For H2A it follows the usual form of ionizsation as if you had a monoprotic acid.
..........H2A ==> H^+ + HA^-
I.......0.135M....0......0
C.........-x......x......x
E........0.135-x..x.......x
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.
For NaHA.
(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-.
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:
(a) H2A
(b) NaHA
(c) Na2A
2 answers
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
3.9E-2=x^2/(.135-x)
Using the quadratic formula:
x^2+.039x-.005265=0
x=.06725
[H^+]=[HA^-]=.06725 M
pH=-log(.06725)=1.17
Finding [H2A]=.135 M-.06725 M=.06775 M
Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M
For H2A, I did
3.9E-2=x^2/(.135-x)
Using the quadratic formula:
x^2+.039x-.005265=0
x=.06725
[H^+]=[HA^-]=.06725 M
pH=-log(.06725)=1.17
Finding [H2A]=.135 M-.06725 M=.06775 M
Finding [A^2-]=Ka2*[HA^-]/[H^]=5.9E-8 M
0 answers