Look at k1 vs k2. k2 is about 1000 times weaker than k1 so the pH is essentially determined by k1.
.........H2A ==> H^+ + HA^-
I........0.08....0......0
C........-x......x......x
E......0.08-x....x......x
Substitute the E line into Ka1 expression and solve for x = H^+ and convert to pH. You MAY need to use the quadratic.
(H2A) = 0.08-x = ?
Then k2 = (H^+)(A^2-)/(HA^-) but note from k1 above that (H^+) = (HA^-) which makes (A^2-) = k2.
For the diprotic weak acid H2A, Ka1 = 2.0 × 10^-6 and Ka2 = 6.7 × 10^-9.
What is the pH of a 0.0800 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
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