PARAGRAPH 2
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol⁻¹), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Kₐ = 2.0 × 10⁻⁵) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured.
(Consider heat capacity of all solutions as 4.2 J g⁻¹ K⁻¹ and density of all solutions as 1.0 g mL⁻¹)
The pH of the solution after Expt. 2 is
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 °C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (-57.0 kJ mol⁻¹), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (Kₐ = 2.0 × 10⁻⁵) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6 °C was measured.
(Consider heat capacity of all solutions as 4.2 J g⁻¹ K⁻¹ and density of all solutions as 1.0 g mL⁻¹)
The pH of the solution after Expt. 2 is
- (A)2.8
- (B)4.7✓ Correct
- (C)5.0
- (D)7.0
Answer: (B)
Solution
After Expt. 2:
This is a buffer solution with equal concentrations of weak acid and its conjugate base.
Using Henderson-Hasselbalch equation:
pH = pKₐ + log([A⁻]/[HA])
pKₐ = -log(2.0 × 10⁻⁵) = 4.7
[CH₃COO⁻] = 0.1/0.2 = 0.5 M
[CH₃COOH] = 0.1/0.2 = 0.5 M
pH = 4.7 + log(0.5/0.5) = 4.7 + 0 = 4.7
Correct answer: (B).
- Initial: 0.2 mol CH₃COOH, 0.1 mol NaOH
- After neutralization: 0.1 mol CH₃COONa (salt), 0.1 mol CH₃COOH (excess acid)
- Total volume = 200 mL
This is a buffer solution with equal concentrations of weak acid and its conjugate base.
Using Henderson-Hasselbalch equation:
pH = pKₐ + log([A⁻]/[HA])
pKₐ = -log(2.0 × 10⁻⁵) = 4.7
[CH₃COO⁻] = 0.1/0.2 = 0.5 M
[CH₃COOH] = 0.1/0.2 = 0.5 M
pH = 4.7 + log(0.5/0.5) = 4.7 + 0 = 4.7
Correct answer: (B).
JEE Advanced · 2015 · 24 May 2015 · Paper 2 · One or more correct options