Equilibrium NEET PYQs (2005–2026) with Solutions

10 solved Equilibrium PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumIonic Equilibrium

   In a qualitative analysis, Bi³⁺ is detected by appearance of precipitate of BiO(OH)(s). Calculate pH when the following equilibrium is established at 298 K: BiO(OH)(s) ⇌ BiO⁺(aq) + OH⁻(aq), K = 4 × 10⁻¹⁰ (Given: log2 = 0.3010)

   1. AOption A: 8.714
   2. BOption B: 4.699
   3. COption C: 5.286
   4. DOption D: 9.301

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   Correct answer: (D) 9.301

   Explanation

   K=[BiO⁺][OH⁻]=s²=4×10⁻¹⁰ ⇒ s=[OH⁻]=2×10⁻⁵ M. Thus [H⁺]=10⁻¹⁴/(2×10⁻⁵)=0.5×10⁻⁹. Therefore pH=−log(0.5×10⁻⁹)=9+log2=9.301. Hence option (D) is correct.

2. Question 2 (NEET 2026)

   NEET 2026MediumSolubility product and molar solubility

   The correct order of solubility of the given salts in water at 298 K is Salt | $K_{sp}$ at 298 K $\mathrm{AgBr}$ | $5.0 \times 10^{-13}$ $\mathrm{Zn(OH)_2}$ | $1.0 \times 10^{-15}$ $\mathrm{Hg_2Cl_2}$ | $1.3 \times 10^{-18}$

   1. AOption A: $\mathrm{Hg_2Cl_2 > Zn(OH)_2 > AgBr}$
   2. BOption B: $\mathrm{AgBr > Zn(OH)_2 > Hg_2Cl_2}$
   3. COption C: $\mathrm{Hg_2Cl_2 > AgBr > Zn(OH)_2}$
   4. DOption D: $\mathrm{Zn(OH)_2 > AgBr > Hg_2Cl_2}$

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   Correct answer: (D) $\mathrm{Zn(OH)_2 > AgBr > Hg_2Cl_2}$

   Explanation

   The salts are of different types, so their $K_{sp}$ values cannot be compared directly. Work out the molar solubility $s$ of each. $\mathrm{AgBr \rightleftharpoons Ag^+ + Br^-}$: $K_{sp} = s^2$, so $s = \sqrt{5.0 \times 10^{-13}} \approx 7.1 \times 10^{-7}\ \mathrm{mol\ L^{-1}}$. $\mathrm{Zn(OH)_2 \rightleftharpoons Zn^{2+} + 2OH^-}$: $K_{sp} = 4s^3$, so $s = \left(\frac{1.0 \times 10^{-15}}{4}\right)^{1/3} \approx 6.3 \times 10^{-6}\ \mathrm{mol\ L^{-1}}$. $\mathrm{Hg_2Cl_2 \rightleftharpoons Hg_2^{2+} + 2Cl^-}$: $K_{sp} = 4s^3$, so $s = \left(\frac{1.3 \times 10^{-18}}{4}\right)^{1/3} \approx 6.9 \times 10^{-7}\ \mathrm{mol\ L^{-1}}$. So the solubility order is $\mathrm{Zn(OH)_2} > \mathrm{AgBr} > \mathrm{Hg_2Cl_2}$, option (D).

3. Question 3 (NEET 2026)

   NEET 2026MediumBuffer Solutions & Solubility Product

   At 298 K, a certain buffer solution contains equal concentrations of X⁻ and HX. K$_b$ for X⁻ is 10⁻¹⁰. What is the pH of this buffer solution?

   1. AOption A: 2
   2. BOption B: 4
   3. COption C: 6
   4. DOption D: 10

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   Correct answer: (B) 4

   Explanation

   Ka=Kw/Kb=10⁻¹⁴/10⁻¹⁰=10⁻⁴, hence pKa=4. Since [X⁻]=[HX], pH=pKa+log([X⁻]/[HX])=4. Hence option (B) is correct.

4. Question 4 (NEET 2026)

   NEET 2026EasyAcids, Bases & pH

   Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change observed at alkaline pH close to the equivalence point is:

   1. AOption A: Pinkish red to yellow
   2. BOption B: Yellow to pinkish red
   3. COption C: Pink to colourless
   4. DOption D: Colourless to pink

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   Correct answer: (D) Colourless to pink

   Explanation

   Phenolphthalein is colourless in acidic medium and pink in alkaline medium. Near the alkaline endpoint, the colour changes from colourless to pink. Hence option (D) is correct.

5. Question 5 (NEET 2026)

   NEET 2026MediumEquilibrium Constant & Le Chatelier Principle

   Given below are certain reactions. Identify the reaction for which Kp ≠ Kc.

   1. AOption A: H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
   2. BOption B: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
   3. COption C: H₂(g) + I₂(g) ⇌ 2HI(g)
   4. DOption D: N₂(g) + O₂(g) ⇌ 2NO(g)

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   Correct answer: (B) N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

   Explanation

   Kp=Kc(RT)^Δng. Kp≠Kc when Δng≠0. For reaction (2), Δng=2−4=−2, so Kp≠Kc. For the other reactions, Δng=0, therefore Kp=Kc. Hence option (B) is correct.

6. Question 6 (NEET 2025)

   NEET 2025MediumIonic Equilibrium

   Phosphoric acid ionizes in three steps with their ionization constant values $K_{a1}$, $K_{a2}$ and $K_{a3}$, respectively, while $K$ is the overall ionization constant. Which of the following statements are true? A. $\log K = \log K_{a1} + \log K_{a2} + \log K_{a3}$ B. $H_3PO_4$ is a stronger acid than $H_2PO_4^{-}$ and $HPO_4^{2-}$. C. $K_{a1} > K_{a2} > K_{a3}$ D. $K_{a1} = \dfrac{K_{a3} + K_{a2}}{2}$ Choose the correct answer from the options given below:

   1. 1Option 1: A and B only
   2. 2Option 2: A and C only
   3. 3Option 3: B, C and D only
   4. 4Option 4: A, B and C only

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   Correct answer: D

   Explanation

   For stepwise ionization of phosphoric acid: $$K = K_{a1}K_{a2}K_{a3}$$ Taking logarithm: $$\log K = \log K_{a1} + \log K_{a2} + \log K_{a3}$$ Hence statement A is true. $H_3PO_4$is stronger than its conjugate acids$H_2PO_4^{-}$and$HPO_4^{2-}$, so statement B is true. For polyprotic acids:$$K_{a1} > K_{a2} > K_{a3}$$ Hence statement C is true. Statement D is incorrect because there is no such relation between the dissociation constants. Therefore, the correct option is (4): A, B and C only.

7. Question 7 (NEET 2025)

   NEET 2025MediumEquilibrium Constant & Le Chatelier Principle

   Higher yield of NO in $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$ can be obtained at [ΔH of the reaction = +180.7 kJ mol⁻¹] A. higher temperature B. lower temperature C. higher concentration of $N_2$ D. higher concentration of $O_2$ Choose the correct answer from the options given below:

   1. AOption A: A, D only
   2. BOption B: B, C only
   3. COption C: B, C, D only
   4. DOption D: A, C, D only

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   Correct answer: (D) A, C, D only

   Explanation

   The forward reaction is endothermic because ΔH is positive. According to Le Chatelier’s principle, increasing temperature favors the forward reaction and increases NO yield. Increasing the concentration of reactants $$N_2$$ or $$O_2$$ also shifts equilibrium towards product formation. Hence statements A, C and D are correct.

8. Question 8 (NEET 2024)

   NEET 2024MediumEquilibrium Constant & Le Chatelier Principle

   For the reaction: $2A \rightleftharpoons B + C$ $K_c = 4\times10^{-3}$ At a given time, the composition of reaction mixture is: $[A]=[B]=[C]=2\times10^{-3}\,M$ Then, which of the following is correct?

   1. AOption A: Reaction has a tendency to go in backward direction.
   2. BOption B: Reaction has gone to completion in forward direction.
   3. COption C: Reaction is at equilibrium.
   4. DOption D: Reaction has a tendency to go in forward direction.

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   Correct answer: (A) Reaction has a tendency to go in backward direction.

   Explanation

   For the reaction: $$2A \rightleftharpoons B+C$$ Reaction quotient: $$Q_c=\frac{[B][C]}{[A]^2}$$ Substituting: $$Q_c=\frac{(2\times10^{-3})(2\times10^{-3})}{(2\times10^{-3})^2}$$ $$Q_c=1$$ Given: $$K_c=4\times10^{-3}$$ Since: $$Q_c>K_c$$ The reaction mixture contains excess products compared to equilibrium, therefore the reaction shifts in the **backward direction** to attain equilibrium. Hence, the correct answer is: **(1) Reaction has a tendency to go in backward direction.**

9. Question 9 (NEET 2024)

   NEET 2024MediumEquilibrium Constant & Le Chatelier Principle

   In which of the following equilibria, Kₚ and K$_c$ are NOT equal?

   1. AOption A: Option (1)
   2. BOption B: Option (2)
   3. COption C: Option (3)
   4. DOption D: Option (4)

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   Correct answer: (C) Option (3)

   Explanation

   Relation between equilibrium constants: $$K_p = K_c(RT)^{\Delta n}$$ When: $$\Delta n = 0$$ then: $$K_p = K_c$$ For: $$PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$$ $$\Delta n = (1+1)-1 = 1$$ Thus $$K_p \neq K_c$$. Hence, option (C) is correct.

10. Question 10 (NEET 2023)

    NEET 2023MediumChemical equilibrium

    The equilibrium concentrations of the species in the reaction A + B ⇌ C + D are 2, 3, 10 and 6 mol L⁻¹, respectively at 300 K. ΔG° for the reaction is (R = 2 cal/mol K)

    1. AOption A: 1372.60 cal
    2. BOption B: −137.26 cal
    3. COption C: −1381.80 cal
    4. DOption D: −13.73 cal

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    Correct answer: (C) −1381.80 cal

    Explanation

    For the reaction: A + B ⇌ C + D Kc = [C][D]/[A][B] = (10 × 6)/(2 × 3) = 10 Using: ΔG° = −RT ln K = −(2)(300)ln(10) = −600 × 2.303 = −1381.80 cal