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Chemistry · chapter PYQs

Equilibrium, on the exam screen.

8 questions from NEET 2026, Re-NEET 2026, NEET 2020, 8 minutes, marked +4 / −1. See them with solutions.

Equilibrium · PYQs

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Q1/8

  1. Question 1 · NEET 2026 · Q50 · +4 / −1

    In a qualitative analysis, Bi3+\mathrm{Bi}^{3+} is detected by appearance of precipitate of BiO(OH)(s)\mathrm{BiO}(\mathrm{OH})(s). Calculate pH when the following equilibrium exists at 298 K. BiO(OH)(s)⇌BiO+(aq)+OH−(aq)\mathrm{BiO}(\mathrm{OH})(\text{s}) \rightleftharpoons \mathrm{BiO}^{+}(\text{aq}) + \mathrm{OH}^{-}(\text{aq}), K=4×10−10K = 4\times10^{-10}. (Given: log⁡2=0.3010\log 2 = 0.3010)

    Answer and explanation

    Answer: (4)

    For the dissolution, K=s2K = s^2 since a solid gives equal amounts of BiO+\mathrm{BiO}^{+} and OH−\mathrm{OH}^{-}. So s=4×10−10=2×10−5 M=[OH−]s = \sqrt{4\times10^{-10}} = 2\times10^{-5}\ \text{M} = [\mathrm{OH}^{-}]. Then pOH=5−log⁡2=4.699\text{pOH} = 5 - \log 2 = 4.699 and pH=14−4.699=9.301\text{pH} = 14 - 4.699 = 9.301.

  2. Question 2 · NEET 2026 · Q78 · +4 / −1

    At 298 K, a certain buffer solution contains equal concentrations of X−X^- and HX, KbK_b for X−X^- is 10−1010^{-10}. What is the pH of this buffer solution?

    Answer and explanation

    Answer: (2)

    For the conjugate pair, Ka(HX)=KwKb=10−1410−10=10−4K_a(HX) = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}, so pKa = 4. Using Henderson-Hasselbalch, pH=pKa+log⁡[X−][HX]pH = pK_a + \log\frac{[X^-]}{[HX]}; since [X−X^-] = [HX], the log term is zero, giving pH = 4.

  3. Question 3 · NEET 2026 · Q83 · +4 / −1

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

    Answer and explanation

    Answer: (4)

    Phenolphthalein is colourless in acidic medium and turns pink in basic medium. As the solution shifts from acidic toward alkaline near the equivalence point, the observed change is colourless to pink (4). Options mentioning yellow refer to methyl orange, not phenolphthalein.

  4. Question 4 · NEET 2026 · Q87 · +4 / −1

    Given below are certain reactions. Identify the reaction for which KP≠KCK_P \neq K_C.

    Answer and explanation

    Answer: (2)

    Since KP=KC(RT)ΔngK_P = K_C(RT)^{\Delta n_g}, the two are equal only when Δng=0\Delta n_g = 0. For N2+3H2⇌2NH3\mathrm{N}_{2} + 3\mathrm{H}_{2} \rightleftharpoons 2\mathrm{NH}_{3}, Δng=2−4=−2≠0\Delta n_g = 2 - 4 = -2 \neq 0, so KP≠KCK_P \neq K_C (2). The other three reactions have equal moles of gas on both sides (Δng=0\Delta n_g = 0), giving KP=KCK_P = K_C.

  5. Question 5 · Re-NEET 2026 · Q52 · +4 / −1

    The correct order of solubility of the given salts in water at 298 K is (Given KspK_{sp} at 298 K: AgBr\mathrm{AgBr} = 5.0×10−135.0 \times 10^{-13}; Zn(OH)2\mathrm{Zn}(\mathrm{OH})_{2} = 1.0×10−151.0 \times 10^{-15}; Hg2Cl2\mathrm{Hg}_{2}\mathrm{Cl}_{2} = 1.3×10−181.3 \times 10^{-18})

    Answer and explanation

    Answer: (4)

    KspK_{sp} values can only be compared directly when the salts share the same stoichiometry, so each must first be converted to molar solubility s. For the 1:1 salt AgBr\mathrm{AgBr}, Ksp=s2K_{sp} = s^2 gives s≈7.1×10−7s \approx 7.1 \times 10^{-7}. For the 1:2 salts, Ksp=4s3K_{sp} = 4s^3: Zn(OH)2\mathrm{Zn}(\mathrm{OH})_{2} gives s≈6.3×10−6s \approx 6.3 \times 10^{-6} and Hg2Cl2\mathrm{Hg}_{2}\mathrm{Cl}_{2} gives s≈6.9×10−7s \approx 6.9 \times 10^{-7}. Ranking these, Zn(OH)2>AgBr>Hg2Cl2\mathrm{Zn}(\mathrm{OH})_{2} > \mathrm{AgBr} > \mathrm{Hg}_{2}\mathrm{Cl}_{2}. NCERT Class 11, Equilibrium (solubility equilibria).

  6. Question 6 · NEET 2020 · Q95 · +4 / −1

    HCl\mathrm{HCl} was passed through a solution of CaCl2\mathrm{CaCl}_{2}, MgCl2\mathrm{MgCl}_{2} and NaCl\mathrm{NaCl}. Which of the following compound(s) crystallise(s)?

    Answer and explanation

    Answer: (1)

    Passing HCl\mathrm{HCl} gas increases the concentration of Cl−\mathrm{Cl^-} ions and lowers the solubility of the chlorides by the common-ion effect. Under these conditions, NaCl\mathrm{NaCl} becomes much less soluble and crystallises, while MgCl2\mathrm{MgCl_2} and CaCl2\mathrm{CaCl_2} remain in solution.

  7. Question 7 · NEET 2020 · Q96 · +4 / −1

    Find out the solubility of Ni(OH)₂ in 0.1 M NaOH. Given that the ionic product of Ni(OH)₂ is 2 × 10⁻¹⁵.

    Answer and explanation

    Answer: (4)

    The solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_2 in 0.1 M NaOH can be calculated using the solubility product Ksp=2×10−15K_{sp} = 2 \times 10^{-15}. Let s be the solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_2. The concentration of OH−\mathrm{OH}^- from NaOH is 0.1 M, and from Ni(OH)2\mathrm{Ni}(\mathrm{OH})_2 is 2s. Thus, Ksp=s×(0.1+2s)2≈s×(0.1)2=2×10−15⇒s=2×10−13 MK_{sp} = s \times (0.1 + 2s)^2 \approx s \times (0.1)^2 = 2 \times 10^{-15} ⇒ s = 2 \times 10^{-13}\ M. Chapter: Equilibrium.

  8. Question 8 · NEET 2020 · Q125 · +4 / −1

    Hydrolysis of sucrose is given by: sucrose + H₂O ⇌ glucose + fructose. If the equilibrium constant (Kc) is 2 × 10¹³ at 300 K, the value of ΔᵣG° at the same temperature will be

    Answer and explanation

    Answer: (4)

    The relationship between the equilibrium constant KcK_c and the standard Gibbs free energy change ΔrG∘\Delta_r G^\circ is given by ΔrG∘=−RTln⁡Kc\Delta_r G^\circ = -RT \ln K_c. Substituting the values, ΔrG∘=−8.314 Jmol−1 K−1×300 K×ln⁡(2×1013)\Delta_r G^\circ = -8.314\ J mol^{-1}\ K^{-1} \times 300\ K \times \ln(2 \times 10^{13}). Chapter: Equilibrium.