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Enthalpy of CuSO4 solution and strong-acid/strong-base neutralization: NEET MCQs with answers

49 questions on this topic of Principles Related to Practical Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Consider the following steps in the calorimetric procedure to find the enthalpy of neutralization: (p) Record the steady temperature of the acid solution. (q) Record the steady temperature of the base solution. (r) Pour the base rapidly into the acid and stir constantly. (s) Measure the maximum temperature reached. If the initial temperature of 100 mL acid was 24.8 °C and 100 mL base was 25.2 °C, what is the correct initial temperature (T₁) to be used in the calculation?

  1. 24.8 °C
  2. 25.2 °C
  3. 50.0 °C
  4. 25.0 °C
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Answer: (D) 25.0 °C

When equal volumes of acid and base at slightly different initial temperatures are mixed, the baseline initial temperature T₁ of the combined system prior to reaction is taken as the arithmetic mean of the two individual temperatures: T₁ = (24.8 + 25.2) / 2 = 25.0 °C.

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2. In an experiment to determine the enthalpy of solution of anhydrous CuSO₄, 0.02 mol of CuSO₄(s) is dissolved in 100 g of water. The temperature rises by 3.0 °C. Taking the specific heat capacity of the solution as 4.2 J g⁻¹ °C⁻¹ and neglecting the heat capacity of the calorimeter and the mass of solute, what is the calculated molar enthalpy of solution of CuSO₄?

  1. +63.0 kJ mol⁻¹
  2. -63.0 kJ mol⁻¹
  3. -12.6 kJ mol⁻¹
  4. -31.5 kJ mol⁻¹
Show answer and explanation

Answer: (B) -63.0 kJ mol⁻¹

Heat evolved q = m × c × ΔT = 100 g × 4.2 J g⁻¹ °C⁻¹ × 3.0 °C = 1260 J = 1.26 kJ. Since temperature increases, the process is exothermic. ΔH_sol = -q / n = -1.26 kJ / 0.02 mol = -63.0 kJ mol⁻¹.

hard

3. Using Hess's law and practical thermochemical cycles, consider the dissolution processes:1. CuSO₄(s) + 800 H₂O(l) → CuSO₄(800 H₂O), ΔH₁ = -66.4 kJ mol⁻¹2. CuSO₄·5H₂O(s) + 795 H₂O(l) → CuSO₄(800 H₂O), ΔH₂ = +11.7 kJ mol⁻¹What is the enthalpy change for the hydration of one mole of anhydrous copper(II) sulphate?

  1. +78.1 kJ mol⁻¹
  2. -78.1 kJ mol⁻¹
  3. -54.7 kJ mol⁻¹
  4. +54.7 kJ mol⁻¹
Show answer and explanation

Answer: (B) -78.1 kJ mol⁻¹

The hydration reaction is: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s). By Hess's law: ΔH_hydration = ΔH₁ - ΔH₂ = -66.4 kJ mol⁻¹ - (+11.7 kJ mol⁻¹) = -78.1 kJ mol⁻¹.

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4. Statement I: The enthalpy of neutralization of 1 mol of H₂SO₄ by 2 mol of NaOH is approximately -114.6 kJ.Statement II: Enthalpy of neutralization is conventionally reported per mole of water formed during the acid-base reaction.

  1. Both Statement I and Statement II are correct, and Statement II is the correct explanation of Statement I.
  2. Both Statement I and Statement II are correct, but Statement II is not the correct explanation of Statement I.
  3. Statement I is correct but Statement II is incorrect.
  4. Statement I is incorrect but Statement II is correct.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct, and Statement II is the correct explanation of Statement I.

Neutralization of 1 mol of diprotic H₂SO₄ by 2 mol of NaOH produces 2 moles of H₂O: 2H⁺ + 2OH⁻ → 2H₂O. Since the standard enthalpy per mole of water formed is -57.3 kJ mol⁻¹, the total enthalpy change for 2 moles of water formed is 2 × (-57.3) = -114.6 kJ. Thus, Statement II correctly explains Statement I.

easy

5. Which of the following is the primary experimental source of error leading to a lower numerical value of the enthalpy of neutralization than 57.3 kJ mol⁻¹ in a simple school laboratory setup?

  1. Incomplete dissociation of HCl and NaOH in dilute aqueous solutions.
  2. Heat loss from the open calorimeter cup and thermometer stem to the surrounding environment during stirring.
  3. Formation of NaCl which has a strongly endothermic lattice energy in water.
  4. Reaction of dissolved sodium ions with water molecules to form a basic complex.
Show answer and explanation

Answer: (B) Heat loss from the open calorimeter cup and thermometer stem to the surrounding environment during stirring.

HCl and NaOH are strong electrolytes and fully dissociate in dilute solutions. The primary reason student measurements systematically yield less than 57.3 kJ mol⁻¹ of heat is heat leakage to the surroundings through uninsulated calorimeter walls and evaporation during stirring before the maximum temperature is reached.

hard

6. A student dissolves 0.050 mol of anhydrous CuSO₄ in 200 g of water contained in a calorimeter with water equivalent 10 g. The temperature rises by 3.60 °C. Taking the specific heat capacity of water as 4.184 J g⁻¹ °C⁻¹, calculate the experimental molar enthalpy of solution of anhydrous CuSO₄.

  1. +31.6 kJ mol⁻¹
  2. +63.3 kJ mol⁻¹
  3. -31.6 kJ mol⁻¹
  4. -63.3 kJ mol⁻¹
Show answer and explanation

Answer: (D) -63.3 kJ mol⁻¹

Total mass effective = m_water + W = 200 + 10 = 210 g. Heat evolved q = m × c × ΔT = 210 g × 4.184 J g⁻¹ °C⁻¹ × 3.60 °C = 3163.1 J = 3.163 kJ. Molar enthalpy of solution ΔH_sol = -q / n = -3.163 kJ / 0.050 mol = -63.26 kJ mol⁻¹ ≈ -63.3 kJ mol⁻¹.

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7. In calorimetric experiments, the water equivalent of the calorimeter must be determined so that:

  1. The heat absorbed by the calorimeter itself is accounted for
  2. The calorimeter does not melt
  3. The thermometer is calibrated
  4. The solution becomes neutral
Show answer and explanation

Answer: (A) The heat absorbed by the calorimeter itself is accounted for

The calorimeter absorbs part of the heat; its water equivalent lets this be included in the heat balance.

NCERT Class 11

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8. Consider the following statements regarding the enthalpy of solution of anhydrous CuSO₄ and hydrated CuSO₄·5H₂O: Statement I: The dissolution of anhydrous CuSO₄ in excess water is an exothermic process. Statement II: The hydration enthalpy of anhydrous Cu²⁺ and SO₄²⁻ ions exceeds the lattice energy of anhydrous CuSO₄. In light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Dissolution of anhydrous CuSO₄ releases heat (exothermic, ΔH < 0) because the total energy released during ion hydration is greater in magnitude than the lattice energy required to break up the solid crystal. Therefore, Statement II correctly explains Statement I.

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9. During the measurement of enthalpy of neutralization of 100 mL of 1.0 M HCl with 100 mL of 1.0 M NaOH, the temperature of the mixture was recorded at 30-second intervals after mixing. Which graph represents the standard cooling correction method used to obtain the true maximum temperature (T₂)?

  1. Extrapolating the linear cooling curve backwards to the time of mixing.
  2. Taking the absolute highest point on the thermometer irrespective of time.
  3. Extrapolating the heating curve forward to infinite time.
  4. Averaging the initial temperature of acid and base and adding 5.7 °C.
Show answer and explanation

Answer: (A) Extrapolating the linear cooling curve backwards to the time of mixing.

Because calorimetry in a simple bench calorimeter is non-adiabatic, some heat is lost to the surroundings while the reaction proceeds. To correct for radiation and conduction heat losses, temperature readings taken during the cooling period are plotted against time and extrapolated backwards to the exact time of mixing to determine the true corrected maximum temperature T₂.

easy

10. In a laboratory determination of the enthalpy of neutralization of HCl by NaOH, which of the following practices leads to a significant systematic error?

  1. Using a glass beaker without a lid instead of a well-lagged polystyrene calorimeter
  2. Recording initial temperatures of both acid and base before mixing
  3. Using magnetic or wire stirring to ensure rapid and uniform mixing
  4. Employing dilute solutions (e.g., 0.5 M or 1.0 M) of both reactants
Show answer and explanation

Answer: (A) Using a glass beaker without a lid instead of a well-lagged polystyrene calorimeter

An open glass beaker has high thermal conductivity and allows heat loss to the surrounding atmosphere by convection and radiation, which systematically underestimates the observed temperature rise and enthalpy of neutralization.

hard

11. When 200 mL of 1 M HCl is completely neutralized by 100 mL of 2 M KOH, the temperature rise is ΔT₁. When 100 mL of 1 M HCl is completely neutralized by 50 mL of 2 M KOH in the same calorimeter, the temperature rise is ΔT₂. Assuming negligible heat capacity of the calorimeter, the ratio ΔT₁ : ΔT₂ is:

  1. 2 : 1
  2. 1 : 1
  3. 1 : 2
  4. 4 : 1
Show answer and explanation

Answer: (B) 1 : 1

In experiment 1, moles of reaction = 0.2 mol; total volume = 300 mL. In experiment 2, moles of reaction = 0.1 mol; total volume = 150 mL. Temperature rise ΔT is proportional to (moles of reaction / total volume). For both cases, the ratio (0.2/300) = (0.1/150) = 1/1500 mol mL⁻¹. Hence, ΔT₁ : ΔT₂ = 1 : 1.

easy

12. In the laboratory determination of the enthalpy of solution of hydrated copper(II) sulphate, CuSO₄·5H₂O, which of the following observations is correct?

  1. The final temperature of the solution is higher than the initial temperature, indicating an exothermic process
  2. The final temperature of the solution is lower than the initial temperature, indicating an endothermic process
  3. There is no measurable temperature change because hydration and lattice dissociation cancel out
  4. A bright red precipitate forms at the bottom of the calorimeter cup during dissolution
Show answer and explanation

Answer: (B) The final temperature of the solution is lower than the initial temperature, indicating an endothermic process

Dissolution of hydrated copper(II) sulphate (CuSO₄·5H₂O) requires more energy to overcome its crystal lattice than the hydration energy released, making it an endothermic process. As heat is absorbed from the aqueous medium, the temperature of the reaction mixture decreases.

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