JEE Chemistry Ionic Equilibrium MCQs Set 27

Chemistry Objective Questions and Answers: Ionic Equilibrium

Access targeted multiple-choice questions for Ionic Equilibrium designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Ionic Equilibrium MCQs with Answers

View or download the dedicated Ionic Equilibrium MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: The first and second dissociation constants of H\(_2\)A are \(1.0 \times 10^{-5}\) and \(5.0 \times 10^{-10}\). The overall dissociation constant is:
(a) \(5.0 \times 10^{-5}\)
(b) \(5.0 \times 10^{15}\)
(c) \(0.2 \times 10^{5}\)
(d) \(5.0 \times 10^{-15}\)
Show Answer & Explanation

Answer: (d) \(5.0 \times 10^{-15}\)

Explanation:
1. \(K = K_1K_2 = 1.0 \times 10^{-5} \times 5.0 \times 10^{-10} = 5.0 \times 10^{-15}\).

Teacher's Note:
1. Overall constants multiply.

Question 2: How many litres of water must be added to 1 L of HCl solution of pH 1 to obtain a solution of pH 2?
(a) 9.0
(b) 0.1
(c) 0.9
(d) 2.0
Show Answer & Explanation

Answer: (a) 9.0

Explanation:
1. Concentration falls from 0.1 M to 0.01 M, a tenfold dilution.
2. Final volume 10 L, so 9 L is added.

Teacher's Note:
1. Each pH unit is a factor of 10 in concentration.

Question 3: The \(pK_a\) of a weak acid HA is 3.2 and the \(pK_b\) of a weak base BOH is 3.4. The pH of a solution of salt AB is:
(a) 1.0
(b) 6.9
(c) 7.0
(d) 7.2
Show Answer & Explanation

Answer: (b) 6.9

Explanation:
1. \(pH = 7 + \frac12(3.2 - 3.4) = 6.9\).

Teacher's Note:
1. The acid is slightly stronger than the base is weak.

Question 4: Which of FeCl\(_3\), Pb(CH\(_3\)COO)\(_2\), Al(CN)\(_3\) and CH\(_3\)COOK is the most basic in aqueous solution?
(a) Pb(CH\(_3\)COO)\(_2\)
(b) FeCl\(_3\)
(c) Al(CN)\(_3\)
(d) CH\(_3\)COOK
Show Answer & Explanation

Answer: (d) CH\(_3\)COOK

Explanation:
1. Potassium acetate comes from a strong base and a weak acid, giving a clearly basic solution.
2. FeCl\(_3\) is acidic, and Pb and Al salts have acidic cations that offset the basic anions.

Teacher's Note:
1. Spectator K\(^+\) leaves the anion's hydrolysis unopposed.

Question 5: The pH of 0.1 M sodium acetate (\(K_a = 1.0 \times 10^{-5}\)), assuming small hydrolysis, is:
(a) 5.0
(b) 8.0
(c) 9.0
(d) 6.0
Show Answer & Explanation

Answer: (c) 9.0

Explanation:
1. \(pH = \frac12(14 + 5 - 1) = 9\).

Teacher's Note:
1. Salt of a weak acid is basic.

Question 6: The conjugate base of hydrazoic acid (HN\(_3\)) is:
(a) N\(^{3-}\)
(b) HN\(_3^-\)
(c) N\(_2^-\)
(d) N\(_3^-\)
Show Answer & Explanation

Answer: (d) N\(_3^-\)

Explanation:
1. Removing one proton from HN\(_3\) gives the azide ion N\(_3^-\).

Teacher's Note:
1. Remove H\(^+\), keep the three nitrogens.

Question 7: Which mixture has pH exactly 1? (1) 100 mL M/10 HCl + 100 mL M/10 NaOH (2) 75 mL M/5 HCl + 25 mL M/5 NaOH (3) 60 mL M/10 HCl + 40 mL M/10 NaOH (4) 55 mL M/10 HCl + 45 mL M/10 NaOH
(a) Mixture 1
(b) Mixture 4
(c) Mixture 2
(d) Mixture 3
Show Answer & Explanation

Answer: (c) Mixture 2

Explanation:
1. Mixture 2: 15 mmol H\(^+\) - 5 mmol OH\(^-\) = 10 mmol in 100 mL = 0.1 M, pH = 1.
2. The others give pH 7, 1.7 and 2.

Teacher's Note:
1. Compute the excess acid per mixture.

Question 8: Which pairs of solutions among NH\(_4\)OH (S1), (NH\(_4\))\(_2\)SO\(_4\) (S2), HCl (S3), KOH (S4) can give a buffer?
(a) S3 + S4 only
(b) S1 + S4 only
(c) S1 + S2, S1 + S3 (S1 in excess) and S2 + S4 (S2 in excess)
(d) S2 + S3 only
Show Answer & Explanation

Answer: (c) S1 + S2, S1 + S3 (S1 in excess) and S2 + S4 (S2 in excess)

Explanation:
1. NH\(_4\)OH with its salt (NH\(_4\))\(_2\)SO\(_4\) gives a basic buffer.
2. NH\(_4\)OH with limited HCl forms NH\(_4^+\) and leaves NH\(_3\).
3. (NH\(_4\))\(_2\)SO\(_4\) with limited KOH forms NH\(_3\) and leaves NH\(_4^+\).

Teacher's Note:
1. Look for combinations that leave both NH\(_3\) and NH\(_4^+\).

Question 9: The pH of a solution containing 4 g CH\(_3\)COOH and 4.1 g CH\(_3\)COONa in 100 mL (\(K_a = 1.8 \times 10^{-5}\)) is:
(a) 4.62
(b) 4.74
(c) 5.0
(d) 4.22
Show Answer & Explanation

Answer: (a) 4.62

Explanation:
1. Acid = 4/60 = 0.0667 mol; salt = 4.1/82 = 0.05 mol.
2. \(pH = 4.745 + \log(0.05/0.0667) = 4.745 - 0.125 = 4.62\).

Teacher's Note:
1. Henderson's equation.

Question 10: The pH of 5 mL of 0.1 M BOH (\(K_b = 1.8 \times 10^{-5}\)) mixed with 25 mL of 0.1 M BCl is:
(a) 8.56
(b) 4.74
(c) 9.26
(d) 5.44
Show Answer & Explanation

Answer: (a) 8.56

Explanation:
1. BOH = 0.5 mmol, BCl = 2.5 mmol.
2. \(pOH = pK_b + \log(2.5/0.5) = 4.745 + 0.70 = 5.44\), so pH = 8.56.

Teacher's Note:
1. Basic buffer: use pOH.

Question 11: 50 mL of 0.2 M HA (\(K_a = 10^{-5}\)) is mixed with 50 mL of NaA to give a buffer of pH 4. The concentration of the NaA solution should be:
(a) 0.1 M
(b) 0.002 M
(c) 0.02 M
(d) 0.2 M
Show Answer & Explanation

Answer: (c) 0.02 M

Explanation:
1. After mixing, [HA] = 0.1 M.
2. \(4 = 5 + \log([A^-]/0.1)\) gives \([A^-] = 0.01\) M after mixing, i.e. 0.02 M before mixing.

Teacher's Note:
1. Allow for the doubling of volume.

Question 12: A 0.5 L buffer is 0.2 M NH\(_4\)Cl and 0.2 M NH\(_3\) (\(pK_b\) = 4.74). The pH before addition, after adding 0.05 mol NaOH and after adding 0.05 mol HCl are respectively:
(a) 9.26, 9.74, 8.78
(b) 9.26, 10.0, 8.5
(c) 9.26, 8.78, 9.74
(d) 9.26, 9.26, 9.26
Show Answer & Explanation

Answer: (a) 9.26, 9.74, 8.78

Explanation:
1. Initially \(pOH = 4.74\), pH = 9.26.
2. With OH\(^-\): NH\(_4^+\) = 0.05 mol, NH\(_3\) = 0.15 mol, \(pOH = 4.74 + \log(1/3)\) so pH = 9.74.
3. With H\(^+\): NH\(_4^+\) = 0.15 mol, NH\(_3\) = 0.05 mol, pH = 8.78.

Teacher's Note:
1. Base converts NH\(_4^+\) to NH\(_3\) and raises pH.

Question 13: A solution is 0.1 M in CH\(_3\)COOH and 0.1 M in CH\(_3\)COONa. Which of the following will change its pH significantly?
(a) addition of a small amount of NaOH
(b) addition of a small amount of HCl
(c) none of these
(d) addition of a small amount of water
Show Answer & Explanation

Answer: (c) none of these

Explanation:
1. It is a buffer, so small amounts of acid, base or water do not change the pH significantly.

Teacher's Note:
1. Buffers resist small additions.

Question 14: Which can be added to one litre of water to act as a buffer?
(a) 1 mol NH\(_4\)OH and 1 mol NaOH
(b) 1 mol CH\(_3\)COOH and 0.5 mol NaOH
(c) 1 mol CH\(_3\)COOH and 1 mol HCl
(d) 1 mol NH\(_4\)Cl and 1 mol HCl
Show Answer & Explanation

Answer: (b) 1 mol CH\(_3\)COOH and 0.5 mol NaOH

Explanation:
1. Half-neutralised acetic acid leaves 0.5 mol CH\(_3\)COOH and 0.5 mol CH\(_3\)COO\(^-\), a buffer.
2. The other mixtures contain two bases, two acids, or an acid with a strong acid.

Teacher's Note:
1. Need both a weak acid and its conjugate base.

Question 15: In which volume ratio should 0.1 M NH\(_4\)OH and 0.1 M HCl be mixed to obtain a buffer?
(a) 1 : 2
(b) no ratio gives a buffer
(c) 1 : 1
(d) 2 : 1
Show Answer & Explanation

Answer: (d) 2 : 1

Explanation:
1. A buffer needs excess NH\(_4\)OH after partial neutralisation.
2. At 2 : 1, half of the base is neutralised, leaving equal NH\(_3\) and NH\(_4^+\).

Teacher's Note:
1. The weak base must be in excess.

Question 16: The pH of a mixture containing 0.1 M X\(^-\) and 0.2 M HX (\(pK_b\) of X\(^-\) = 4) is:
(a) 4 + log 2
(b) 4 - log 2
(c) 10 + log 2
(d) 10 - log 2
Show Answer & Explanation

Answer: (d) 10 - log 2

Explanation:
1. \(pK_a = 14 - 4 = 10\).
2. \(pH = 10 + \log(0.1/0.2) = 10 - \log 2\).

Teacher's Note:
1. Convert \(pK_b\) to \(pK_a\) first.

Question 17: \(K_a\) of HCN is \(5 \times 10^{-10}\). To keep pH constant at 9, the volume of 5 M KCN to be added to 10 mL of 2 M HCN is (\(\log 2 = 0.3\)):
(a) 10 mL
(b) 4 mL
(c) 2 mL
(d) 8 mL
Show Answer & Explanation

Answer: (c) 2 mL

Explanation:
1. \(pK_a = 9.3\), so \(\log([CN^-]/[HCN]) = -0.3\) and the ratio is 0.5.
2. CN\(^-\) needed = 0.5 x 20 mmol = 10 mmol, i.e. 2 mL of 5 M.

Teacher's Note:
1. The ratio is independent of total volume.

Question 18: To prepare a buffer of pH 8.26 in 500 mL of 0.01 M NH\(_4\)OH, the amount of (NH\(_4\))\(_2\)SO\(_4\) to be added is (\(pK_a\) of NH\(_4^+\) = 9.26):
(a) 0.025 mol
(b) 0.05 mol
(c) 0.005 mol
(d) 0.10 mol
Show Answer & Explanation

Answer: (a) 0.025 mol

Explanation:
1. \(8.26 = 9.26 + \log([NH_3]/[NH_4^+])\) gives \([NH_4^+] = 10[NH_3]\).
2. NH\(_3\) = 0.005 mol, so NH\(_4^+\) = 0.05 mol.
3. Each (NH\(_4\))\(_2\)SO\(_4\) gives 2 NH\(_4^+\), so 0.025 mol is needed.

Teacher's Note:
1. Remember two NH\(_4^+\) per formula unit.

Question 19: A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has pH 5. At the end point, 27 mL of the base is required. The \(K_a\) of the acid is:
(a) \(1.8 \times 10^{-5}\)
(b) \(8 \times 10^{-5}\)
(c) \(8 \times 10^{-6}\)
(d) \(1.8 \times 10^{-6}\)
Show Answer & Explanation

Answer: (c) \(8 \times 10^{-6}\)

Explanation:
1. Total acid = 2.7 mmol; base added = 1.2 mmol, leaving 1.5 mmol HA.
2. \(5 = pK_a + \log(1.2/1.5)\), so \(pK_a = 5.097\).
3. \(K_a = 8 \times 10^{-6}\).

Teacher's Note:
1. The end-point volume gives the total acid.

Question 20: 1 M benzoic acid (\(pK_a = 4.2\)) and 1 M sodium benzoate solutions are used to prepare 93 mL of buffer of pH 4.5 (\(\log 2 = 0.3\)). The volume of benzoic acid required is:
(a) 62 mL
(b) 31 mL
(c) 93 mL
(d) 46.5 mL
Show Answer & Explanation

Answer: (b) 31 mL

Explanation:
1. \(4.5 = 4.2 + \log(V_s/V_a)\) gives \(V_s/V_a = 2\).
2. \(V_a + V_s = 93\), so \(V_a = 31\) mL.

Teacher's Note:
1. Equal molarity means volume ratio equals mole ratio.

Question 21: The \(pK_a\) of a weak acid HA is 4.5. The pOH of an aqueous buffered solution in which 50% of the acid is ionised is:
(a) 9.5
(b) 4.5
(c) 2.5
(d) 7.0
Show Answer & Explanation

Answer: (a) 9.5

Explanation:
1. At 50% ionisation \([A^-] = [HA]\), so pH = 4.5.
2. pOH = 14 - 4.5 = 9.5.

Teacher's Note:
1. Half ionised means pH equals \(pK_a\).

Question 22: Addition of NaOH to a weak acid HA (\(K_a = 10^{-5}\)) gives a buffer of pH 6. The ratio of salt to acid is:
(a) 1 : 10
(b) 10 : 1
(c) 5 : 4
(d) 4 : 5
Show Answer & Explanation

Answer: (b) 10 : 1

Explanation:
1. \(6 = 5 + \log([salt]/[acid])\), so the ratio is 10.

Teacher's Note:
1. Henderson's equation.

Question 23: 50 mL of 0.2 M ammonia is treated with 25 mL of 0.2 M HCl (\(pK_b\) = 4.75). The pH of the mixture is:
(a) 9.25
(b) 3.75
(c) 8.25
(d) 4.75
Show Answer & Explanation

Answer: (a) 9.25

Explanation:
1. NH\(_3\) = 10 mmol; HCl = 5 mmol, leaving 5 mmol NH\(_3\) and forming 5 mmol NH\(_4^+\).
2. \(pOH = pK_b = 4.75\), so pH = 9.25.

Teacher's Note:
1. Half neutralisation of a weak base.

Question 24: 20 mL of 0.1 M H\(_2\)SO\(_4\) is added to 30 mL of 0.2 M NH\(_4\)OH (\(pK_b\) = 4.7). The pH of the mixture is:
(a) 9.4
(b) 5.0
(c) 9.0
(d) 5.2
Show Answer & Explanation

Answer: (c) 9.0

Explanation:
1. H\(^+\) = 4 mmol; NH\(_4\)OH = 6 mmol, leaving 2 mmol NH\(_3\) and forming 4 mmol NH\(_4^+\).
2. \(pOH = 4.7 + \log(4/2) = 5.0\), so pH = 9.0.

Teacher's Note:
1. Count both protons of H\(_2\)SO\(_4\).

Question 25: A saturated solution of PbCl\(_2\) contains \(2 \times 10^{-3}\) mol per litre. Its \(K_{sp}\) is:
(a) \(4 \times 10^{-6}\)
(b) \(1.6 \times 10^{-8}\)
(c) \(8 \times 10^{-6}\)
(d) \(3.2 \times 10^{-8}\)
Show Answer & Explanation

Answer: (d) \(3.2 \times 10^{-8}\)

Explanation:
1. \(K_{sp} = 4s^3 = 4 \times (2 \times 10^{-3})^3 = 3.2 \times 10^{-8}\).

Teacher's Note:
1. PbCl\(_2\) is an AB\(_2\) salt.

Question 26: \(K_{sp}\) at 25\(^\circ\)C of AgCl, AgBr and AgI are \(3 \times 10^{-10}\), \(7.7 \times 10^{-13}\) and \(1.5 \times 10^{-16}\). The decreasing order of solubility is:
(a) AgI > AgBr > AgCl
(b) AgCl > AgI > AgBr
(c) AgCl > AgBr > AgI
(d) AgBr > AgCl > AgI
Show Answer & Explanation

Answer: (c) AgCl > AgBr > AgI

Explanation:
1. All are 1:1 salts, so the order of solubility follows that of \(K_{sp}\).

Teacher's Note:
1. Same type of salt: compare \(K_{sp}\).

Free study material for Chemistry

Multiple Choice Questions (MCQs) for JEE Chemistry Ionic Equilibrium

Download Multiple Choice Questions: Ionic Equilibrium (JEE Chemistry)

Test your conceptual understanding of Ionic Equilibrium with these targeted multiple-choice questions. Designed in alignment with the latest JEE curriculum for JEE Chemistry, these problem sets build accuracy and prepare students for objective exams.

Concept Clarification for Ionic Equilibrium

Cross-reference your completed choices with comprehensive NCERT solutions for JEE Chemistry to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for JEE Chemistry

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Chemistry Ionic Equilibrium MCQs Set 27?

You can get most exhaustive JEE Chemistry Ionic Equilibrium MCQs Set 27 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Ionic Equilibrium MCQs Set 27 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Ionic Equilibrium MCQs Set 27, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Ionic Equilibrium MCQs Set 27?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Ionic Equilibrium MCQs Set 27 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.