Calculate the Ka given thermodynamic data

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Problem #1: At 25 °C the enthalpy change, ΔH°, for the ionization of trichloroacetic acid is +6.3 kJ/mol and the entropy change, ΔS°, is +0.0084 kJ/mol-K. What is the pKa of trichloroacetic acid?

Solution:

1) The equations to use are these:

ΔG° = ΔH° - TΔS°

ΔG° = - RT ln K

2) Therefore:

ΔH° - TΔS° = - RT ln K

3) Remember this:

R = 8.31447 has the units J/mol-K, not kJ/mol-K

4) Substituting:

6300 - (298) (8.4) = - (8.31447) (298) ln K

3796.8 divided by - (8.31447) (298) = ln K

ln K = -1.532381450328817

K = 0.216

pKa = 0.664

This page gives a Ka value of 0.23 rather than the 0.216 calculated in this problem. To make it closer, we could round off the 0.216 to 0.22, but I'll just leave it the way it is.


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