Practice Chemical Equilibrium MCQs for MDCAT (Medical & Dental) Chemistry — topic-wise sets with solved answers.
Q1. A chemist measures the equilibrium concentrations for the H2 + I2 ℑ 2HI reaction at 444°C as [H2]=0.1 M, [I2]=0.1 M, and [HI]=0.4 M. What is the value of Kc?
Answer: 16.0
Explanation: Kc equals [HI]^2 / ([H2][I2]); substituting values gives (0.4)^2 / (0.1 * 0.1) = 16. Option A fails by forgetting to square the HI concentration.
Q2. During the thermal decomposition of PCl5 into PCl3 and Cl2 in a closed vessel, what is the mathematical relationship between Kp and Kc?
Answer: Kp = Kc(RT)
Explanation: Kp = Kc(RT)^Δn; here Δn = 2 - 1 = 1, so Kp = Kc(RT). Option A assumes Δn is zero, which only happens when moles are equal.
Q3. A laboratory technician prepares a solution of Sodium Hydroxide (NaOH) with a concentration of 0.001 M at 25°C. What is the pH of this solution?
Answer: 11
Explanation: For NaOH, [OH-] = 0.001 M, so pOH = 3; since pH + pOH = 14, pH = 11. Option D is the pOH, not the pH.
Q4. In a 1.0 dm³ flask, the equilibrium mixture for N2O4 ℑ 2NO2 contains 0.1 moles of N2O4 and 0.4 moles of NO2. Calculate the Kc value.
Answer: 1.6 mol dm⁻³
Explanation: In 1 dm³, moles equal molarity. Kc = [NO2]^2 / [N2O4] = (0.4)^2 / 0.1 = 1.6. Option B fails by not squaring the product concentration.
Q5. The molar solubility of Lead (II) chloride (PbCl2) in water is 0.01 mol/dm³ at 25°C. What is the calculated solubility product (Ksp) for this salt?
Answer: 4 × 10⁻⁶
Explanation: Ksp for PbCl2 = 4s³; substituting s = 0.01 gives 4 * (10⁻²)³ = 4 × 10⁻⁶. Option A incorrectly uses the s² formula for AB type salts.
Q6. An acidic buffer is prepared using 0.1 M sodium acetate and 0.01 M acetic acid (pKa = 4.74). What is the pH of the resulting buffer?
Answer: 5.74
Explanation: pH = pKa + log([Salt]/[Acid]); pH = 4.74 + log(0.1/0.01) = 4.74 + 1 = 5.74. Option A is just the pKa without the log ratio.
Q7. According to the PTB textbook, what is the value of the ionic product of water (Kw) at exactly 25°C?
Answer: 1.0 × 10⁻ⁱ⁴ mol²dm⁻⁶
Explanation: At 25°C, Kw = [H+][OH-] = 10⁻¹⁴. Option B is the pKw value, which is the negative log, not the product itself.
Q8. For the reaction A + B ℑ C + D, Kc is 10. If a mixture contains [A]=1, [B]=1, [C]=2, [D]=2, what is the predicted direction?
Answer: The reaction proceeds in the forward direction
Explanation: Qc = [C][D] / ([A][B]) = (2*2)/(1*1) = 4. Since Qc (4) < Kc (10), the reaction proceeds forward. Option A is incorrect because Qc ≠ Kc.
Q9. Calculate the molar solubility of AgCl (Ksp = 1.0 × 10⁻ⁱ⁰) in a 0.1 M solution of Sodium Chloride (NaCl) at 25°C.
Answer: 1.0 × 10⁻⁹ mol/dm³
Explanation: Common ion effect: [Cl-] ≈ 0.1 M from NaCl. Ksp = [Ag+][Cl-] → 10⁻ⁱ⁰ = [Ag+](0.1) → [Ag+] = 10⁻⁹. Option A ignores the NaCl concentration.
Q10. A student finds that a hydrochloric acid solution has a concentration of 0.0001 M. What is the pOH of this solution at 25°C?
Answer: 10
Explanation: HCl is a strong acid; [H+] = 10⁻⁴ M. pH = 4, so pOH = 14 - 4 = 10. Option A is the pH value, not the pOH.
Q11. If the equilibrium constant (Kc) for the formation of HI from H2 and I2 is 0.25 at a specific temperature, what is Kc for the dissociation of HI?
Answer: 4.0
Explanation: Kc for the reverse reaction is the reciprocal of the forward Kc; 1 / 0.25 = 4. Option A is simply the forward constant.
Q12. A saturated solution of Mg(OH)2 has a pH of 10.0 at 25°C. Based on this information, calculate the solubility product constant (Ksp).
Answer: 5 × 10⁻ⁱ
Explanation: pH 10 means pOH = 4, so [OH-] = 10⁻⁴ M. For Mg(OH)2, [Mg2+] = 0.5 × 10⁻⁴. Ksp = [Mg2+][OH-]² = (0.5 × 10⁻⁴)(10⁻⁴)² = 5 × 10⁻ⁱ. Option A is a common calculation error.
Q13. For which of the following gaseous reversible reactions is the value of Kp numerically equal to the value of Kc?
Answer: H2 + I2 ℑ 2HI
Explanation: Kp = Kc(RT)^Δn. For Kp to equal Kc, Δn (moles of gaseous products minus reactants) must be 0. Option B results in Kp = Kc(RT).
Q14. A 0.1 M solution of a weak monobasic acid has a Ka of 1.0 × 10⁻⁵. What is the pH of this solution?
Answer: 3
Explanation: For a weak acid, [H+] = √(Ka × C) = √(10⁻⁵ × 0.1) = 10⁻³. Thus pH = 3. Option A assumes the acid is strong and fully dissociates.
Q15. In the equilibrium X ℑ CO2 + OH-, 0.44g of CO2 and 0.01 moles of OH- are present in 1 dm³. Calculate Kc (Molar mass of CO2 = 44).
Answer: 1.0 × 10⁻⁴
Explanation: Molality of CO2 = 0.44g / 44g/mol = 0.01 mol. In 1 dm³, [CO2] = 0.01 M. [OH-] = 0.01 M. Kc = [CO2][OH-] = 10⁻⁴. Option A misses the molar mass conversion.
Q16. What are the units of the equilibrium constant Kc for the synthetic reaction A + B ℑ C, where all species are in aqueous phase?
Answer: dm³ mol⁻¹
Explanation: Kc units are (mol/dm³)^Δn. Here Δn = 1 - (1+1) = -1, so units are dm³/mol. Option A is for reactions where Δn = 0.
Q17. In a Haber process mixture at 12 atm, there are 1 mole of N2, 3 moles of H2, and 2 moles of NH3. Calculate the partial pressure of NH3.
Answer: 4 atm
Explanation: Partial pressure = mole fraction × total pressure. Total moles = 1+3+2 = 6. P(NH3) = (2/6) × 12 atm = 4 atm. Option B incorrectly uses the mole ratio 2/4.
Q18. Calculate the percentage ionization of a 0.1 M Ammonia (NH3) solution, given that its base dissociation constant Kb is 1.8 × 10⁻⁵.
Answer: 1.34%
Explanation: Degree of ionization α = √(Kb/C) = √(1.8 × 10⁻⁵ / 0.1) = √(1.8 × 10⁻⁴) ≈ 0.0134 or 1.34%. Option A is the value of Kb itself.
Q19. If 50 cm³ of 0.1 M HCl is mixed with 50 cm³ of 0.1 M NaOH at 25°C, what is the pH of the final mixture?
Answer: pH = 7
Explanation: Mixing equal volumes of a strong acid and strong base of the same molarity results in complete neutralization, forming a neutral solution (pH 7).
Q20. A specific weak acid has a Ka value of 1.0 × 10⁻⁵. What is the pKb value of its conjugate base at 25°C?
Answer: 9
Explanation: pKa = -log(Ka) = -log(10⁻⁵) = 5. pKb = 14 - pKa = 14 - 5 = 9. Option A is the pKa value.
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