JEE Chemistry Ionic Equilibrium MCQs Set 12

Chemistry Objective Questions and Answers: Ionic Equilibrium

Review structured MCQ sets for JEE Chemistry Ionic Equilibrium. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Ionic Equilibrium MCQs with Answers

Navigate directly to the 50 objective questions for Ionic Equilibrium using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The pH of a solution made by mixing 50.0 mL of 0.2 M \(\ce{NH4Cl}\) and 75.0 mL of 0.1 M \(\ce{NaOH}\) (\(K_b=1.8\times10^{-5}\)) is
(a) 4.27
(b) 9.26
(c) 9.73
(d) 8.78
Show Answer & Explanation

Answer: (c) 9.73

Explanation:
1. \(\ce{NH4+}=10\) mmol and \(\ce{NaOH}=7.5\) mmol, giving 7.5 mmol \(\ce{NH3}\) and 2.5 mmol \(\ce{NH4+}\).
2. \(\text{pOH}=4.74+\log(2.5/7.5)=4.27\).
3. \(\text{pH}=9.73\).

Teacher's Note:
1. Strong base converts ammonium into ammonia.
2. More base than salt gives pOH below pKb.

Question 2: A buffer contains 0.02 mol propionic acid and 0.015 mol sodium propionate in 1.00 L (\(K_a=1.34\times10^{-5}\)). Its pH is
(a) 4.87
(b) 4.75
(c) 4.60
(d) 5.00
Show Answer & Explanation

Answer: (b) 4.75

Explanation:
1. \(\text{p}K_a=4.873\).
2. \(\text{pH}=4.873+\log(0.015/0.020)=4.873-0.125\).
3. \(\text{pH}=4.75\).

Teacher's Note:
1. Salt is less than acid, so the pH is below pKa.
2. Use the Henderson equation.

Question 3: To 10 mL of the buffer in the previous problem (0.02 M acid, 0.015 M salt, \(K_a=1.34\times10^{-5}\)) \(1.0\times10^{-5}\) mol of \(\ce{HCl}\) is added. The new pH is
(a) 4.70
(b) 2.00
(c) 4.75
(d) 4.80
Show Answer & Explanation

Answer: (a) 4.70

Explanation:
1. Moles in 10 mL: acid \(=2.0\times10^{-4}\), salt \(=1.5\times10^{-4}\).
2. HCl converts salt to acid: acid \(=2.1\times10^{-4}\), salt \(=1.4\times10^{-4}\).
3. \(\text{pH}=4.873+\log(1.4/2.1)=4.70\).

Teacher's Note:
1. The buffer absorbs strong acid with only a small pH change.
2. Convert the concentrations to moles in the sample.

Question 4: To 10 mL of the same buffer (0.02 M propionic acid, 0.015 M sodium propionate, \(K_a=1.34\times10^{-5}\)) \(1.0\times10^{-5}\) mol of \(\ce{NaOH}\) is added. The new pH is
(a) 4.70
(b) 4.80
(c) 9.00
(d) 4.75
Show Answer & Explanation

Answer: (b) 4.80

Explanation:
1. Moles in 10 mL: acid \(=2.0\times10^{-4}\), salt \(=1.5\times10^{-4}\).
2. NaOH converts acid to salt: acid \(=1.9\times10^{-4}\), salt \(=1.6\times10^{-4}\).
3. \(\text{pH}=4.873+\log(1.6/1.9)=4.80\).

Teacher's Note:
1. Base raises the pH slightly.
2. A buffer resists change but does not stop it.

Question 5: The \([H^+]\) of a solution that is 0.01 M in chloroacetic acid and 0.002 M in sodium chloroacetate (\(K_a=1.5\times10^{-3}\)) is
(a) \(7.5\times10^{-3}\ \text{M}\)
(b) \(3.87\times10^{-3}\ \text{M}\)
(c) \(1.5\times10^{-3}\ \text{M}\)
(d) \(2.5\times10^{-3}\ \text{M}\)
Show Answer & Explanation

Answer: (d) \(2.5\times10^{-3}\ \text{M}\)

Explanation:
1. Let \(x=[H^+]\): \(\dfrac{x(0.002+x)}{0.01-x}=1.5\times10^{-3}\).
2. \(x^2+0.0035x-1.5\times10^{-5}=0\), so \(x=2.5\times10^{-3}\ \text{M}\).

Teacher's Note:
1. The buffer formula fails because the acid is fairly strong.
2. Solve the quadratic including the dissociation of the acid.

Question 6: A solution has \([H^+]=4\times10^{-3}\) M. For thymol blue, pH is 2.0 when half of the indicator is in the unionised form. The percentage of indicator in the unionised form at this \([H^+]\) is
(a) 71.4%
(b) 50%
(c) 28.6%
(d) 10%
Show Answer & Explanation

Answer: (c) 28.6%

Explanation:
1. Half unionised at pH 2 means \(\text{p}K_{in}=2\), so \(K_{in}=10^{-2}\).
2. \(\dfrac{[In^-]}{[HIn]}=\dfrac{K_{in}}{[H^+]}=\dfrac{10^{-2}}{4\times10^{-3}}=2.5\).
3. Fraction unionised \(=1/(1+2.5)=28.6\%\).

Teacher's Note:
1. At half ionisation pH equals pKin.
2. Convert the ratio into a percentage of the total.

Question 7: An indicator is a weak acid with \(K_{in}=4\times10^{-4}\). For which of the neutralisations (a) NaOH + CH3COOH, (b) HCl + NH3, (c) HCl + NaOH would it be useful?
(a) only (b) and (c)
(b) only (a)
(c) only (a) and (b)
(d) all of (a), (b) and (c)
Show Answer & Explanation

Answer: (a) only (b) and (c)

Explanation:
1. The indicator changes colour near \(\text{p}K_{in}=3.4\), roughly pH 2.4 to 4.4.
2. (c) The jump for a strong acid and strong base covers pH 4 to 10, which includes 3.4 to 4.4. (b) The equivalence point of a strong acid with a weak base is acidic (about 5), and the steep part reaches down to pH 3 to 4.
3. (a) Weak acid with strong base ends in the basic region, where this indicator is already fully in one colour, so it is not useful.

Teacher's Note:
1. Choose an indicator whose range lies on the steep part of the titration curve.
2. Match acidic indicators to acidic equivalence points.

Question 8: Bromophenol blue is an indicator with \(K_a=6\times10^{-5}\). The percentage of this indicator in its basic form at pH 5 is
(a) 85.7%
(b) 50%
(c) 14.3%
(d) 60%
Show Answer & Explanation

Answer: (a) 85.7%

Explanation:
1. \(\dfrac{[In^-]}{[HIn]}=\dfrac{K_a}{[H^+]}=\dfrac{6\times10^{-5}}{10^{-5}}=6\).
2. Basic fraction \(=6/7=85.7\%\).

Teacher's Note:
1. The base-to-acid ratio equals Ka over [H+].
2. Basic form is the ionised form.

Question 9: An indicator has \(K_a=3\times10^{-5}\), with a red acid form and a blue basic form. By how much must the pH change to change the indicator from 75% red to 75% blue?
(a) 1.908
(b) 0.954
(c) 0.301
(d) 0.477
Show Answer & Explanation

Answer: (b) 0.954

Explanation:
1. \(\text{pH}=\text{p}K_a+\log\dfrac{[\text{blue}]}{[\text{red}]}\).
2. At 75% red the ratio is \(1/3\), and at 75% blue it is \(3\).
3. \(\Delta\text{pH}=\log3-\log(1/3)=2\log3=0.954\).

Teacher's Note:
1. Both ratios are symmetric around the pKa.
2. The change is independent of the actual Ka value.

Question 10: In the titration of a weak acid HX with NaOH, the pH is 5.8 after 10.0 mL of NaOH and 6.402 after 20.0 mL. The \(\text{p}K_a\) of HX is
(a) 7.00
(b) 5.80
(c) 6.40
(d) 6.10
Show Answer & Explanation

Answer: (d) 6.10

Explanation:
1. With equivalence volume \(V_e\): \(\text{pH}=\text{p}K_a+\log\dfrac{V}{V_e-V}\).
2. Subtracting the two equations: \(0.602=\log\dfrac{2(V_e-10)}{V_e-20}\), so \(4=\dfrac{2(V_e-10)}{V_e-20}\) and \(V_e=30\) mL.
3. \(\text{p}K_a=5.8-\log(10/20)=6.10\), so \(K_a=7.9\times10^{-7}\).

Teacher's Note:
1. The unknown equivalence volume can be eliminated using two points.
2. pKa is the pH at half equivalence, 15 mL here.

Question 11: 0.515 g of phenol (\(\ce{C6H5OH}\), molar mass 94) is dissolved in exactly 100 mL of water and titrated with 0.123 M NaOH (\(K_a=1.3\times10^{-10}\)). The pH at the equivalence point is
(a) 7.00
(b) 11.22
(c) 8.73
(d) 5.28
Show Answer & Explanation

Answer: (b) 11.22

Explanation:
1. Moles of phenol \(=0.515/94=5.48\) mmol, so the volume of NaOH needed is \(44.5\) mL and the final volume is \(144.5\) mL.
2. \([\text{C}_6\text{H}_5\text{O}^-]=5.48/144.5=0.0379\ \text{M}\) and \(K_h=K_w/K_a=7.7\times10^{-5}\). Solving \(\dfrac{x^2}{0.0379-x}=7.7\times10^{-5}\) gives \([OH^-]=1.67\times10^{-3}\ \text{M}\).
3. \(\text{pOH}=2.78\), so \(\text{pH}=11.22\) and \([Na^+]=0.0379\ \text{M}\).

Teacher's Note:
1. Phenoxide is a much stronger base than acetate because phenol is a weaker acid.
2. Equivalence point of a weak acid with strong base is above 7.

Question 12: A weak acid (50.0 mL) is titrated with 0.1 M NaOH. The pH is 4.16 after 10.0 mL and 4.76 after 25.0 mL of base. The \(K_a\) of the acid and the pH at the equivalence point are
(a) \(K_a=5.7\times10^{-10}\) and pH \(=8.73\)
(b) \(K_a=1.74\times10^{-5}\) and pH \(=7.00\)
(c) \(K_a=1.74\times10^{-5}\) and pH \(=8.73\)
(d) \(K_a=1.74\times10^{-5}\) and pH \(=5.27\)
Show Answer & Explanation

Answer: (c) \(K_a=1.74\times10^{-5}\) and pH \(=8.73\)

Explanation:
1. Half way point at 25 mL (equivalence at 50 mL) gives \(\text{p}K_a=4.76\), so \(K_a=1.74\times10^{-5}\).
2. Salt concentration at equivalence \(=0.05\ \text{M}\).
3. \([OH^-]=\sqrt{\dfrac{K_w}{K_a}C}=5.36\times10^{-6}\ \text{M}\), so \(\text{pOH}=5.27\) and \(\text{pH}=8.73\).

Teacher's Note:
1. pH at half neutralisation equals pKa.
2. The 10 mL point is consistent: 4.76 + log(10/40) = 4.16.

Question 13: 50 mL of 0.1 M acetic acid (\(K_a=2\times10^-5\)) is titrated with 0.1 M NaOH. The pH after adding 20 mL of NaOH is
(a) 4.52
(b) 3.52
(c) 5.30
(d) 4.70
Show Answer & Explanation

Answer: (a) 4.52

Explanation:
1. \(\ce{CH3COOH}=5\) mmol and \(\ce{NaOH}=2\) mmol, leaving 3 mmol acid and 2 mmol acetate.
2. \(\text{pH}=4.70+\log(2/3)=4.52\).

Teacher's Note:
1. Identify the stage of titration before choosing a formula.
2. Buffer region uses the Henderson equation; equivalence uses salt hydrolysis.

Question 14: 50 mL of 0.1 M acetic acid (\(K_a=2\times10^-5\)) is titrated with 0.1 M NaOH. The pH after adding 40 mL of NaOH is
(a) 5.30
(b) 4.10
(c) 5.00
(d) 4.70
Show Answer & Explanation

Answer: (a) 5.30

Explanation:
1. \(\ce{CH3COOH}=5\) mmol and \(\ce{NaOH}=4\) mmol, leaving 1 mmol acid and 4 mmol acetate.
2. \(\text{pH}=4.70+\log4=5.30\).

Teacher's Note:
1. Identify the stage of titration before choosing a formula.
2. Buffer region uses the Henderson equation; equivalence uses salt hydrolysis.

Question 15: 50 mL of 0.1 M acetic acid (\(K_a=2\times10^-5\)) is titrated with 0.1 M NaOH. The pH after adding 50 mL of NaOH is
(a) 5.30
(b) 7.00
(c) 9.30
(d) 8.70
Show Answer & Explanation

Answer: (d) 8.70

Explanation:
1. Equivalence: 5 mmol of acetate in 100 mL, so \(C=0.05\ \text{M}\).
2. \([OH^-]=\sqrt{\dfrac{10^{-14}}{2\times10^{-5}}\times0.05}=5\times10^{-6}\ \text{M}\), so \(\text{pOH}=5.30\) and \(\text{pH}=8.70\).

Teacher's Note:
1. Identify the stage of titration before choosing a formula.
2. Buffer region uses the Henderson equation; equivalence uses salt hydrolysis.

Question 16: The salts MX and \(\ce{QX2}\) each have \(K_{sp}=4.0\times10^{-18}\). Which is more soluble and what are the solubilities?
(a) MX is more soluble: \(2\times10^{-9}\ \text{M}\) against \(10^{-6}\ \text{M}\)
(b) \(\ce{QX2}\) is more soluble: \(2\times10^{-9}\ \text{M}\) against \(10^{-6}\ \text{M}\)
(c) \(\ce{QX2}\) is more soluble: \(10^{-6}\ \text{M}\) against \(2\times10^{-9}\ \text{M}\)
(d) both are equally soluble at \(2\times10^{-9}\ \text{M}\)
Show Answer & Explanation

Answer: (c) \(\ce{QX2}\) is more soluble: \(10^{-6}\ \text{M}\) against \(2\times10^{-9}\ \text{M}\)

Explanation:
1. For MX, \(S=\sqrt{K_{sp}}=2\times10^{-9}\ \text{M}\).
2. For \(\ce{QX2}\), \(K_{sp}=4S^3\), so \(S=(10^{-18})^{1/3}=10^{-6}\ \text{M}\).
3. \(\ce{QX2}\) is more soluble.

Teacher's Note:
1. Salts of different types cannot be compared by Ksp alone.
2. Calculate the molar solubility of each.

Question 17: The solubility of \(\ce{PbSO4}\) in water is 0.038 g/L (molar mass 303). Its solubility product is
(a) \(1.6\times10^{-4}\)
(b) \(1.25\times10^{-4}\)
(c) \(1.6\times10^{-8}\)
(d) \(3.2\times10^{-8}\)
Show Answer & Explanation

Answer: (c) \(1.6\times10^{-8}\)

Explanation:
1. \(S=0.038/303=1.25\times10^{-4}\ \text{mol L}^{-1}\).
2. \(K_{sp}=S^2=1.57\times10^{-8}\approx1.6\times10^{-8}\).

Teacher's Note:
1. For a 1:1 salt Ksp is the square of the solubility.
2. Convert g per L into mol per L first.

Question 18: The solubility of \(\ce{Mg(OH)2}\) in water (\(K_{sp}=1.2\times10^{-11}\)) is
(a) \(3.46\times10^{-6}\ \text{M}\)
(b) \(1.2\times10^{-11}\ \text{M}\)
(c) \(2.29\times10^{-4}\ \text{M}\)
(d) \(1.44\times10^{-4}\ \text{M}\)
Show Answer & Explanation

Answer: (d) \(1.44\times10^{-4}\ \text{M}\)

Explanation:
1. \(K_{sp}=4S^3\).
2. \(S=(1.2\times10^{-11}/4)^{1/3}=(3\times10^{-12})^{1/3}=1.44\times10^{-4}\ \text{M}\).

Teacher's Note:
1. Mg(OH)2 gives one Mg2+ and two OH-.
2. Cube root, not square root.

Question 19: The number of moles of \(\ce{CuI}\) (\(K_{sp}=5\times10^{-12}\)) that dissolve in 1.0 L of 0.10 M \(\ce{NaI}\) is
(a) \(5\times10^{-11}\)
(b) \(5\times10^{-12}\)
(c) \(2.24\times10^{-6}\)
(d) \(5\times10^{-13}\)
Show Answer & Explanation

Answer: (a) \(5\times10^{-11}\)

Explanation:
1. \([I^-]\approx0.10\ \text{M}\) from NaI.
2. \(S=[Cu^+]=K_{sp}/[I^-]=5\times10^{-12}/0.10=5\times10^{-11}\ \text{mol L}^{-1}\).

Teacher's Note:
1. Common ion greatly lowers the solubility.
2. The iodide from the sparingly soluble salt is negligible.

Question 20: A saturated solution of \(\ce{CaF2}\) contains \(4.1\times10^{-4}\) M fluoride ion. Its \(K_{sp}\) (neglecting hydrolysis) is
(a) \(1.7\times10^{-7}\)
(b) \(1.7\times10^{-10}\)
(c) \(6.9\times10^{-11}\)
(d) \(3.45\times10^{-11}\)
Show Answer & Explanation

Answer: (d) \(3.45\times10^{-11}\)

Explanation:
1. \([Ca^{2+}]=\tfrac12[F^-]=2.05\times10^{-4}\ \text{M}\).
2. \(K_{sp}=[Ca^{2+}][F^-]^2=2.05\times10^{-4}\times(4.1\times10^{-4})^2=3.45\times10^{-11}\).

Teacher's Note:
1. Do not set the calcium concentration equal to the fluoride concentration.
2. Square the fluoride concentration.

Question 21: The solubility of \(\ce{Fe(OH)3}\) (\(K_{sp}=1.0\times10^{-36}\)) in a solution of pH 8.0 is
(a) \(1\times10^{-18}\ \text{M}\)
(b) \(1\times10^{-36}\ \text{M}\)
(c) \(1\times10^{-24}\ \text{M}\)
(d) \(1\times10^{-30}\ \text{M}\)
Show Answer & Explanation

Answer: (a) \(1\times10^{-18}\ \text{M}\)

Explanation:
1. \([OH^-]=10^{-6}\ \text{M}\) at pH 8.
2. \(S=K_{sp}/[OH^-]^3=10^{-36}/10^{-18}=10^{-18}\ \text{M}\).

Teacher's Note:
1. pH fixes the hydroxide concentration.
2. Cube the hydroxide concentration.

Question 22: The solubility of \(\ce{Ag2CrO4}\) in water is 0.044 g/L (molar mass 332). Its \(K_{sp}\) is
(a) \(1.3\times10^{-4}\)
(b) \(1.7\times10^{-8}\)
(c) \(9.3\times10^{-12}\)
(d) \(3.1\times10^{-12}\)
Show Answer & Explanation

Answer: (c) \(9.3\times10^{-12}\)

Explanation:
1. \(S=0.044/332=1.33\times10^{-4}\ \text{M}\).
2. \(K_{sp}=4S^3=4\times(1.33\times10^{-4})^3=9.3\times10^{-12}\).

Teacher's Note:
1. Silver chromate is a 2:1 salt.
2. Ksp = 4 S cubed.

Question 23: The solubility of \(\ce{A2X3}\) in pure water (\(K_{sp}=1.1\times10^{-23}\)), assuming no reaction of the ions with water, is
(a) \(3.3\times10^{-8}\ \text{M}\)
(b) \(1\times10^{-4}\ \text{M}\)
(c) \(1\times10^{-5}\ \text{M}\)
(d) \(1.1\times10^{-5}\ \text{M}\)
Show Answer & Explanation

Answer: (c) \(1\times10^{-5}\ \text{M}\)

Explanation:
1. \(K_{sp}=(2S)^2(3S)^3=108S^5\).
2. \(S=(1.1\times10^{-23}/108)^{1/5}=(1.02\times10^{-25})^{1/5}\approx1.0\times10^{-5}\ \text{M}\).

Teacher's Note:
1. A2X3 gives 2 A and 3 X ions per formula unit.
2. Take the fifth root.

Question 24: The solubility of \(\ce{AgCl}\) (\(K_{sp}=1\times10^{-10}\)) in 0.1 M \(\ce{BaCl2}\) is
(a) \(1\times10^{-9}\ \text{M}\)
(b) \(1\times10^{-5}\ \text{M}\)
(c) \(2.5\times10^{-10}\ \text{M}\)
(d) \(5\times10^{-10}\ \text{M}\)
Show Answer & Explanation

Answer: (d) \(5\times10^{-10}\ \text{M}\)

Explanation:
1. \([Cl^-]=2\times0.1=0.2\ \text{M}\).
2. \(S=[Ag^+]=K_{sp}/[Cl^-]=10^{-10}/0.2=5\times10^{-10}\ \text{M}\).

Teacher's Note:
1. Barium chloride provides two chloride ions per formula unit.
2. Common ion effect.

Question 25: A solution has \([Mg^{2+}]=0.0010\ \text{M}\) and \(K_{sp}\) of \(\ce{Mg(OH)2}=1.2\times10^{-11}\). Will \(\ce{Mg(OH)2}\) precipitate if \([OH^-]\) is (i) \(10^{-5}\) M and (ii) \(10^{-3}\) M?
(a) precipitate in (i), no precipitate in (ii)
(b) no precipitate in (i), precipitate in (ii)
(c) no precipitate in either
(d) precipitate in both
Show Answer & Explanation

Answer: (b) no precipitate in (i), precipitate in (ii)

Explanation:
1. (i) \(Q=10^{-3}\times(10^{-5})^2=10^{-13}<1.2\times10^{-11}\), so no precipitate.
2. (ii) \(Q=10^{-3}\times(10^{-3})^2=10^{-9}>1.2\times10^{-11}\), so it precipitates.

Teacher's Note:
1. Compare the ionic product with Ksp.
2. A precipitate forms only if Q exceeds Ksp.

Question 26: The solubility of \(\ce{AgCN}\) (\(K_{sp}=4\times10^{-16}\)) in a buffer of pH 3 is (take \(K_a\) of HCN \(=9\times10^{-10}\))
(a) \(4\times10^{-3}\ \text{M}\)
(b) \(2\times10^{-8}\ \text{M}\)
(c) \(2.1\times10^{-5}\ \text{M}\)
(d) \(2\times10^{-9}\ \text{M}\)
Show Answer & Explanation

Answer: (c) \(2.1\times10^{-5}\ \text{M}\)

Explanation:
1. The cyanide formed combines with \(\ce{H+}\): \(S=[Ag^+]=[CN^-]+[HCN]=[CN^-](1+[H^+]/K_a)\).
2. \(1+10^{-3}/9\times10^{-10}=1.11\times10^{6}\), so \(K_{sp}=[Ag^+][CN^-]\) gives \(S^2=K_{sp}\times1.11\times10^{6}=4.44\times10^{-10}\).
3. \(S=2.1\times10^{-5}\ \text{M}\).

Teacher's Note:
1. The salt of a weak acid is more soluble in acid.
2. Include the protonated cyanide in the total solubility.

Free study material for Chemistry

Practice MCQs for JEE Chemistry Ionic Equilibrium

About Ionic Equilibrium MCQs for JEE Chemistry

Review structured objective questions for JEE Chemistry Ionic Equilibrium. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

How to Verify Your MCQ Answers

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

Enhance Speed with Online MCQ Tests

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Ionic Equilibrium MCQs Set 12?

You can get most exhaustive JEE Chemistry Ionic Equilibrium MCQs Set 12 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 12 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 12, 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 12?

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 12 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.