JEE Chemistry Ionic Equilibrium MCQs Set 26

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Chapter-wise Objective Questions: Ionic Equilibrium

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Multiple Choice Questions

Question 1: The self-ionisation constant of pure formic acid, \(K = [HCOOH_2^+][HCOO^-]\), is \(10^{-6}\) at room temperature. Its density is 1.15 g/cm\(^3\) (molar mass 46). The percentage of formic acid molecules converted to formate ion is:
(a) 0.008%
(b) 0.002%
(c) 0.006%
(d) 0.004%
Show Answer & Explanation

Answer: (d) 0.004%

Explanation:
1. \([HCOO^-] = 10^{-3}\) M.
2. Molarity of pure formic acid = 1150/46 = 25 M.
3. Percentage = \(10^{-3}/25 \times 100 = 0.004\)%.

Teacher's Note:
1. Compare with the molarity of the pure liquid.

Question 2: The \(pK_a\) of an acid HA is 6. The equilibrium constant for \(A^- + H_3O^+ \rightleftharpoons HA + H_2O\) is:
(a) \(1 \times 10^{6}\)
(b) \(1 \times 10^{-6}\)
(c) \(1 \times 10^{-8}\)
(d) \(1 \times 10^{8}\)
Show Answer & Explanation

Answer: (a) \(1 \times 10^{6}\)

Explanation:
1. This is the reverse of ionisation of HA, so \(K = 1/K_a = 10^{6}\).

Teacher's Note:
1. Reverse reaction means reciprocal constant.

Question 3: \(10^{-6}\) M HCl is diluted 100 times. Its pH is:
(a) 6.98
(b) 7.02
(c) 8.0
(d) 6.0
Show Answer & Explanation

Answer: (a) 6.98

Explanation:
1. After dilution the acid gives \(10^{-8}\) M H\(^+\), and water contributes.
2. \(x^2 - 10^{-8}x - 10^{-14} = 0\) gives \([H^+] = 1.05 \times 10^{-7}\) M, pH = 6.98.

Teacher's Note:
1. An acid solution can never reach pH above 7.

Question 4: Which solution has a pH exactly equal to 8?
(a) \(10^{-8}\) M NaOH
(b) a solution containing exactly \(10^{-8}\) M H\(^+\)
(c) \(2 \times 10^{-6}\) M Ba(OH)\(_2\)
(d) \(10^{-8}\) M HCl
Show Answer & Explanation

Answer: (b) a solution containing exactly \(10^{-8}\) M H\(^+\)

Explanation:
1. By definition, \([H^+] = 10^{-8}\) M means pH = 8.
2. The others are affected by water's own ionisation or give different [OH\(^-\)].

Teacher's Note:
1. The definition of pH.

Question 5: 10 mL of a strong acid of pH 2 is mixed with 990 mL of a strong acid of pH 4. The pH of the mixture is:
(a) 3.3
(b) 3.7
(c) cannot be found without the formulae
(d) 3.0
Show Answer & Explanation

Answer: (b) 3.7

Explanation:
1. H\(^+\) = 10 x 0.01 + 990 x 0.0001 = 0.1 + 0.099 = 0.199 mmol in 1000 mL.
2. \([H^+] = 1.99 \times 10^{-4}\) M, pH = 3.7.

Teacher's Note:
1. Add moles, divide by total volume.

Question 6: The percentage dissociation of a 0.024 M solution of a weak acid HA (\(K_a = 2 \times 10^{-3}\)) is:
(a) 0.29%
(b) 25%
(c) 0.25%
(d) 29%
Show Answer & Explanation

Answer: (b) 25%

Explanation:
1. \(\alpha\) is large, so solve \(0.024\alpha^2/(1-\alpha) = 2 \times 10^{-3}\).
2. \(\alpha = 0.25\), i.e. 25%.

Teacher's Note:
1. Do not use the approximation when \(\alpha\) is above 10%.

Question 7: For a weak base BOH, \(K_b = 10^{-4}\). The pH of \(10^{-4}\) M BOH is (\(\log 6.2 = 0.79\)):
(a) none of these
(b) 9.79
(c) 8
(d) 10
Show Answer & Explanation

Answer: (b) 9.79

Explanation:
1. Solve \(x^2/(10^{-4} - x) = 10^{-4}\): \(x = 6.18 \times 10^{-5}\) M.
2. pOH = 4.21, so pH = 9.79.

Teacher's Note:
1. Use the quadratic for a base that is highly ionised.

Question 8: Four 0.1 M solutions of NaW, NaX, NaY and NaZ have pH 7, 9, 10 and 11 respectively. The strongest acid is:
(a) HY
(b) HW
(c) HX
(d) HZ
Show Answer & Explanation

Answer: (b) HW

Explanation:
1. The lower the pH of the sodium salt solution, the weaker the conjugate base and so the stronger the acid.
2. NaW has pH 7 (no hydrolysis), so HW is a strong acid.

Teacher's Note:
1. Stronger acid means weaker conjugate base.

Question 9: For HA + B\(^-\) \(\rightleftharpoons\) HB + A\(^-\) with \(K_{eq} = 10^{-4}\), how many of these statements are true? (i) HB is a stronger acid than HA (ii) HA is stronger than HB (iii) equal strength (iv) B\(^-\) is a stronger base than A\(^-\) (v) A\(^-\) is a stronger base than B\(^-\) (vi) B\(^-\) and HB are a conjugate pair (vii) A\(^-\) is the conjugate base of HA (viii) HA can be HSO\(_4^-\) and HB can be HCOOH (ix) A\(^-\) can be F\(^-\) and B\(^-\) can be CN\(^-\)
(a) 3
(b) 5
(c) 4
(d) 6
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. \(K_{eq} = K_a(HA)/K_a(HB) = 10^{-4}\), so HB is the stronger acid and A\(^-\) the stronger base.
2. True: (i), (v), (vi), (vii).
3. (viii) and (ix) would need HA stronger than HB, which contradicts the data.

Teacher's Note:
1. Equilibrium favours the weaker acid.

Question 10: What is the pOH of an aqueous solution with \([H^+] = 10^{-2}\) M and \(K_w = 2 \times 10^{-12}\)?
(a) 12.0
(b) 2.0
(c) 10.0
(d) 9.7
Show Answer & Explanation

Answer: (d) 9.7

Explanation:
1. \([OH^-] = 2 \times 10^{-12}/10^{-2} = 2 \times 10^{-10}\) M.
2. pOH = 10 - log 2 = 9.7.

Teacher's Note:
1. Use the given \(K_w\), not \(10^{-14}\).

Question 11: The percentage ionisation of water at a certain temperature is \(3.6 \times 10^{-7}\)%. Taking [H\(_2\)O] = 55.55 M, \(K_w\) and pH of water at this temperature are:
(a) \(4 \times 10^{-14}\) and 6.7
(b) \(2 \times 10^{-7}\) and 6.7
(c) \(4 \times 10^{-14}\) and 7.4
(d) \(10^{-14}\) and 7
Show Answer & Explanation

Answer: (a) \(4 \times 10^{-14}\) and 6.7

Explanation:
1. \([H^+] = 55.55 \times 3.6 \times 10^{-9} = 2 \times 10^{-7}\) M.
2. \(K_w = (2 \times 10^{-7})^2 = 4 \times 10^{-14}\) and pH = 6.7.

Teacher's Note:
1. Ionisation percent applies to the 55.55 M water.

Question 12: 0.1 mol HCl is dissolved in water of volume V. As V tends to infinity, the pH of the solution approaches:
(a) 7
(b) infinity
(c) 0
(d) 14
Show Answer & Explanation

Answer: (a) 7

Explanation:
1. As the acid is infinitely diluted, H\(^+\) from HCl tends to zero and water's own \(10^{-7}\) M dominates.
2. The pH tends to 7.

Teacher's Note:
1. An acid solution can never become basic by dilution.

Question 13: The value of \(pK_a(H_2O) + pK_a(H_3O^+)\) is:
(a) 7
(b) 14
(c) -1.74
(d) 15.7
Show Answer & Explanation

Answer: (b) 14

Explanation:
1. \(K_a(H_2O) = K_w/55.5\), \(pK_a = 15.74\).
2. \(K_a(H_3O^+) = 55.5\), \(pK_a = -1.74\).
3. The sum is \(pK_w = 14\).

Teacher's Note:
1. The two are a conjugate pair.

Question 14: The concentration of CH\(_3\)COOH in a solution obtained by dissolving 0.01 mol of ammonium acetate in 1 L of water (\(K_a = K_b = 1.8 \times 10^{-5}\)), multiplied by \(9 \times 10^5\), is closest to:
(a) 100
(b) 25
(c) 5
(d) 50
Show Answer & Explanation

Answer: (d) 50

Explanation:
1. \(K_h = K_w/(K_aK_b) = 3.1 \times 10^{-5}\), so \(h/(1-h) = 5.56 \times 10^{-3}\) and \(h = 5.5 \times 10^{-3}\).
2. \([CH_3COOH] = 0.01 \times 5.5 \times 10^{-3} = 5.5 \times 10^{-5}\) M, and \(\times 9 \times 10^5 = 50\).

Teacher's Note:
1. Cation and anion hydrolyse to the same extent.

Question 15: Which statements are correct? (A) All Bronsted bases are Lewis bases. (B) Not all Bronsted acids are Lewis acids. (C) Most cations are acids and most anions are bases. (D) All Bronsted bases are Arrhenius bases.
(a) B, C and D only
(b) A, B and C only
(c) A and B only
(d) all four
Show Answer & Explanation

Answer: (b) A, B and C only

Explanation:
1. A Bronsted base has a lone pair, so it is a Lewis base (A true).
2. A Bronsted acid donates H\(^+\) and does not necessarily accept electron pairs (B true).
3. Cations tend to be acidic and anions basic (C true).
4. NH\(_3\) is a Bronsted base but not an Arrhenius base (D false).

Teacher's Note:
1. Lewis, Bronsted and Arrhenius definitions overlap but are not identical.

Question 16: Select the incorrect statements: (A) NH\(_4^+\) is a strong acid. (B) CH\(_3\)COO\(^-\) is a weak base. (C) H\(^-\) is a weak base. (D) HS\(^-\) is both a weak acid and a weak base.
(a) B and D only
(b) A only
(c) C only
(d) A and C only
Show Answer & Explanation

Answer: (d) A and C only

Explanation:
1. NH\(_4^+\) is a weak acid, so (A) is incorrect.
2. The hydride ion is an extremely strong base, so (C) is incorrect.
3. (B) and (D) are correct.

Teacher's Note:
1. Strong bases have very weak conjugate acids.

Question 17: For pure water, which are correct? (A) pK\(_w\) increases with a decrease in temperature. (B) The absolute dissociation constant decreases with a decrease in temperature. (C) The degree of dissociation of water increases with a decrease in temperature. (D) Both pH and pOH decrease with a rise in temperature.
(a) A, B and C
(b) A, B and D
(c) A and C
(d) B, C and D
Show Answer & Explanation

Answer: (b) A, B and D

Explanation:
1. Ionisation is endothermic, so lower temperature gives smaller \(K_w\) (A, B true) and a lower degree of dissociation (C false).
2. Higher temperature lowers both pH and pOH (D true).

Teacher's Note:
1. Ionisation of water increases with temperature.

Question 18: One litre of a strong acid solution has \(10^{-5}\) mol of H\(^+\). Which are correct? (A) pOH = 9 (B) percentage ionisation of water = \(1.8 \times 10^{-9}\)% (C) OH\(^-\) ions per mL = \(6.022 \times 10^{15}\) (D) [H\(^+\)] from water = \(10^{-9}\) M
(a) B and C
(b) A, B and C
(c) A and C
(d) A, B and D
Show Answer & Explanation

Answer: (d) A, B and D

Explanation:
1. pH = 5, so pOH = 9 and \([OH^-] = 10^{-9}\) M, which equals the H\(^+\) from water.
2. Percentage ionisation = \(10^{-9}/55.5 \times 100 = 1.8 \times 10^{-9}\)%.
3. OH\(^-\) per mL = \(10^{-12} \times 6.022 \times 10^{23} = 6.022 \times 10^{11}\), so (C) is false.

Teacher's Note:
1. Check the unit conversion for 1 mL.

Question 19: Equal volumes of 0.2 M HCl and 0.2 M Sr(OH)\(_2\) are mixed. Which are correct? (A) [Sr\(^{2+}\)] = 0.1 M (B) [Cl\(^-\)] = 0.1 M (C) pH = 13 (D) the solution is neutral
(a) B and C only
(b) A, B and C
(c) A and D
(d) A, B and D
Show Answer & Explanation

Answer: (b) A, B and C

Explanation:
1. After mixing, [Sr\(^{2+}\)] = 0.1 M and [Cl\(^-\)] = 0.1 M.
2. OH\(^-\) = 0.2 x 2 = 0.4 against H\(^+\) 0.2 (before mixing), leaving 0.2 M in the doubled volume, i.e. 0.1 M, pH = 13.
3. The solution is basic.

Teacher's Note:
1. Find the excess OH\(^-\).

Question 20: When 0.1 M CH\(_3\)COOH is diluted at 25\(^\circ\)C, which are correct? (A) [H\(^+\)] increases (B) pH increases (C) the number of H\(^+\) ions increases (D) K\(_a\) increases
(a) B and D
(b) A and C
(c) A and D
(d) B and C
Show Answer & Explanation

Answer: (d) B and C

Explanation:
1. Dilution lowers [H\(^+\)], so pH rises (B true, A false).
2. The degree of ionisation increases, so the number of H\(^+\) ions increases (C true).
3. K\(_a\) depends only on temperature.

Teacher's Note:
1. Degree of ionisation grows on dilution.

Question 21: The degree of hydrolysis of the salt of a strong acid and a weak base: (A) is independent of dilution (B) increases with dilution (C) increases with decrease in K\(_b\) (D) decreases with a decrease in temperature. Which are correct?
(a) B, C and D
(b) A and C
(c) B only
(d) A, B and D
Show Answer & Explanation

Answer: (a) B, C and D

Explanation:
1. \(h = \sqrt{K_w/(K_bC)}\) rises on dilution and for weaker bases.
2. Hydrolysis is endothermic, so it decreases at lower temperature.

Teacher's Note:
1. Use the formula for \(h\).

Question 22: A strong acid solution of pH 4 (1 L) is added to 7/3 L of water (\(\log 3.7 = 0.568\)). The pH of the resulting solution is:
(a) 4.4
(b) 4.365
(c) 4.52
(d) 4.432
Show Answer & Explanation

Answer: (c) 4.52

Explanation:
1. Total volume = 10/3 L, so \([H^+] = 10^{-4} \times 3/10 = 3 \times 10^{-5}\) M.
2. pH = 5 - log 3 = 4.52.

Teacher's Note:
1. This acid is not dilute enough for water's contribution to matter.

Question 23: In the hydrolysis of CH\(_3\)COONH\(_4\) (salt of a weak acid and a weak base), the degrees of hydrolysis of the cation and the anion are:
(a) exactly the same always
(b) zero
(c) different, but very close when they hydrolyse in the presence of each other
(d) different and can never be treated as similar
Show Answer & Explanation

Answer: (c) different, but very close when they hydrolyse in the presence of each other

Explanation:
1. Each ion's hydrolysis removes the other ion's product (H\(^+\) with OH\(^-\)), driving both forward.
2. This coupling makes the two degrees of hydrolysis nearly equal.

Teacher's Note:
1. The combination of H\(^+\) and OH\(^-\) couples both reactions.

Question 24: For 0.1 M CH\(_3\)COONH\(_4\) with \(K_a(CH_3COOH) = K_b(NH_3) = 2 \times 10^{-5}\), the degrees of hydrolysis of the cation and anion are:
(a) very different
(b) cannot be said
(c) slightly different
(d) exactly the same
Show Answer & Explanation

Answer: (d) exactly the same

Explanation:
1. With \(K_a = K_b\), the hydrolysis constants of NH\(_4^+\) and CH\(_3\)COO\(^-\) are equal.
2. Their concentrations are also equal, so their degrees of hydrolysis are identical (and pH = 7).

Teacher's Note:
1. Symmetry gives equal degrees.

Question 25: The conjugate base of H\(_2\)PO\(_4^-\) is:
(a) PO\(_4^{3-}\)
(b) P\(_2\)O\(_5\)
(c) H\(_3\)PO\(_4\)
(d) HPO\(_4^{2-}\)
Show Answer & Explanation

Answer: (d) HPO\(_4^{2-}\)

Explanation:
1. Removing one proton from H\(_2\)PO\(_4^-\) gives HPO\(_4^{2-}\).

Teacher's Note:
1. Conjugate base means minus one H\(^+\).

Question 26: The conjugate base of OH\(^-\) is:
(a) O\(_2\)
(b) O\(^{2-}\)
(c) H\(_2\)O
(d) O\(^-\)
Show Answer & Explanation

Answer: (b) O\(^{2-}\)

Explanation:
1. Removing H\(^+\) from OH\(^-\) leaves O\(^{2-}\).

Teacher's Note:
1. Remove a proton.

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