In: Chemistry
What fraction of ethane-1,2-dithiol is in each form (H2A, HA-, A2-) at pH=9.25? Please do show all work. Thank you!
ethane-1,2-dithiol pKa values
pKa1 = 8.43
pKa2 = 9.32
pH = pKa1 + log [HA-] /[H2A]
9.25 = 8.43 + log[HA-] /[H2A]
log [HA-] /[H2A]= 0.82
[HA-] /[H2A] = 6.61
fraction of [HA-] = [HA-] / ([H2A] + [HA-])
= [H-A] / ([HA-] + [HA-]/ 6.61)
=1 / 1 + 1/6.61
= 0.869
fraction of [HA-] = 0.869
fraction of [H2A] = 0.131
pH = pKa1 + log [A2-] /[HA-]
9.25 = 9.32 + log[A2-] /[HA-]
-0.07 = log[A2-] /[HA-]
[A2-] /[HA-] = 0.851
[A-2] = 0.851 x 0.869 =0.734
fraction of A-2 = 0.740