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← JEE Advanced 2015 · Paper 2
ChemistryMultiple correct+4 / -2Question 40 of 60

JEE Advanced 2015 · Paper 2, Chemistry question 40

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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=2.0×10−5K_a=2.0\times10^{-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

Choose one or more options
Answer and worked solution

Official answer: B

Composition. 0.2 mol of CH₃COOH and 0.1 mol of NaOH leave 0.1 mol of acid and 0.1 mol of acetate: a buffer with equal amounts.

pH.
pH=pKa=−log⁡(2×10−5)=4.7.\mathrm{pH}=\mathrm pK_a=-\log(2\times10^{-5})=4.7.

(B)