Calculate Keq for the equibria based on Ka?

Question

i) CH3COOH(ka = 10-5) + NaHCO3 -><- CH3COONa + H2CO3(ka = 10-7)

ii) phenol(ka = 10-9) + NaHCO3 -><- C6H5ONa + H2CO3

Based on these calculations, can you suggest why NaHCO3 cannot be used in extracting phenols from organic mixtures?

part b) Phenols can be extracted into aqueous solutions with Na2CO3(ka of HCO3- = 10-10)

Answer 1

Na+ is a spectator ion so we can leave it out:

CH3COOH + HCO3- <=> CH3COO- + H2CO3

Keq= [CH3COO-][H2CO3] / [CH3COOH][HCO3-] (1)

For acetic acid you have
CH3COOH <=> CH3COO- + H+,
so Ka(acet)= [CH3COO-][H+] / [CH3COOH] =>
[CH3COO-]/[CH3COOH]= Ka(acet)/[H+] (2)

For carbonic acid:
H2CO3 <=> H+ +HCO3-
Ka1=[HCO3-][H+] / [H2CO3] =>
[H2CO3]/[HCO3-] = [H+]/Ka1 (3)

We substitute the ratios in (1) based on (2), (3) so we get
Keq= (Ka(acet)/[H+]) * ([H+]/Ka1)= Ka(acet)/Ka1 = 10^-5/10^-7 =>
Keq=10^2 =100

Similarly for phenol we get:
Keq=Ka(phenol)/Ka1 = 10^-9/10^-7 =10^-2 =0.01

In order to extract phenol into the aqueous phase you need to make it in the form of the salt so that it becomes more soluble in water. The Keq is very small (0.01<1, thus reactants are favoured) which means that NaHCO3 is not basic enough to react to the desired degree with phenol.

Na2CO3 is a much stronger base than NaHCO3 since the conjugate acid is weaker (Ka=10^-10 compared to 10^-7). Or if you prefer, you calculate the Keq as above:
Keq= 10^-9/10^-10 =10>1 which favours the products (formation of the phenol salt).