Question 1

The ionic product of water (Kw) at 25°C is 1.0 × 10⁻¹⁴. What is the pH of a neutral solution at this temperature?

Accepted Answer

At neutrality, [H⁺] = [OH⁻], and since [H⁺] × [OH⁻] = Kw = 10⁻¹⁴, we get [H⁺] = 10⁻⁷ M, giving pH = 7.

Question 2

A weak acid HA has a Ka value of 1.8 × 10⁻⁵. Which of the following correctly describes the strength of this acid relative to strong acids?

Accepted Answer

Weak acids have Ka < 1 and ionize only partially; the small Ka value (10⁻⁵) confirms incomplete ionization compared to strong acids which fully ionize.

Question 3

The solubility product (Ksp) of AgCl at 25°C is 1.8 × 10⁻¹⁰. If the concentration of Cl⁻ ions in a saturated solution is 1.34 × 10⁻⁵ M, what is the concentration of Ag⁺ ions?

Accepted Answer

Using Ksp = [Ag⁺][Cl⁻], we have 1.8 × 10⁻¹⁰ = [Ag⁺] × 1.34 × 10⁻⁵, giving [Ag⁺] ≈ 1.5 × 10⁻⁴ M.

Question 4

Which of the following salts will hydrolyze in water to produce an acidic solution?

Accepted Answer

NH₄Cl is the salt of a weak base (NH₃) and strong acid (HCl); the NH₄⁺ ion hydrolyzes to release H⁺ ions, making the solution acidic.

Question 5

The Ka of acetic acid is 1.8 × 10⁻⁵ and its Kb is 5.6 × 10⁻¹⁰. What is the relationship between Ka, Kb, and Kw for the acetate ion (CH₃COO⁻)?

Accepted Answer

For a conjugate acid-base pair, Kb of the conjugate base = Kw / Ka of the conjugate acid; thus Kb(acetate) = 10⁻¹⁴ / 1.8 × 10⁻⁵.

Question 6

The ionic product of water (Kw) at 25°C is 1.0 × 10⁻¹⁴. What is the pH of a neutral solution at this temperature?

Accepted Answer

At neutrality, [H⁺] = [OH⁻] = 10⁻⁷ M, so pH = -log(10⁻⁷) = 7.

Question 7

A weak acid HA has a Ka value of 1.8 × 10⁻⁵. Which statement correctly describes its ionization behavior?

Accepted Answer

Weak acids partially ionize and reach equilibrium; Ka determines the extent of ionization, not complete dissociation.

Question 8

The pKa of acetic acid is 4.74. What is the Ka value?

Accepted Answer

Ka = 10⁻pKa = 10⁻⁴·⁷⁴ ≈ 1.82 × 10⁻⁵.

Question 9

For a weak base ammonia (NH₃) with Kb = 1.8 × 10⁻⁵, what is the pOH of a 0.1 M NH₃ solution (assuming [OH⁻] = √(Kb × C))?

Accepted Answer

[OH⁻] = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) ≈ 1.34 × 10⁻³, so pOH = -log(1.34 × 10⁻³) ≈ 3.42.

Question 10

A salt solution of NaA (where A⁻ is a weak conjugate base) shows basic pH. This is due to which process?

Accepted Answer

The conjugate base A⁻ hydrolyzes: A⁻ + H₂O ⇌ HA + OH⁻, producing OH⁻ ions and making the solution basic.

Question 11

The common ion effect in the solution of a weak acid HA with added salt NaA will:

Accepted Answer

Adding A⁻ ions shifts equilibrium left (Le Chatelier's principle), decreasing further ionization of HA.

Question 12

At 25°C, if the pH of a solution is 3.5, what is the [OH⁻] concentration (Kw = 1.0 × 10⁻¹⁴)?

Accepted Answer

[H⁺] = 10⁻³·⁵ ≈ 3.16 × 10⁻⁴ M; [OH⁻] = Kw/[H⁺] = 10⁻¹⁴/(3.16 × 10⁻⁴) ≈ 3.16 × 10⁻¹¹ M.

Question 13

Which of the following salts will produce an acidic solution in water?

Accepted Answer

AlCl₃ undergoes hydrolysis: Al³⁺ + H₂O ⇌ Al(OH)²⁺ + H⁺, producing H⁺ and making the solution acidic.

Question 14

A weak acid HA has Ka = 1.8 × 10⁻⁵. If 0.1 M solution of this acid is prepared, what is the degree of ionization (α)? (Assume α << 1)

Accepted Answer

Using Ka = Cα², we get α = √(Ka/C) = √(1.8 × 10⁻⁵/0.1) = √(1.8 × 10⁻⁴) ≈ 0.0134.

Question 15

The solubility product (Ksp) of AgCl at 25°C is 1.8 × 10⁻¹⁰. What is the solubility of AgCl in a 0.1 M NaCl solution?

Accepted Answer

In presence of common ion Cl⁻ (0.1 M), Ksp = [Ag⁺][Cl⁻]; 1.8 × 10⁻¹⁰ = s × 0.1, so s = 1.8 × 10⁻⁹ mol/L.

Question 16

A buffer solution is prepared by mixing 50 mL of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵) and 50 mL of 0.1 M sodium acetate. What is the pH of this buffer? (Use log 1.8 ≈ 0.26)

Accepted Answer

Using Henderson-Hasselbalch equation: pH = pKa + log([CH₃COO⁻]/[CH₃COOH]); pKa = 4.74, and ratio = 1, so pH = 4.74.

Agniveer Army Technical Ionic Equilibrium

Ionic Equilibrium — Practice Questions