25 questions in 25 minutes on the NTA computer-based test interface, all from Equilibrium. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Equilibrium, of which 20 are real NEET previous-year questions and 14 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Equilibrium mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 20 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Equilibrium?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
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Chemistry
- Q1An aqueous solution of NH₄Cl (salt of weak base and strong acid) is:
- Basic
- Neutral
- Amphoteric
- Acidic
Answer: (D) Acidic
NH₄⁺ hydrolyses to give H₃O⁺ (NH₄⁺ + H₂O ⇌ NH₄OH + H⁺), making the solution acidic.
- Q2The Kₛₚ values 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 these salts will precipitate last if AgNO₃ solution is added to a solution containing equal moles of NaCl, NaBr, NaI and Na₂CrO₄?
- AgBr
- Ag₂CrO₄
- AgCl
- AgI
Answer: (B) Ag₂CrO₄
Converting each Kₛₚ to molar solubility: AgCl S = 1.34 × 10⁻⁵, AgBr S = 0.71 × 10⁻⁶, AgI S = 0.9 × 10⁻⁸, Ag₂CrO₄ S = sqrt[3]Kₛₚ/4 = 0.65 × 10⁻⁴. Ag₂CrO₄ has the highest solubility, so it needs the most Ag⁺ and precipitates last.
- Q3Which one of the following conditions will favour the maximum formation of the product in the reaction A₂(g) + B₂(g) leftharpoons X₂(g), Δᵣ H = -X kJ?
- High temperature and low pressure
- Low temperature and high pressure
- Low temperature and low pressure
- High temperature and high pressure
Answer: (B) Low temperature and high pressure
The reaction is exothermic, so low temperature shifts it forward. It goes from 2 gas moles to 1 (Δ n_g = -1), so high pressure shifts it toward fewer moles (product). Low temperature and high pressure maximise product.
- Q4Find the solubility of Ni(OH)₂ in 0.1 M NaOH, given that the solubility product of Ni(OH)₂ is 2 × 10⁻¹⁵:
- 2 × 10⁻¹³ M
- 2 × 10⁻⁸ M
- 1 × 10⁸ M
- 1 × 10⁻¹³ M
Answer: (A) 2 × 10⁻¹³ M
NaOH provides [OH⁻] = 0.1 M (common ion). Kₛₚ = [Ni²⁺][OH⁻]² = S(0.1)². So S = (2 × 10⁻¹⁵)/(0.01) = 2 × 10⁻¹³ M.
- Q5The solubility of a saturated solution of calcium fluoride is 2 × 10⁻⁴ mol/L. Its solubility product is:
- 12 × 10⁻²
- 14 × 10⁻⁴
- 22 × 10⁻²
- 32 × 10⁻¹²
Answer: (D) 32 × 10⁻¹²
CaF₂ leftharpoons Ca²⁺ + 2F⁻, Kₛₚ = (S)(2S)² = 4S³ = 4(2 × 10⁻⁴)³ = 4 × 8 × 10⁻¹² = 32 × 10⁻¹².
- Q6The dissociation constants for acetic acid and HCN at 25°C are 1.5 × 10⁻⁵ and 4.5 × 10⁻¹⁰ respectively. The equilibrium constant for the equilibrium CN⁻ + CH₃COOH leftharpoons HCN + CH₃COO⁻ would be:
- 3.0 × 10⁵
- 3.0 × 10⁴
- 3.0 × 10⁻⁴
- 3.0 × 10⁻⁵
Answer: (B) 3.0 × 10⁴
This is acetic acid donating H⁺ to CN⁻. K = (Kₐ(acetic))/(Kₐ(HCN)) = (1.5 × 10⁻⁵)/(4.5 × 10⁻¹⁰) = 3.33 × 10⁴ ≈ 3.0 × 10⁴.
- Q7In a buffer solution containing equal concentrations of B⁻ and HB, the K_b for B⁻ is 10⁻¹⁰. The pH of the buffer solution is:
- 4
- 6
- 10
- 7
Answer: (A) 4
For this basic buffer pOH = pK_b + log([salt])/([base]). With equal concentrations the log term is 0, so pOH = pK_b = -log(10⁻¹⁰) = 10. Then pH = 14 – 10 = 4.
- Q8The equilibrium constant Kc depends on which of the following?
- Presence of a catalyst
- Initial concentrations of reactants
- Total pressure of the system
- Temperature only
Answer: (D) Temperature only
Kc is a function of temperature alone; it is independent of initial concentrations, catalyst and pressure.
- Q9In which of the following equilibria are K_c and Kₚ NOT equal?
- H₂(g) + I₂(g) leftharpoons 2HI(g)
- SO₂(g) + NO₂(g) leftharpoons SO₃(g) + NO(g)
- 2NO(g) leftharpoons N₂(g) + O₂(g)
- 2C(s) + O₂(g) leftharpoons 2CO₂(g)
Answer: (D) 2C(s) + O₂(g) leftharpoons 2CO₂(g)
Kₚ = K_c(RT)^(Δ n), equal only when Δ n_(gas) = 0. In (A), (B), (C) gaseous moles are balanced (Δ n = 0). In (D), carbon is solid, so Δ n_(gas) = 2 – 1 = +1 ≠ 0, hence Kₚ ≠ K_c.
- Q10For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kp and Kc is:
- Kp = Kc(RT)²
- Kp = Kc(RT)⁻²
- Kp = Kc
- Kp = Kc(RT)⁻¹
Answer: (B) Kp = Kc(RT)⁻²
Δn = 2 − (1+3) = −2, so Kp = Kc(RT)^Δn = Kc(RT)⁻².
- Q11Solubility of MX₂ type electrolytes is 0.5 × 10⁻⁴ mol/L, then the Kₛₚ of the electrolyte is:
- 5 × 10⁻¹²
- 1 × 10⁻¹³
- 5 × 10⁻¹³
- 25 × 10⁻¹⁰
Answer: (C) 5 × 10⁻¹³
MX₂ leftharpoons M²⁺ + 2X⁻, Kₛₚ = (S)(2S)² = 4S³ = 4(0.5 × 10⁻⁴)³ = 4 × 1.25 × 10⁻¹³ = 5 × 10⁻¹³.
- Q12Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. The H⁺ ion concentration in the mixture is:
- 3.7 × 10⁻³ M
- 1.11 × 10⁻³ M
- 1.11 × 10⁻⁴ M
- 3.7 × 10⁻⁴ M
Answer: (D) 3.7 × 10⁻⁴ M
[H⁺] values are 10⁻³, 10⁻⁴, 10⁻⁵. Mixing equal volumes V each: [H⁺] = (10⁻³ + 10⁻⁴ + 10⁻⁵)/(3) = (1.11 × 10⁻³)/(3) = 3.7 × 10⁻⁴ M.
- Q13The pK_b of dimethylamine and pKₐ of acetic acid are 3.27 and 4.77 respectively at T(K). The pH of dimethyl ammonium acetate solution is:
- 7.75
- 5.50
- 6.25
- 8.50
Answer: (A) 7.75
For a salt of a weak acid and weak base, pH = 7 + (1)/(2)(pKₐ – pK_b) = 7 + (1)/(2)(4.77 – 3.27) = 7 + 0.75 = 7.75.
- Q14The pOH of a solution at 25°C that contains 1 × 10⁻¹⁰ M of hydronium ion is:
- 9.00
- 4.00
- 7.00
- 1.00
Answer: (B) 4.00
pH = -log(10⁻¹⁰) = 10. Since pH + pOH = 14, pOH = 14 – 10 = 4.
- Q15Adding a catalyst to a system at equilibrium:
- Increases the value of K
- Shifts equilibrium toward fewer moles
- Shifts equilibrium forward
- Has no effect on the equilibrium position
Answer: (D) Has no effect on the equilibrium position
A catalyst speeds up forward and backward reactions equally, so equilibrium is reached faster but the position and K are unchanged.
- Q16KMnO₄ can be prepared from K₂MnO₄ per the reaction 3MnO₄²⁻ + 2H₂O leftharpoons 2MnO₄⁻ + MnO₂ + 4OH⁻. The reaction can go to completion by adding:
- SO₂
- CO₂
- HCl
- KOH
Answer: (B) CO₂
Removing OH⁻ from the right shifts the equilibrium forward. CO₂ is acidic and consumes OH⁻ (CO₂ + 2OH⁻ → CO₃²⁻ + H₂O) without introducing a strong acid that would reverse the reaction. HCl would over-acidify and reverse it; KOH adds OH⁻.
- Q17If the equilibrium constant for N₂(g) + O₂(g) leftharpoons 2NO(g) is K, the equilibrium constant for (1)/(2)N₂(g) + (1)/(2)O₂(g) leftharpoons NO(g) is:
- K²
- K
- K^(1/2)
- (1)/(2)K
Answer: (C) K^(1/2)
Halving all coefficients raises the equilibrium constant to the power ½. So the new constant is K’ = K^(1/2) = √(K).
- Q18The conjugate acid of NH₂⁻ is:
- NH₂OH
- N₂H₄
- NH₄⁺
- NH₃
Answer: (D) NH₃
The conjugate acid is formed by adding one H⁺. NH₂⁻ + H⁺ → NH₃. So the conjugate acid of NH₂⁻ is NH₃.
- Q19For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the correct expression for Kc is:
- [SO₃]²/([SO₂]²[O₂])
- [SO₂]²[O₂]/[SO₃]²
- [SO₃]/([SO₂][O₂])
- 2[SO₃]/(2[SO₂]+[O₂])
Answer: (A) [SO₃]²/([SO₂]²[O₂])
Kc = products over reactants, each raised to its stoichiometric coefficient: [SO₃]²/([SO₂]²[O₂]).
- Q20A 20 litre container at 400 K contains CO₂(g) at a pressure of 0.4 atm and an excess of solid SrO. The container volume is decreased by a piston. The maximum volume of the container when the pressure of CO₂ attains its maximum value will be, given SrCO₃(s) ⇌ SrO(s) + CO₂(g), Kₚ = 1.6 atm:
- 4 L
- 5 L
- 2 L
- 10 L
Answer: (B) 5 L
The maximum partial pressure CO₂ can reach is Kₚ = 1.6 atm (above this SrCO₃ forms). Compressing the gas (Boyle’s law, isothermal): p₁V₁ = p₂V₂, so 0.4 × 20 = 1.6 × V₂, giving V₂ = 8/1.6 = 5 L.
- Q21MY and NY₃, two nearly insoluble salts, have the same Kₛₚ value of 6.2 × 10⁻¹³ at room temperature. Which statement is true regarding MY and NY₃?
- The addition of KY to a solution of MY and NY₃ has no effect on their solubilities
- The salts MY and NY₃ are more soluble in 0.5 M KY than in pure water
- The molar solubility of MY in water is less than that of NY₃
- The molar solubilities of MY and NY₃ in water are identical
Answer: (C) The molar solubility of MY in water is less than that of NY₃
For MY: Kₛₚ = S², so S_(MY) = √(6.2 × 10⁻¹³) = 7.9 × 10⁻⁷ M. For NY₃: Kₛₚ = 27S⁴, so S_(NY₃) = (6.2 × 10⁻¹³/27)^(1/4) = 3.9 × 10⁻⁴ M. Thus S_(MY) < S_(NY₃).
- Q22What 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?
- 0.10 M
- 0.12 M
- 0.0050 M
- 0.40 M
Answer: (A) 0.10 M
Milliequivalents of HCl = 20 × 0.050 = 1. Milliequivalents of OH⁻ from Ba(OH)₂ = 2 × 30 × 0.10 = 6. Excess OH⁻ = 6 – 1 = 5 meq in total volume 50 mL, so [OH⁻] = 5/50 = 0.10 M.
- Q23A weak acid HA has a Kₐ of 1.00 × 10⁻⁵. If 0.100 mole of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to:
- 0.100%
- 1.00%
- 99.0%
- 99.9%
Answer: (B) 1.00%
C = 0.100 M. α = √(Kₐ/C) = √(10⁻⁵/0.1) = √(10⁻⁴) = 10⁻². Percentage dissociated = 10⁻² × 100 = 1.0%.
- Q24What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
- 12.65
- 1.04
- 7.0
- 2.0
Answer: (A) 12.65
NaOH is in excess. After mixing equal volumes, net [OH⁻] = (0.1 – 0.01)/(2) = 0.045 M. pOH = -log(0.045) = 1.35, so pH = 14 – 1.35 = 12.65.
- Q25The hydrogen ion concentration of a 10⁻⁸ M HCl aqueous solution at 298 K (K_w = 10⁻¹⁴) is:
- 9.525 × 10⁻⁸ M
- 1.0 × 10⁻⁸ M
- 1.0525 × 10⁻⁷ M
- 1.0 × 10⁻⁶ M
Answer: (C) 1.0525 × 10⁻⁷ M
At such low acid concentration water’s contribution matters. Charge/mass balance: [H⁺] = 10⁻⁸ + [OH⁻] and [H⁺][OH⁻] = 10⁻¹⁴. Solving x² – 10⁻⁸x – 10⁻¹⁴ = 0 gives x = 1.05 × 10⁻⁷ M.