4. a. Given that the pKa for phenol is 10.0, estimate the pH values at which 1/3 and 2/3 of the phenol present in a 1 mM aqueous solution is deprotonated.
b. What is the value for the free energy change at the biochemistsâ standard state ( âG°Ⲡ) for the dissociation reaction of phenol? Remember that the biochemistsâ standard state just means to use pH=7 instead of assuming 1 M H+.
The answers are below: Can someone explain this out in more detail, and tell me where the numbers are coming from?
a. When 1/3 of the phenol is protonated, then [phenol]/[phenolate] = 1â2 . Given that the pKa is 10.0, Ka =[phenolate][H+]/[phenol], so [H+] = Ka [phenol]/[phenolate] = 1â2 Ka = 5Ã10-11.
Therefore pH = -log [H+] = -log (5Ã10-11) â 10.3
When 2/3 of the phenol is protonated, then [phenol]/[phenolate] = 2.
[H+] = Ka [phenol]/[phenolate] = 2 Ka = 2Ã10-10. Therefore pH = -log [H+] = -log (2Ã10-10) â 9.7
Ka = 10-10.
b. âG°Ⲡ= âG° + RT ln Q where Q represents the reaction quotient with non-standard concentrations. âG°=-RTlnK. SoâG°â²=-RTlnK+RTlnQ=RTln(Q/K).
For the dissociation reaction of phenol, Q = [phenolate][H+]/[phenol].
At the biochemistsâ standard state of pH=7, [H+] = 1Ã10-7 while the other species still have a 1 M standard state.
So Q = 1Ã10-7 .
Therefore âG°Ⲡ= RT ln(Q/K) = RT ln(1Ã10-7 /1Ã10-10) = RT ln 1000 = (8.314 J mol-1 K-1)(298 K) (6.91) âG°Ⲡ= +17,100 J mol-1 = +17.1 kJ mol-1
The positive value for âG°Ⲡmeans that the dissociation reaction is not favored at pH = 7. In fact, phenol is barely (about 1/1000, or 0.1%) dissociated at that pH.