Posted by Jo on Saturday, July 21, 2012 at 12:04am.
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-.
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
Related Questions
Chemistry - Given a diprotic acid, H2A, with two ionization constants of Ka1 = 4...
Chemistry - Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3...
chemistry - For the diprotic weak acid H2A Ka1=2.1x10^-5 and Ka2=5.5x10^-7. What...
AP CHEMISTRY - Phosphoric acid is triprotic acid (three ionizable hydrogens). ...
chemistry - A 0.15M solution of weak triprotic acid is adjusted (base added) so ...
Chemistry - a 0.451 L solution of 1.71 ascorbic acid (ka1=7.9E-5, ka2=1.6E-12) ...
AP Chemistry - 18.0 ml of 0.122 M diprotic acid was titrated with 0.1017 M NaOH...
chemistry - For 6.0*10^-2 M H2CO3, a weak diprotic acid, calculate the following...
Chemistry - Two Concept Questions for you: 1) Why is the concentration of a ...
AP Chemistry - help with acid/bases If the pH at 50.0 mL of NaOH added is 4.0 ...
For Further Reading
