JEE Advance - Chemistry (2015 - Paper 2 Offline - No. 5)
Paragraph
When 100 mL of 1.0 M KCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7o 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 constant (-57.0 kJ/mol), 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 $$\times$$ 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
The pH of the solution after Expt. 2 is
When 100 mL of 1.0 M KCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7o 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 constant (-57.0 kJ/mol), 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 $$\times$$ 10-5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6o C was measured. (Consider heat capacity of all solutions as 4.2 J/gK and density of all solutions as 1.0 g m/L)
Question
The pH of the solution after Expt. 2 is
2.8
4.7
5.0
7.0
Explanation
In Expt. 2, the final solution is a buffer as it contains equimolar amounts of acid and salt.
$$pH = p{K_a} + \log {{(salt)} \over {(acid)}}$$ ..... (1)
$$p{K_a} = - \log (2 \times {10^{ - 5}})$$
$$ = - 0.3010 + 5$$
$$ = 4.699 \approx 4.7$$
$$[Salt] = [C{H_3}COONa] = {{100} \over {200}} \times 2 = 1\,M$$
$$[Acid) = [C{H_3}COOH] = {{200 - 100} \over {200}} \times 2 = {{100} \over {200}} \times 2 = 1\,M$$
Substituting the values in Eq. (1), we get
$$pH = 4.7 + \log {1 \over 1} = 4.7$$
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