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 5.7ºC 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 (K a = 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 5.6ºC 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 )
(i) Enthalpy of dissociation (in kJ mol –1 ) of acetic acid obtained from the Expt. 2 is :
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) Let the heat capacity of insulated beaker be C.
Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g
⇒ Total heat capacity = (C + 200 × 4.2) J/K
Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1
⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ
⇒ 5.7 × 1000 = (C + 200 × 4.2) × Δ T.
5.7 × 1000 = (C + 200 + 4.2) × 5.7
⇒ (C + 200 × 4.2) = 1000
In second experiment, 
Total mass of aqueous content = 200 g
⇒ Total heat capacity = (C + 200 × 4.2) = 1000
⇒ Heat released = 1000 × 5.6 = 5600 J.
Overall, only 0.1 mol of CH 3 COOH undergo neutralization.
⇒ Δ H neutralization of CH 3 COOH = 
= – 56000 J/mol = – 56 KJ/mol.
⇒ Δ H ionization of CH 3 COOH = 57 – 56 = 1 KJ/mol
(ii) Final solution contain 0.1 mole of CH 3 COOH and CH 3 COONa each.
Hence it is a buffer solution.
pH = pK a + 
= 5–log 2 + log 
= 4.7
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems