JEE Chemistry Ionic Equilibrium MCQs Set 33

Chemistry Objective Questions and Answers: Ionic Equilibrium

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Multiple Choice Questions

Question 1: What is the Kb of a weak base that produces one OH− ion per molecule, if a 0.05 M solution is 2.5% ionised?
(a) 7.8 × 10−8
(b) 1.6 × 10−6
(c) 3.1 × 10−5
(d) 1.2 × 10−3
Show Answer & Explanation

Answer: (c) 3.1 × 10−5

Explanation:
1. Kb = Cα2 = 0.05 × (0.025)2 = 3.1 × 10−5.
2. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 2: 0.2 g of benzoic acid (C6H5COOH, molar mass 122) is titrated with 0.12 M Ba(OH)2. The volume of Ba(OH)2 needed to reach the equivalence point is:
(a) 6.83 mL
(b) 13.6 mL
(c) 17.6 mL
(d) 35.2 mL
Show Answer & Explanation

Answer: (a) 6.83 mL

Explanation:
1. Moles of acid = 0.2/122 = 1.64 × 10−3.
2. Ba(OH)2 supplies 2 OH−, so moles of base = 0.82 × 10−3.
3. V = 0.82 × 10−3/0.12 = 6.83 mL.
4. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 3: The degree of dissociation of water at 25°C is 1.8 × 10−7 % and its density is 1.0 g/cm3. The ionisation constant of water is:
(a) 1.0 × 10−15
(b) 1.0 × 10−14
(c) 1.8 × 10−16
(d) 1.0 × 10−6
Show Answer & Explanation

Answer: (c) 1.8 × 10−16

Explanation:
1. [H2O] = 1000/18 = 55.5 M and α = 1.8 × 10−9.
2. K = Cα2 = 55.5 × (1.8 × 10−9)2 = 1.8 × 10−16.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 4: A certain weak acid has a dissociation constant of 1.0 × 10−4. The equilibrium constant for its reaction with a strong base is:
(a) 1.0 × 10−4
(b) 1.0 × 10−10
(c) 1.0 × 1010
(d) 1.0 × 1014
Show Answer & Explanation

Answer: (c) 1.0 × 1010

Explanation:
1. HA + OH− ⇌ A− + H2O has K = Ka/Kw.
2. K = 10−4/10−14 = 1010.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 5: Calculate the pH of a 10−3 M solution of Ba(OH)2 assuming complete ionisation (Kw = 1 × 10−14).
(a) 12.30
(b) 11.30
(c) 10.00
(d) 9.00
Show Answer & Explanation

Answer: (b) 11.30

Explanation:
1. [OH−] = 2 × 10−3 M, so pOH = 3 − log 2 = 2.70.
2. pH = 14 − 2.70 = 11.30.
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 6: 0.01 mole of sodium hydroxide is added to 10 litres of pure water. By how many pH units does the pH of the water change?
(a) 4
(b) 7
(c) 5
(d) 8
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. [OH−] = 0.01/10 = 10−3 M, so pOH = 3 and pH = 11.
2. Pure water had pH 7, so the change is 11 − 7 = 4 units.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 7: The degree of dissociation of acetic acid in a 0.1 M solution is 1.32 × 10−2. Find its pKa:
(a) 5.75
(b) 3.75
(c) 4.00
(d) 4.75
Show Answer & Explanation

Answer: (d) 4.75

Explanation:
1. Ka = Cα2 = 0.1 × (1.32 × 10−2)2 = 1.74 × 10−5.
2. pKa = 4.76 ≈ 4.75.
3. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 8: Calculate the pH of a 10−5 M HCl solution if 1 mL of it is diluted to 1000 mL (Kw = 1 × 10−14).
(a) 7.98
(b) 6.98
(c) 7.0
(d) 5
Show Answer & Explanation

Answer: (b) 6.98

Explanation:
1. Diluted concentration = 10−8 M, so water's own H+ cannot be ignored.
2. [H+] = 10−8 + x where x(x + 10−8) = 10−14, giving x = 9.5 × 10−8.
3. Total [H+] = 1.05 × 10−7 M, so pH = 6.98.
4. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 9: Calculate the degree of hydrolysis of a 0.01 M solution of KF (Ka of HF = 6.6 × 10−4):
(a) 3.87 × 10−6
(b) 3.87 × 10−5
(c) 3.87 × 10−2
(d) None of these
Show Answer & Explanation

Answer: (b) 3.87 × 10−5

Explanation:
1. Kh = Kw/Ka = 10−14/(6.6 × 10−4) = 1.5 × 10−11.
2. h = √(Kh/C) = √(1.5 × 10−9) = 3.9 × 10−5 (closest: 3.87 × 10−5).
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 10: For a polybasic acid H3A, the dissociation constants for the successive steps are Ka1, Ka2 and Ka3. The observed trend is:
(a) Ka1 > Ka2 > Ka3
(b) Ka1 = Ka2 = Ka3
(c) Ka1 < Ka2 < Ka3
(d) Ka1 = Ka2 + Ka3
Show Answer & Explanation

Answer: (a) Ka1 > Ka2 > Ka3

Explanation:
1. Removing H+ from an already negative ion is harder because of the increased electrostatic attraction.
2. So Ka1 > Ka2 > Ka3.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 11: Equimolar solutions of HF, HCOOH and HCN at 298 K have Ka values 6.8 × 10−4, 1.8 × 10−4 and 4.8 × 10−9 respectively. The order of their acidic strength is:
(a) HF > HCN > HCOOH
(b) HF > HCOOH > HCN
(c) HCN > HF > HCOOH
(d) HCOOH > HCN > HF
Show Answer & Explanation

Answer: (b) HF > HCOOH > HCN

Explanation:
1. A larger Ka means a stronger acid.
2. 6.8 × 10−4 > 1.8 × 10−4 > 4.8 × 10−9.
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 12: NH4CN is the salt of a weak acid HCN (Ka = 6.2 × 10−10) and a weak base NH4OH (Kb = 1.8 × 10−5). A molar solution of NH4CN will be:
(a) Neutral
(b) Strongly acidic
(c) Strongly basic
(d) Weakly basic
Show Answer & Explanation

Answer: (d) Weakly basic

Explanation:
1. pH = 7 + ½(pKa − pKb) = 7 + ½(9.21 − 4.74) = 9.2.
2. The base is stronger than the acid, so the solution is only weakly basic.
3. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 13: The solubility product of BaCl2 is 4 × 10−9. What is its solubility in mol/L?
(a) 4 × 10−3
(b) 3.2 × 10−9
(c) 10−3
(d) 1 × 10−9
Show Answer & Explanation

Answer: (c) 10−3

Explanation:
1. Ksp = 4s3 = 4 × 10−9.
2. s3 = 10−9, so s = 10−3 M.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 14: The dissociation constant of CH3COOH and NH4OH in aqueous solution is 10−5 each. If the pH of a CH3COOH solution is 3, what will be the pH of an NH4OH solution of the same concentration?
(a) 3.0
(b) 4.0
(c) 10.0
(d) 11.0
Show Answer & Explanation

Answer: (d) 11.0

Explanation:
1. For the acid, [H+]2 = KaC gives 10−6 = 10−5C, so C = 0.1 M.
2. For the base, [OH−] = √(10−5 × 0.1) = 10−3 M, so pOH = 3.
3. pH = 11.
4. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 15: What concentration of SO42− is required to start the precipitation of BaSO4 in a solution containing 1 × 10−4 mol L−1 of Ba2+ (Ksp of BaSO4 = 4 × 10−10)?
(a) 4 × 10−10 M
(b) 2 × 10−10 M
(c) 4 × 10−6 M
(d) 2 × 10−3 M
Show Answer & Explanation

Answer: (c) 4 × 10−6 M

Explanation:
1. Precipitation starts when the ionic product equals Ksp.
2. [SO42−] = 4 × 10−10/10−4 = 4 × 10−6 M.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 16: The solubility product of AgCl is 1.5625 × 10−10 at 25°C. Its solubility in grams per litre will be (Ag = 108, Cl = 35.5):
(a) 143.5
(b) 108
(c) 1.57 × 10−8
(d) 1.79 × 10−3
Show Answer & Explanation

Answer: (d) 1.79 × 10−3

Explanation:
1. s = √Ksp = 1.25 × 10−5 mol L−1.
2. Mass = 1.25 × 10−5 × 143.5 = 1.79 × 10−3 g L−1.
3. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 17: The solubility product of radium sulphate is 4 × 10−11. What will be the solubility of Ra2+ in 0.10 M Na2SO4?
(a) 4 × 10−10 M
(b) 2 × 10−5 M
(c) 4 × 10−5 M
(d) 2 × 10−10 M
Show Answer & Explanation

Answer: (a) 4 × 10−10 M

Explanation:
1. [SO42−] ≈ 0.10 M from Na2SO4.
2. [Ra2+] = 4 × 10−11/0.10 = 4 × 10−10 M.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 18: Which of the following species can act as an acid as well as a base?
(a) SO42−
(b) HSO4−
(c) PO43−
(d) OH−
Show Answer & Explanation

Answer: (b) HSO4−

Explanation:
1. HSO4− can donate H+ (giving SO42−) and can accept H+ (giving H2SO4).
2. SO42−, PO43− and OH− can only accept H+ (OH− is not a practical acid).
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 19: In which of the following is silver chloride most easily soluble?
(a) 0.1 mol dm−3 AgNO3 solution
(b) 0.1 mol dm−3 HCl solution
(c) H2O
(d) Aqueous NH3
Show Answer & Explanation

Answer: (d) Aqueous NH3

Explanation:
1. AgNO3 and HCl add common ions (Ag+, Cl−) and lower the solubility.
2. Aqueous NH3 forms the soluble complex [Ag(NH3)2]+, pulling the equilibrium forward.
3. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 20: The pH of 10−6 M HCl(aq) is:
(a) just less than 6
(b) exactly equal to 6
(c) just greater than 6
(d) just less than 7
Show Answer & Explanation

Answer: (a) just less than 6

Explanation:
1. Water also contributes H+ (about 10−8 M here), so [H+] is slightly above 10−6 M.
2. pH ≈ 5.996, i.e. just less than 6.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 21: How much water must be added to 300 mL of 0.2 M CH3COOH (Ka = 1.8 × 10−5) for the degree of dissociation to double?
(a) 600 mL
(b) 900 mL
(c) 1200 mL
(d) 1500 mL
Show Answer & Explanation

Answer: (b) 900 mL

Explanation:
1. α = √(Ka/C), so α doubles when C falls to one quarter.
2. Final volume = 4 × 300 = 1200 mL, so water added = 1200 − 300 = 900 mL.
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 22: Selenious acid (H2SeO3), a diprotic acid, has Ka1 = 10−6 and Ka2 = 10−8. The approximate pH of 0.01 M NaHSeO3 is given by:
(a) pH = 7 + pKa1/2 + log C/2
(b) pH = 7 − pKa1/2 − log C/2
(c) pH = (pKa1 + pKa2)/2
(d) pH = 7 + pKa1/2 − pKa2/2
Show Answer & Explanation

Answer: (c) pH = (pKa1 + pKa2)/2

Explanation:
1. HSeO3− is amphiprotic (it can both donate and accept H+).
2. For such an ion pH = ½(pKa1 + pKa2) = ½(6 + 8) = 7.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 23: The solubility of AgCl in 0.2 M NaCl is x and that in 0.1 M AgNO3 is y. Which of the following is correct?
(a) x = y
(b) x > y
(c) x < y
(d) Data insufficient
Show Answer & Explanation

Answer: (c) x < y

Explanation:
1. In 0.2 M NaCl: x = Ksp/0.2. In 0.1 M AgNO3: y = Ksp/0.1.
2. The common ion concentration is larger in NaCl, so x is smaller: x < y.
3. Answer: (c)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 24: What is the pH of a saturated solution of Cu(OH)2? (Ksp = 3.2 × 10−20)
(a) 6.4
(b) 7.6
(c) 7.3
(d) 7.9
Show Answer & Explanation

Answer: (b) 7.6

Explanation:
1. Ksp = 4s3 gives s3 = 8 × 10−21, so s = 2 × 10−7 M.
2. [OH−] = 2s = 4 × 10−7 M, so pOH = 6.4 and pH = 7.6.
3. Answer: (b)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 25: An acid-base indicator has KHIn = 3.0 × 10−5. The acid form is red and the basic form is blue. The change in [H+] required to change the indicator from 75% red to 75% blue is:
(a) 8 × 10−5 M
(b) 9 × 10−5 M
(c) 10−5 M
(d) 3 × 10−4 M
Show Answer & Explanation

Answer: (a) 8 × 10−5 M

Explanation:
1. [H+] = KHIn[HIn]/[In−]. At 75% red this is 3 × 10−5 × 3 = 9 × 10−5 M.
2. At 75% blue it is 3 × 10−5/3 = 10−5 M.
3. Change = 9 × 10−5 − 1 × 10−5 = 8 × 10−5 M.
4. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 26: pKa of NH4+ is 9.26. Hence the effective pH range for an NH4OH − NH4Cl buffer is:
(a) 8.26 to 10.26
(b) 4.74 to 5.74
(c) 3.74 to 5.74
(d) 8.26 to 9.26
Show Answer & Explanation

Answer: (a) 8.26 to 10.26

Explanation:
1. A buffer works for pH = pKa ± 1.
2. pH = 9.26 ± 1 = 8.26 to 10.26.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 27: The pH of a 0.10 M CH3COOH solution increases when which of the following is added?
(a) NaHSO4
(b) HClO4
(c) KNO3
(d) K2CO3
Show Answer & Explanation

Answer: (d) K2CO3

Explanation:
1. NaHSO4 and HClO4 add H+, lowering the pH; KNO3 is neutral.
2. K2CO3 is basic: CO32− consumes H+, so the pH rises.
3. Answer: (d)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 28: H2CO3 and NaHCO3 constitute the buffer system in blood and maintain its pH close to 7.4. An excess of acid entering the blood stream is removed by:
(a) HCO3−
(b) H2CO3
(c) H+
(d) CO32−
Show Answer & Explanation

Answer: (a) HCO3−

Explanation:
1. The basic component of the buffer, HCO3−, neutralises added H+.
2. HCO3− + H+ → H2CO3.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

Question 29: The pKa for acid A is greater than the pKa for acid B. The stronger acid is:
(a) B
(b) A
(c) Both A and B
(d) None of these
Show Answer & Explanation

Answer: (a) B

Explanation:
1. A smaller pKa means a larger Ka.
2. So acid B, with the smaller pKa, is stronger.
3. Answer: (a)

Teacher's Note:
1. Write the governing relation first, then substitute. Check that the final value is reasonable.

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