Equilibrium NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Equilibrium question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Equilibrium previous year questions
59 questions. Tap an option to check your answer.
Q1NEET 2021
The pKb of dimethylamine and pKa of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is :
Not quite — the answer is C.
For a salt of a weak acid and weak base, pH = 7 + (pKa − pKb)/2 = 7 + (4.77 − 3.27)/2 = 7 + 0.75 = 7.75. The solution is slightly basic because dimethylamine (pKb 3.27) is a stronger base than acetic acid is an acid.
Q2NEET 2021
Statement I : Acid strength increases in the order given as HF << HCl << HBr << HI. Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases. In the light of the above statements, choose the correct answer from the options given below.
Not quite — the answer is A.
Acid strength of hydrogen halides increases HF < HCl < HBr < HI. This is because the halogen size increases down the group, the H–X bond becomes longer and weaker, and H⁺ is released more easily. Both statements are true.
Q3NEET 2020 · Q92
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.
Not quite — the answer is A.
In 0.1 M NaOH, [OH⁻] ≈ 0.1 M from the common ion, so s × (0.1)² = 2×10⁻¹⁵, giving s = 2×10⁻¹⁵/10⁻² = 2×10⁻¹³ M. Forgetting to square [OH⁻] gives 2×10⁻¹⁴, and ignoring the common ion leads to the much larger 10⁻⁵–10⁻⁸ range.
Q4NEET 2019 · Q62
pH of a saturated solution of Ca(OH)₂ is 9. The solubility product (Ksp) of Ca(OH)₂ is:
Not quite — the answer is A.
pOH = 5, so [OH⁻] = 10⁻⁵ M and [Ca²⁺] = 0.5 × 10⁻⁵ M. Ksp = [Ca²⁺][OH⁻]² = 0.5 × 10⁻⁵ × 10⁻¹⁰ = 0.5 × 10⁻¹⁵. Taking s = 10⁻⁵ and computing 4s³ wrongly treats [OH⁻] as the solubility.
Q5NEET 2019 · Q71
Conjugate base for Brönsted acids H₂O and HF are:
Not quite — the answer is C.
A conjugate base is what remains after the acid loses a proton: H₂O → OH⁻ and HF → F⁻. H₃O⁺ and H₂F⁺ are conjugate acids formed by gaining a proton.
Q6NEET 2019 · Q77
Which will make basic buffer?
Not quite — the answer is C.
10 mmol HCl neutralises 10 mmol of the 20 mmol NH₄OH, leaving 10 mmol NH₄OH with 10 mmol NH₄Cl: a weak base with its salt, i.e. a basic buffer. Excess NaOH with acetate is just a strong-base solution, and equal amounts of acid and base give only a salt.
Q7NEET 2018 · Q74
Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: a. 60 mL M/10 HCl + 40 mL M/10 NaOH b. 55 mL M/10 HCl + 45 mL M/10 NaOH c. 75 mL M/5 HCl + 25 mL M/5 NaOH d. 100 mL M/10 HCl + 100 mL M/10 NaOH. pH of which one of them will be equal to 1?
Not quite — the answer is D.
For c: HCl = 75 × 0.2 = 15 mmol, NaOH = 25 × 0.2 = 5 mmol; excess H⁺ = 10 mmol in 100 mL = 0.1 M, so pH = 1. Mixture a leaves 0.02 M H⁺, b leaves 0.01 M H⁺, and d is exactly neutral.
Q8NEET 2018 · Q77
The solubility of BaSO₄ in water is 2.42 × 10⁻³ g L⁻¹ at 298 K. The value of its solubility product (K_sp) will be (Given molar mass of BaSO₄ = 233 g mol⁻¹)
Not quite — the answer is C.
Molar solubility s = 2.42 × 10⁻³/233 = 1.04 × 10⁻⁵ mol L⁻¹. For BaSO₄ ⇌ Ba²⁺ + SO₄²⁻, K_sp = s² = 1.08 × 10⁻¹⁰ mol² L⁻². Squaring the gram solubility without converting to moles gives a wrong exponent.
Q9NEET 2018 · Q83
Which one of the following conditions will favour maximum formation of the product in the reaction, A₂(g) + B₂(g) ⇌ X₂(g) ΔᵣH = −X kJ?
Not quite — the answer is C.
The forward reaction is exothermic, so low temperature favours the product. It also reduces gas moles from 2 to 1, so high pressure shifts equilibrium forward (Le Chatelier). High temperature would favour the reverse reaction.
Q10NEET 2017 · Q27
The equilibrium constants of the following are: N₂ + 3H₂ ⇌ 2NH₃ K₁; N₂ + O₂ ⇌ 2NO K₂; H₂ + ½O₂ → H₂O K₃. The equilibrium constant (K) of the reaction: 2NH₃ + 5/2 O₂ ⇌ 2NO + 3H₂O, will be:
Not quite — the answer is B.
The required reaction = (reverse of the NH₃ formation) + (NO formation) + 3 × (H₂O formation): 2NH₃ → N₂ + 3H₂ (1/K₁), N₂ + O₂ → 2NO (K₂), 3H₂ + 3/2 O₂ → 3H₂O (K₃³). Adding reactions multiplies constants, so K = K₂K₃³/K₁.
Q11NEET 2017 · Q41
Concentration of the Ag⁺ ions in a saturated solution of Ag₂C₂O₄ is 2.2 × 10⁻⁴ mol L⁻¹. Solubility product of Ag₂C₂O₄ is:
Not quite — the answer is D.
Ag₂C₂O₄ ⇌ 2Ag⁺ + C₂O₄²⁻, so [C₂O₄²⁻] = ½ × 2.2 × 10⁻⁴ = 1.1 × 10⁻⁴ M. Ksp = [Ag⁺]²[C₂O₄²⁻] = (2.2 × 10⁻⁴)² × 1.1 × 10⁻⁴ = 4.84 × 10⁻⁸ × 1.1 × 10⁻⁴ ≈ 5.3 × 10⁻¹². Forgetting to square [Ag⁺] gives a wrong value.
Q12NEET 2017 · Q43
Which one of the following statements is not correct?
Not quite — the answer is B.
A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached sooner but the equilibrium constant stays the same. The other statements are correct: a catalyst cannot start a reaction that is not feasible, enzymes are biological catalysts, and coenzymes enhance enzyme activity.
Q13NEET 2017 (Gujarati paper) · Q32
The standard equilibrium constant, Kp at 298 K for the reaction, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 5.8 × 10⁵. The value of standard equilibrium constant, if the concentration of gases is expressed in terms of mol/L, will be: [Given: R = 0.08314 L bar K⁻¹ mol⁻¹]
Not quite — the answer is C.
Kp = Kc(RT)^Δn with Δn = 2 − 4 = −2, so Kc = Kp(RT)². RT = 0.08314 × 298 = 24.78, (RT)² = 613.9, and Kc = 5.8 × 10⁵ × 613.9 ≈ 3.56 × 10⁸. Dividing by (RT)² instead would give a much smaller value.
Q14NEET 2017 (Gujarati paper) · Q38
Consider the following reaction for which the change in enthalpy is positive. 2A(g) + B(g) ⇌ C(g) + D(g). Which of the following will not affect the equilibrium?
Not quite — the answer is D.
A catalyst speeds up the forward and backward reactions equally, so equilibrium is reached faster but its position and K do not change. Since Δn = −1, pressure shifts the equilibrium; changing temperature changes K for this endothermic reaction; and changing reactant concentration shifts the position.
Q15NEET 2017 (Gujarati paper) · Q41
For the reaction CO(g) + Cl₂(g) ⇌ COCl₂(g), Kp/Kc is equal to:
Not quite — the answer is A.
Kp = Kc(RT)^Δn, and here Δn = 1 − 2 = −1. So Kp/Kc = (RT)⁻¹ = 1/RT. RT would be the answer if the number of gas moles increased by one.
Q16NEET 2016 Phase 2 · Q54
The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅N⁺H) in a 0.10 M aqueous pyridine solution (K_b for C₅H₅N = 1.7 × 10⁻⁹) is
Not quite — the answer is B.
For a weak base α = √(K_b/C) = √(1.7 × 10⁻⁹/0.10) = √(1.7 × 10⁻⁸) ≈ 1.3 × 10⁻⁴. As a percentage this is 0.013%. Forgetting to convert the fraction to a percentage, or not taking the square root, gives the other values.
Q17NEET 2016 Phase 2 · Q58
The solubility of AgCl(s) with solubility product 1.6 × 10⁻¹⁰ in 0.1 M NaCl solution would be
Not quite — the answer is B.
Because of the common ion, [Cl⁻] ≈ 0.1 M, so s = K_sp/[Cl⁻] = 1.6 × 10⁻¹⁰/0.1 = 1.6 × 10⁻⁹ M. 1.26 × 10⁻⁵ M is the solubility in pure water, which ignores the common-ion effect.
Q18NEET 2016 Phase 1 · Q55
MY and NY₃, two nearly insoluble salts, have the same Ksp values of 6.2 × 10⁻¹³ at room temperature. Which statement would be true in regard to MY and NY₃?
Not quite — the answer is B.
For MY, Ksp = s² so s = √(6.2 × 10⁻¹³) ≈ 7.9 × 10⁻⁷ M. For NY₃, Ksp = 27s⁴, so s = (6.2 × 10⁻¹³/27)^(1/4) ≈ 3.9 × 10⁻⁴ M, which is larger. Equal Ksp does not mean equal solubility when the formulas differ, and adding KY lowers both solubilities by the common ion effect.
Q19NEET 2016 Phase 1 · Q72
Consider the following liquid - vapour equilibrium. Liquid ⇌ Vapour Which of the following relations is correct?
Not quite — the answer is D.
For liquid ⇌ vapour, the vapour pressure P plays the role of the equilibrium constant, and the van't Hoff (Clausius–Clapeyron) equation gives d lnP/dT = ΔHᵥ/RT². The sign is positive because vapour pressure rises with temperature for an endothermic vaporisation, and the derivative is with respect to T, not T².
Q20AIPMT 2015
Which of the following statements is correct for a reversible process in a state of equilibrium?
Not quite — the answer is D.
ΔG = ΔG° + 2.303RT log Q, and at equilibrium ΔG = 0 with Q = K. So ΔG° = −2.303RT log K. ΔG = −2.303RT log K is wrong because ΔG itself is zero at equilibrium.
Q21AIPMT 2015
The Ksp of Ag₂CrO₄, AgCl, AgBr and AgI are respectively, 1.1 × 10⁻¹², 1.8 × 10⁻¹⁰, 5.0 × 10⁻¹³, 8.3 × 10⁻¹⁷. Which one of the following salts will precipitate last if AgNO₃ solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na₂CrO₄?
Not quite — the answer is A.
[Ag⁺] needed to start precipitation (for 1 M anion): AgI 8.3 × 10⁻¹⁷, AgBr 5.0 × 10⁻¹³, AgCl 1.8 × 10⁻¹⁰ M, but Ag₂CrO₄ needs √(1.1 × 10⁻¹²) ≈ 1.05 × 10⁻⁶ M. Ag₂CrO₄ needs the highest Ag⁺, so it precipitates last even though its Ksp is not the largest.
Q22AIPMT 2015
If the value of an equilibrium constant for a particular reaction is 1.6 × 10¹², then at equilibrium the system will contain:
Not quite — the answer is D.
K = [products]/[reactants] = 1.6 × 10¹² is very large, so at equilibrium products far exceed reactants. Similar amounts would correspond to K near 1.
Q23AIPMT 2014
Which of the following salts will give highest pH in water?
Not quite — the answer is B.
Na₂CO₃ is a salt of a strong base and weak acid; CO₃²⁻ hydrolyses to give OH⁻, so its solution is basic. NaCl and KCl give neutral solutions and CuSO₄ is acidic by hydrolysis of Cu²⁺.
Q24AIPMT 2014
Using the Gibbs energy change, ΔG° = +63.3 kJ, for the following reaction, Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq), the Ksp of Ag₂CO₃(s) in water at 25°C is: (R = 8.314 J K⁻¹ mol⁻¹)
Not quite — the answer is A.
ΔG° = −2.303RT log Ksp, so log Ksp = −63 300/(2.303 × 8.314 × 298) = −11.09. Ksp = 10⁻¹¹·⁰⁹ ≈ 8.0 × 10⁻¹².
Q25AIPMT 2014
For the reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat. The equilibrium shifts in forward direction:
Not quite — the answer is C.
The forward reaction lowers the number of gas moles (4 → 2) and releases heat. By Le Chatelier's principle, raising pressure and lowering temperature both favour it; adding NH₃ or removing reactants pushes it backward.
Q26NEET 2013
Identify the correct order of solubility in aqueous medium
Not quite — the answer is D.
Na₂S is an ionic alkali-metal sulphide and is freely soluble. ZnS (Ksp ≈ 10⁻²²) is sparingly soluble, and CuS (Ksp ≈ 10⁻³⁶) is far less soluble. So the order is Na₂S > ZnS > CuS.
Q27NEET 2013
Which of these is least likely to act as a Lewis base ?
Not quite — the answer is C.
A Lewis base donates an electron pair. CO, F⁻ and PF₃ all have lone pairs to donate, but boron in BF₃ has only six valence electrons and accepts a pair, so BF₃ is a Lewis acid.
Q28AIPMT 2012 Mains
Given that the equilibrium constant for the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? SO₃(g) ⇌ SO₂(g) + (1/2)O₂(g)
Not quite — the answer is C.
The required reaction is the given one reversed and halved, so K′ = (1/278)^(1/2) = 1/16.7 ≈ 6.0 × 10⁻². Reversing without taking the square root gives 3.6 × 10⁻³.
Q29AIPMT 2012 Mains
Given the reaction between 2 gases represented by A₂ and B₂ to give the compound AB(g): A₂(g) + B₂(g) ⇌ 2AB(g). At equilibrium, the concentration of A₂ = 3.0 × 10⁻³ M, of B₂ = 4.2 × 10⁻³ M and of AB = 2.8 × 10⁻³ M. If the reaction takes place in a sealed vessel at 527°C, then the value of Kc will be
Not quite — the answer is C.
Kc = [AB]²/([A₂][B₂]) = (2.8 × 10⁻³)²/(3.0 × 10⁻³ × 4.2 × 10⁻³) = 7.84 × 10⁻⁶/1.26 × 10⁻⁵ ≈ 0.62. The temperature is not needed, since Kc is computed directly from concentrations.
Q30AIPMT 2012 Prelims
Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value?
Not quite — the answer is B.
BaCl₂ is a salt of a strong acid and a strong base, so it does not hydrolyse and its solution is neutral (pH ≈ 7). BeCl₂ and AlCl₃ hydrolyse strongly to give acidic solutions, and LiCl is very slightly acidic due to the small, polarising Li⁺. So BaCl₂ has the highest pH.
Q31AIPMT 2012 Prelims
Buffer solutions have constant acidity and alkalinity because:
Not quite — the answer is B.
A buffer contains a weak acid with its conjugate base (or a weak base with its conjugate acid). Added H⁺ or OH⁻ is consumed to form the weakly ionised acid or base, so [H⁺] changes very little. A buffer's pH is not fixed absolutely (it depends on the ratio of components), and it does not hold a large excess of H⁺ or OH⁻.
Q32AIPMT 2012 Prelims
pH of a saturated solution of Ba(OH)₂ is 12. The value of solubility product (Ksp) of Ba(OH)₂ is
Not quite — the answer is C.
pH 12 means pOH = 2 and [OH⁻] = 10⁻² M, so [Ba²⁺] = 5 × 10⁻³ M. Ksp = [Ba²⁺][OH⁻]² = 5 × 10⁻³ × (10⁻²)² = 5 × 10⁻⁷. Writing 4s³ with s = 10⁻² wrongly treats [OH⁻] as the solubility, giving 4 × 10⁻⁶.
Q33AIPMT 2011 Mains
In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag⁺ and Pb²⁺ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl⁻ concentration is 0.10 M. What will the concentrations of Ag⁺ and Pb²⁺ be at equilibrium? (Ksp for AgCl = 1.8 × 10⁻¹⁰, Ksp for PbCl₂ = 1.7 × 10⁻⁵)
Not quite — the answer is A.
At [Cl⁻] = 0.10 M: [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.1 = 1.8 × 10⁻⁹ M and [Pb²⁺] = Ksp/[Cl⁻]² = 1.7 × 10⁻⁵/0.01 = 1.7 × 10⁻³ M. Choices that use [Cl⁻] instead of [Cl⁻]² for PbCl₂ give 1.7 × 10⁻⁴ M, which is wrong.
Q34AIPMT 2011 Prelims
A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of NH₄⁺ is 0.20 M. If the equilibrium constant, Kb for NH₃ equals 1.8 × 10⁻⁵, what is the pH of this solution?
Not quite — the answer is C.
pKb = −log(1.8 × 10⁻⁵) = 4.74. pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.20/0.30) = 4.74 − 0.18 = 4.57, so pH = 14 − 4.57 = 9.43. Swapping the ratio gives pOH 4.92 and pH 9.08, the trap.
Q35AIPMT 2011 Prelims
For the reaction N₂(g) + O₂(g) ⇌ 2NO(g), the equilibrium constant is K₁. The equilibrium constant is K₂ for the reaction 2NO(g) + O₂(g) ⇌ 2NO₂(g). What is K for the reaction NO₂(g) ⇌ (1/2)N₂(g) + O₂(g)?
Not quite — the answer is D.
Adding the two given reactions gives N₂ + 2O₂ ⇌ 2NO₂ with K = K₁K₂. The required reaction is the reverse of this halved, so K = (K₁K₂)^(−1/2) = [1/(K₁K₂)]^(1/2). Halving a reaction takes the square root of K; it does not divide K by 2.
Q36AIPMT 2011 Prelims
The value of ΔH for the reaction X₂(g) + 4Y₂(g) ⇌ 2XY₄(g) is less than zero. Formation of XY₄(g) will be favoured at
Not quite — the answer is A.
The forward reaction reduces gas moles from 5 to 2, so high pressure shifts it forward (Le Chatelier). It is exothermic (ΔH < 0), so low temperature favours products. Hence high pressure and low temperature.
Q37AIPMT 2010 Mains · Q92
The reaction 2A(g) + B(g) ⇌ 3C(g) + D(g) is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
Not quite — the answer is B.
Forming 0.25 M of D consumes 0.25 M B and 0.50 M A and forms 0.75 M C. At equilibrium [A] = 0.50, [B] = 0.75, [C] = 0.75, [D] = 0.25, so K = (0.75)³(0.25)/[(0.50)²(0.75)]. Using the initial 1.00 M values in the denominator ignores the amounts consumed.
Q38AIPMT 2010 Prelims · Q55
In a buffer solution containing equal concentration of B⁻ and HB, the Kb for B⁻ is 10⁻¹⁰. The pH of buffer solution is
Not quite — the answer is A.
For the buffer B⁻/HB, pOH = pK_b + log([HB]/[B⁻]). With equal concentrations pOH = pK_b = 10, so pH = 14 − 10 = 4. Answering 10 confuses pOH with pH.
Q39AIPMT 2010 Prelims · Q57
What is [H⁺] in mol/L of a solution that is 0.20 M in CH₃COONa and 0.10 M in CH₃COOH? Ka for CH₃COOH = 1.8 × 10⁻⁵.
Not quite — the answer is A.
Buffer: [H⁺] = K_a[CH₃COOH]/[CH₃COO⁻] = 1.8 × 10⁻⁵ × 0.10/0.20 = 9.0 × 10⁻⁶ M. 1.8 × 10⁻⁵ would hold only if the acid and salt concentrations were equal.
Q40AIPMT 2010 Prelims · Q66
In which of the following equilibrium Kc and Kp are not equal?
Not quite — the answer is A.
K_p = K_c(RT)^Δn_g. For 2C(s) + O₂(g) ⇌ 2CO(g), only gases count: Δn_g = 2 − 1 = 1, so K_p ≠ K_c. In the other three reactions the moles of gas are the same on both sides (Δn_g = 0).
Q41AIPMT 2010 Prelims · Q67
If pH of a saturated solution of Ba(OH)₂ is 12, the value of its Ksp is
Not quite — the answer is A.
pH 12 → pOH 2 → [OH⁻] = 10⁻² M. Ba(OH)₂ gives s and 2s, so s = 5 × 10⁻³ M. K_sp = s(2s)² = 5 × 10⁻³ × 10⁻⁴ = 5.00 × 10⁻⁷ M³. Using 4s³ with s = 10⁻² gives 4 × 10⁻⁶.
Q42AIPMT 2009 Prelims · Q182
Which of the following molecules acts as a Lewis acid?
Not quite — the answer is D.
In (CH₃)₃B boron has only six valence electrons and an empty p orbital, so it accepts an electron pair: a Lewis acid. The O, P and N compounds each carry a lone pair and act as Lewis bases.
Q43AIPMT 2009 Prelims · Q187
The ionization constant of ammonium hydroxide is 1.77 × 10⁻⁵ at 298 K. Hydrolysis constant of ammonium chloride is:
Not quite — the answer is D.
For the salt of a weak base and strong acid, Kh = Kw/Kb = 1.0 × 10⁻¹⁴/1.77 × 10⁻⁵ = 5.65 × 10⁻¹⁰. Values with 10⁻¹² or 10⁻¹³ come from slips in the exponent.
Q44AIPMT 2009 Prelims · Q197
What is the [OH⁻] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?
Not quite — the answer is D.
HCl gives 20 × 0.050 = 1.0 mmol H⁺; Ba(OH)₂ gives 30 × 0.10 × 2 = 6.0 mmol OH⁻. After neutralisation 5.0 mmol OH⁻ remain in 50 mL: [OH⁻] = 0.10 M. Forgetting that Ba(OH)₂ gives two OH⁻ per formula unit leads to a wrong value.
Q45AIPMT 2008 Prelims · Q53
The value of equilibrium constant of the reaction HI(g) ⇌ ½H₂(g) + ½I₂(g) is 8.0. The equilibrium constant of the reaction H₂(g) + I₂(g) ⇌ 2HI(g) will be
Not quite — the answer is C.
Doubling the given reaction gives 2HI ⇌ H₂ + I₂ with K = 8² = 64. Reversing it gives H₂ + I₂ ⇌ 2HI with K = 1/64. 1/8 only reverses the reaction and forgets that doubling the equation squares K.
Q46AIPMT 2008 Prelims · Q62
Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H⁺ ion concentration in the mixture?
Not quite — the answer is C.
With equal volumes V, total H⁺ = (10⁻³ + 10⁻⁴ + 10⁻⁵)V = 1.11 × 10⁻³ V mol in a total volume 3V. [H⁺] = 1.11 × 10⁻³/3 = 3.7 × 10⁻⁴ M. 1.11 × 10⁻³ M forgets the threefold dilution.
Q47AIPMT 2008 Prelims · Q63
The values of Kp₁ and Kp₂ for the reactions X ⇌ Y + Z ....(1) and A ⇌ 2B ....(2) are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio
Not quite — the answer is D.
With degree of dissociation α: for the first reaction, moles 1 − α, α, α (total 1 + α) give Kp₁ = α²P₁/(1 − α²). For the second reaction, moles 1 − α, 2α (total 1 + α) give Kp₂ = 4α²P₂/(1 − α²). Then Kp₁/Kp₂ = P₁/(4P₂) = 9, so P₁ : P₂ = 36 : 1. 9 : 1 would forget the factor 4 from (2α)².
Q48AIPMT 2008 Prelims · Q64
If the concentration of OH⁻ ions in the reaction Fe(OH)₃(s) ⇌ Fe³⁺(aq) + 3OH⁻(aq) is decreased by 1/4 times, then equilibrium concentration of Fe³⁺ will increase by
Not quite — the answer is D.
K = [Fe³⁺][OH⁻]³ is constant (the solid does not appear). If [OH⁻] becomes one quarter, [OH⁻]³ falls by 4³ = 64, so [Fe³⁺] must increase 64 times. 4 times ignores the cube from the coefficient 3.
Q49AIPMT 2008 Prelims · Q68
The dissociation equilibrium of a gas AB₂ can be represented as, 2AB₂(g) ⇌ 2AB(g) + B₂(g) The degree of dissociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure P is
Not quite — the answer is C.
Starting with 1 mol AB₂: AB₂ = 1 − x, AB = x, B₂ = x/2, total = 1 + x/2 ≈ 1. Kp = (xP)²(xP/2)/(P)² = x³P/2, so x = (2Kp/P)^(1/3). The square-root form would hold for a reaction where x enters only as x², e.g. AB₂ ⇌ AB + B.
Q50AIPMT 2007 Prelims · Q54
A weak acid, HA, has a Ka of 1.00 × 10⁻⁵. If 0.100 mole of this acid dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to
Not quite — the answer is A.
α = √(Ka/C) = √(1.00 × 10⁻⁵/0.100) = 1.00 × 10⁻², i.e. 1.00%. 0.100% would come from Ka/C without the square root.
Q51AIPMT 2007 Prelims · Q82
Which one of the following ionic species has the greatest proton affinity to form stable compound?
Not quite — the answer is A.
Proton affinity is greatest for the strongest base. NH₂⁻ is the conjugate base of NH₃, an extremely weak acid, so it is the strongest base here. F⁻, I⁻ and HS⁻ are conjugate bases of much stronger acids.
Q52AIPMT 2007 Prelims · Q83
The following equilibrium constants are given: N₂ + 3H₂ ⇌ 2NH₃ ; K₁ N₂ + O₂ ⇌ 2NO ; K₂ H₂ + ½O₂ ⇌ H₂O ; K₃ The equilibrium constant for the oxidation of NH₃ by oxygen to give NO is
Not quite — the answer is D.
Target: 2NH₃ + (5/2)O₂ ⇌ 2NO + 3H₂O = (reverse of eq. 1) + (eq. 2) + 3 × (eq. 3). So K = K₂K₃³/K₁. Using K₃ only once ignores that three water molecules form.
Q53AIPMT 2007 Prelims · Q89
Calculate the pOH of a solution at 25°C that contains 1 × 10⁻¹⁰ M of hydronium ions, i.e. H₃O⁺.
Not quite — the answer is A.
pH = −log(1 × 10⁻¹⁰) = 10, and at 25°C pOH = 14 − 10 = 4.000. 10 would be the pH itself, which is not an option.
Q54AIPMT 2006 Prelims · Q69
For the reaction, CH₄(g) + 2O₂(g) ⇌ CO₂(g) + 2H₂O(l), ΔrH = −170.8 kJ mol⁻¹ Which of the following statements is not true ?
Not quite — the answer is B.
The correct expression excludes the pure liquid H₂O and raises O₂ to its coefficient: Kp = p(CO₂)/[p(CH₄)·p(O₂)²]. The given Kp = [CO₂]/[CH₄][O₂] omits the square on O₂, so it is the false statement. Adding a reactant shifts the equilibrium right, and a negative ΔrH means exothermic.
Q55AIPMT 2006 Prelims · Q82
Which of the following pairs constitutes a buffer ?
Not quite — the answer is A.
A buffer needs a weak acid with its conjugate base (or weak base with its conjugate acid). HNO₂ is a weak acid and NaNO₂ supplies NO₂⁻, so this pair is an acidic buffer. The other pairs contain a strong acid or strong base with a neutral salt and cannot resist pH changes.
Q56AIPMT 2006 Prelims · Q83
The hydrogen ion concentration of a 10⁻⁸ M HCl aqueous solution at 298 K (Kw = 10⁻¹⁴) is :
Not quite — the answer is B.
At 10⁻⁸ M the acid's H⁺ is smaller than that from water, so both must be counted: x(x − 10⁻⁸) = 10⁻¹⁴. Solving, x = [10⁻⁸ + √(10⁻¹⁶ + 4 × 10⁻¹⁴)]/2 ≈ 1.05 × 10⁻⁷ M (pH ≈ 6.98). Taking [H⁺] = 10⁻⁸ M would make an acid solution basic (pH 8).
Q57AIPMT 2005 Prelims · Q69
What is the correct relationship between the pHs of isomolar solutions of sodium oxide (pH₁), sodium sulphide (pH₂), sodium selenide (pH₃) and sodium telluride (pH₄) ?
Not quite — the answer is D.
These salts are basic by anion hydrolysis. Acid strength rises H₂O < H₂S < H₂Se < H₂Te, so the conjugate bases weaken O²⁻ > S²⁻ > Se²⁻ > Te²⁻ and pH₁ > pH₂ > pH₃ > pH₄.
Q58AIPMT 2005 Prelims · Q80
Equilibrium constants K₁ and K₂ for the following equilibria : NO(g) + ½O₂(g) ⇌ NO₂(g) and 2NO₂(g) ⇌ 2NO(g) + O₂(g) are related as :
Not quite — the answer is D.
The second equilibrium is the first reversed and doubled. Reversing gives 1/K₁ and doubling squares it, so K₂ = 1/K₁². 1/K₁ forgets the factor of 2.
Q59AIPMT 2005 Prelims · Q89
H₂S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because :
Not quite — the answer is A.
HCl suppresses ionisation of H₂S by the common-ion (H⁺) effect, lowering [S²⁻]. This low [S²⁻] still exceeds the very small Ksp of group II sulphides but not the larger Ksp of group IV sulphides. Group II sulphides have smaller, not larger, Ksp.
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