posted by Jo on .
Given a diprotic acid, H2A, with two ionization constants of Ka1=3.9E-2 and Ka2=5.9E-8.
Calculate the pH and molar concentration of each protonated form for a 0.135 M solution of:
For H2A it follows the usual form of ionizsation as if you had a monoprotic acid.
..........H2A ==> H^+ + HA^-
Ka1 = (H^+)(HA^-)/(H2A)
Substitute and solve for x = (H^+) = (HA^-) and convert to pH. (H2A) = 0.135-x. I strongly suspect you will need t use the quadratic equation to solve this because ka1 is relatively large.
(H^+) = sqrt [(k2(HA^-)+Kw)/(1+(HA^-)/k1)] and convert to pH.
For Na2A, set up an ICE chart for the hydrolysis of A^2-.
I'm not getting the right answers for the pH, [H2A], and [HA^-] in part a (0.135 M solution for H2A)
For H2A, I did
Using the quadratic formula:
Finding [H2A]=.135 M-.06725 M=.06775 M
Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M