Option : (c) Sodium nitrate

(a) Strong acids :

HCl, H₂SO₄

Strong bases : 

NaOH, KOH

(b) Weak acids : 

HCOOH, CH₃COOH

Weak bases : 

NH₄OH, C₂H₅NH₂

(a) Acidic buffer (CH₃COOH + CH₃COONa) 

(b) Basic buffer (NH₄OH + NH₄Cl) 

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

(d) Basic buffer (Cu(OH)₂ + CuCl₂) 

[Note : Cu(OH)₂ being insoluble is not used to prepare a buffer solution.]

There are two types of electrolytes as follows : 

(a) Strong electrolyte : The electrolytes which ionise completely or almost completely are called strong electrolytes. 

For example, 

NaCl,HCl,H₂SO₄,etc.

(b) Weak electrolytes : The electrolytes which dissociate to a less extent are called weak electrolytes. 

For example, 

CH₃COOH, NH₄OH,etc.

Electrolytes : The substances which in their aqueous solutions (or in any polar solvents) dissociate or ionize forming positively charged ions (cations) and negatively charged ions (anions) are called electrolytes. 

For example,

NaCl, HCl, etc.

Common ion : An ion common to two electrolytes is called common ion. This is generally applicable to a mixture of a strong and a weak electrolyte. 

For example,

A solution containing weak electrolyte CH₃COOH and strong electrolyte salt CH₃COONa. 

CH₃COONa → CH₃COO^- + Na⁺; 

CH₃COOH ⇌ CH₃COO^- + H⁺ 

Hence,

CH₃COOH and CH₃COONa have a common ion CH₃COO^-.

Common ion effect : The suppression of the degree of dissociation of a weak electrolyte by the addition of a strong electrolyte having an ion in common with the weak electrolyte is called common ion effect. 

For example, 

CH₃COOH and CH₃COONa have common ion CH₃COO^-

(1) Consider the dissociation or ionisation of a weak acid, CH₃COOH in its solution.

CH₃COOH_(aq) ⇌ CH₃COO^-_(aq) + H⁺_(aq)

The dissociation constant Ka for CH₃COOH will be,

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

K_a is constant for CH₃COOH at constant temperature.

(2) If a strong electrolyte like salt CH₃COONa is added to the solution of CH₃COOH, then on dissociation it gives a common ion CH₃COO^-.

CH₃COONa → CH₃COO^- + Na⁺

(3) Due to common ion CH₃COO^-, overall concentration of CH₃COO^- in the solution is increased, which increases the ratio, \(\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}\).

In order to keep this ratio constant, the concentration of H⁺ is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier’s principle.

(4) Thus the ionisation of a weak acid is suppressed by a common ion.

A weak electrolyte dissociates partially in aqueous solution to produce cations and anions. 

Equilibrium exists between ions thus formed and the undissociated molecules.

BA ⇌ B⁺ + A^-

For such an equilibrium, the dissociation constant K is defined as,

K = \(\frac{[B^+]\times [A^-]}{BA}\)

K is constant for the weak electrolyte at a given temperature.

Now, 

If another electrolyte BC or DA is added to the solution BA, having a common ion either B⁺ or A^-, then the concentration of either B⁺ or A^- is increased. 

However, 

As K is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect. 

For example, 

The dissociation of a weak acid CH₃COOH is suppressed by adding CH₃COONa having common ion CH₃COO^-.

CH₃COOH ⇌ CH₃COO^- + H⁺

CH₃COONa → CH₃COO^- + Na⁺

(B) \(10^{-10}\)

As we know,

\(P^H+P^{OH}=14\)

\(\therefore P^H=14-4\)

\(P^H=10\)

\(-\log[H^+]=10\)

\([H]^+=10^{-10}\)

Option : (b) 10^-10

Correct Option is : (B) Le Chatelier’s principle

The common ion effect is based on Le Chatelier’s principle

Option : (b) Le Chatelier’s principle

Given : 

pH = 6.06, [H₃O⁺] = ? 

PH = -log₁₀[H₃O⁺] 

∴ log₁₀[H₃O⁺] = -pH

∴ [H₃O⁺] = Antilog – pH 

= Antilog – 6.06 

= Antilog \(\bar 7.94\)

= 8.714 × 10^-7 M

∴ [H₃O⁺] = 8.714 × 10^-7 M

Correct answer is

(b) < 10^-7 M

Correct Option is : (a) 12

As we know,

pH + pOH = 14

\(\therefore\) pOH = 14 - pH

pOH = 14 + log \([H^+]\)

pOH = 14 + log \(10^{-2}\)

pOH = 14-2

pOH = 12

Option : (a) 12

(c) less than 7

The pH of 10^-8 M of HCl is less than 7.

Correct Option is (C) 7.40

The pH of human blood of a normal person is equal to 7.4

Option : (c) 7.40 

Correct Option is : (A) a strong acid

The solution of strong acids have small pH value (less than 3). Weak acids have pH value greater than 3 and less than 7.

Neutral solutions have pH value equal to 7 and bases have pH value greater than 7.

Option : (a) a strong acid

Correct answer is

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

Correct Option is : (d) 11.5

NaOH is Strong base . It gives OH ion in solution.

As we know,

pH + pOH = 14

pH = 14 - pOH

pH = 14 + log \([OH^-]\)

pH = 14 + log \((3.162 \times 10^{-3})\)

pH = 14 -3 + log (3.162) + \(log \ 10^{-3}\)

pH = 14 - 3 + log (3.162) 

pH = 11 + 0.4996

pH = 11.5

Option : (d) 11.5

Correct Option is : (C) \(10^{-9} mol/dm^3\)

As we know,

pH + pOH = 14

pOH = 14 - pH

pOH = 14-5

pOH = 9

pOH = - log\([OH^-]\)

\(pOH \implies [OH^-] = 10^{-9} mol/dm^3\)

Option : (c) 10^-9 mol/dm³

(D) \(CH_3COOH\)

Ostwald’s dilution law is applicable for weak acids and weak bases.

Here, \(CH_3COOH\) is weak acid while \(HCl\) and \(H_2SO_4\) are strong acid, and \(NaOH\) is strong base.

Option : (d) CH₃COOH

Correct answer is

(c) CH₃COOH + CH₃COONa in water

(a) 7.4

Blood in human body is highly buffered at pH of 7.4.

Correct Option is : (D) \(Ca^{2+}\)

\(Ca^{2+}\) ion cannot be precipitated by both HCl and \(H_2S\) . While \(Pb^{2+}\) , \(Cu^{2+}\) , and \(Ag^+\) ions belongs to group II Cations they precipitated as metal sulphide in presence of HCl .

Option : (d) Ca²⁺

(A) \(CuSO_4\) reacts with water

Blue vitriot \((CuSO_4\,5H_2O)\) in water is acidic because when \(CuSO_4\) dissolves in water, it forms aqueous solution of strong acid (\(H_2SO_4\)) and weak base (\(Cu(OH)_2\)).

\(CuSO_4+2H_2O\longrightarrow Cu(OH)_2+H_2SO_4\)

\(Cu(OH)_2\) remains in equilibrium but \(H_2SO_4\) fully ionises to give \(H^+\) ions

Therefore, solution of \(CuSO_4\) is acidic in nature.

Option : (a) CuSO₄ reacts with water.

Correct answer is

(a) 7.2 × 10^-7

Given : S = 1.28 × 10^-5 mol dm^-3; K_sp = ?

AgBr dissociates as,

AgBr ⇋ \(Ag^+_{(aq)}\)+ \(Br^-_{(aq)}\)

K_sp = [Ag⁺] [Br^-]

As the solubility of AgBr in water is 1.28 × 10^-5 moles/dm³.

[Ag⁺] = [Br^-] = 1.28 x 10^-5 mol dm³

∴ K_sp = [1.28 × 10^-5] [1.28 × 10^-5]

= 1.638 × 10^-10

∴ Solubility product of AgBr = 1.638 × 10^-10

Given :

Solubility product of PbS = K_sp = 4.2 × 10^-28

Concentration of Pb⁺⁺ = [Pb⁺⁺] = 0.001 M

Concentration of S^-- = [S^--] = ?

For PbS,

PbS_(s) ⇌ Pb⁺⁺+ S^–

∴ K_sp = [Pb⁺⁺] × [S^--]

∴ [S^–] = \(\frac{K_{sp}}{[Pb^{++}]}\)

= \(\frac{4.2\times 10^{-28}}{0.001}\)

= 4.2 × 10^-25 M

To precipitate Pb⁺⁺ as PbS, ionic product must be greater than 4.2 × 10^-28.

Hence, 

[S^--] > 4.2 × 10^-25 M.

∴ Concentration of S^-- required > 4.2 × 10^-25 M.

(c) CuS

CuS is precipitated in an acidic medium while sulphile of \(N i^{+2}\), \(CO^{+2}\) and \(Mn^{+2}\) are precipitated in alkaline medium.

Option : (c) CuS

Given : 

Solubility of AgCl = S = 1.562 × 10^-10 mol dm^-3 

Solubility of AgCl in g dm^-3 = ? 

Molar mass of AgCl = M = 143.5 g mol^-1 

Solubility in gram per dm³,

= solubility in mol dm^-3 × molar mass 

= 1.562 × 10^-10 × 143.5 

= 2.241 × 10^-5 g dm^-3 

∴ Solubility of AgCl = 2.241 × 10^-8 g dm^-3

Correct Option is : (a) 2.6244

\(CH_3COOH\) is a weak acid.

Ka = \(\frac{C \alpha \times C \alpha}{C(1 - \alpha)}\)

Ka = \(\frac{C \alpha}{(1 - \alpha)}\)           ..... \(\alpha << 1\)

\(\implies Ka = C \alpha^2\)

\([H^+] = C \alpha\)         ..... [\(\alpha = \frac{9.5}{100}\)]

\([H^+] = 0.025 \times 0.095\)

\([H^+] = 0.002375\)

\(\therefore \ pH = -log [H^+]\)

\(pH = -log [0.002375]\)

\(pH = 2.6244\)

Option : (a) 2.6244

Option : (d) 2 × 10^-4

The dissociation of acetic acid is represented below : 

CH₃COOH ⇌ CH₃ – COO^- + H⁺

Given : 

Dissociation = 2%, C = 0.1 M 

[H₃O⁺] = ? 

The concentration of hydrogen ion, [H⁺], is given by the following formula : 

[H₃O⁺] = αC 

α = Degree of dissociation = \(\frac{Percent\,dissociation}{100}\)

= \(\frac{2}{100}\)= 0.02 

[H₃O⁺] = Hydrogen ion concentration = ? 

C = Molar concentration of acetic acid 

= 0.1 M = 0.1 mol dm^-3 

∴ [H₃O⁺] = C × α = 0.1 × 0.02 

= 0.002 = 2.0 × 10^-3 mol dm^-3 

∴ Hydrogen ion concentration = 2.0 × 10^-3 mol dm^-3

Option : (c) \( [\mathrm{F}_{\mathrm{e}}^{3+}][OH^-]^3\)

correct Option is : (B) ionic product < solubility product

A solution is called unsaturated if its ionioc product less than solubility product.

Option : (b) ionic product < solubility product

Option : (c) ≥ 4.2 × 10^-25 mol/dm³

Correct Option is : (D) \(Be(OH)_2\)

\(Be(OH)_2\) has has the lowest value of \(K_{sp}\) at ordinary temperature , because the solubility of a hydroxide of group 2 elements increases down the group . Therefore Be(OH) has lowest \(K_{sp}\) value.

Option : (d) Be(OH)₂

Mechanism of action of a basic buffer :

(1) A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. 

Moreover, due to the presence of the common ion, largely supplied by the salt,the dissociation of the base is further suppressed.

(NH₄OH + NH₄Cl) :

NH₄Cl_(aq) → \(NH^+_{4(aq)}\)+ \(CI^-_{(aq)}\) (Complete)

(2) When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water. 

Thus,

The addition of an acid does not change the pH of the buffer.

\(H^+_{(aq)}\) + NH4OH(aq) ⇌ \(NH^+_{4(aq)}\)+ H₂O₍₁₎

This removal of added H⁺ is called reserved basicity.

(3) When a small quantity of a strong base is added, the hydroxide ions combine with \(NH^+_4\) ions to form undissociated NH₄OH. As a result, the hydrogen or hydroxyl ion concentration does not change. 

Thus,

The pH of the solution does not change.

\(OH^-_{(aq)}\)+ \(NH^+_4\)⇌ NH₄OH_(aq)

This removal of added OH^- is called reserved acidity.

Given : 

Percent dissociation = 4.3, 

C = 0.01 M, 

K_b = ?, 

pH = ? 

The degree of dissociation and dissociation constant of NH₄OH are related to each other by the formula :

K_b = α²C

K_b = Dissociation constant of NH₄OH = ?

α = Degree of dissociation of NH₄OH = 4.3%

= 4.3 × 10^-2

C = Molar concentration of NH₄OH = 0.01 M

∴ K_b = (4.3 × 10^-2)² × 0.01

= 18.49 × 10^-4 × 10^-2

∴ K_b = 1.849 × 10^-5

Since NH₄OH is a monoacidic base,

[OH^-] = αC

= 4.3 × 10^-2 × 0.01

= 4.3 × 10^-4 mol dm^-3

pOH = -log₁₀[OH^-]

= -log₁₀4.3 × 10^-4

= -[0.6335 – 4] = 3.3665

pH + pOH = 14

pH = 14 – 3.3665 = 10.6335

K_b = 1.849 × 10^-5, pH = 10.6335

(B) an increase in pH

\(Na_2CO_3+H_2O\longrightarrow 2NaOH+H_2O+CO_2\)

on hydrolysis of \(Na_2CO_3\) gives \(NaOH\) solution, water and Carbon dioxide gas.

We know that, \(NaOH\) is a strong base.

it means the \(P^H\) of pure water increase.

Option : (b) an increase in pH

(c) a proton acceptor

According to Lowry-Bronsted concept,

Acid is a substance which has the tendency to give a proton \((H^+)\) and a base is a substance which has tendency to accept a proton.

Therefore, base is a proton acceptor.

Option : (c) a proton acceptor.

The important property of reserve acidity and reserve basicity of a buffer solution is used to maintain constant pH. Weak acid or weak base along with ions (cations or anions) from salt react with excess of added acid (H⁺) or base [OH^-] and maintain pH constant.

Properties (or advantages) 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.

Pure water ionises to a very less extent.

The ionisation equilibrium is represented as follows,

H₂O₍₁₎ + H₂O₍₁₎ ⇋ H₃O⁺_(aq) + OH^-_(aq)

The equilibrium constant K for the above equilibrium is represented as,

K = \(\frac{[H_3O^+][OH^-]}{[H_2O]^2}\)

∴ K × [H₂O]² = = [H₃O⁺] × [OH^-]

Since K and active mass of pure water [H₂O] are constant we can write,

K x [H₂O] = K_w'

∴ K_w = [H₃O⁺] x [OH^-]

Where K_w is called ionic product of water. 

At 25 °C,

K_w = 1 × 10^-14.