...........H3PO3 ==> H^+ + H2PO3^-
I..........2.0.......0......0
C...........-x.......x......x
E..........2.0-x.....x......x
Ignore the contribution to H^+ of k2 then substitute the E line into k1 expression and solve for x = (H^+)=(H2PO3^-)
Phosphorous acid, H3PO3, is actually a diprotic acid for which Ka1 = 5.0 × 10-2 and Ka2 = 2.0 × 10-7. What are the values of [H+] and [H2PO3-] in a 2.0 M solution of H3PO3?
3 answers
Is this correct?
(x ^ 2) / 2.0 = 5.0 x 10 ^ -2
x = 3.162E-1 = [H +] = [H2PO3-]
pH = -log (3.162E-1) = 0.50
(x ^ 2) / 2.0 = 5.0 x 10 ^ -2
x = 3.162E-1 = [H +] = [H2PO3-]
pH = -log (3.162E-1) = 0.50
The problem doesn't ask for pH. I would redo it as a quadratic.
0.316 = (H^+) with the assumption that
2.0-x = 2.0
But it really is 2.0-x and it doesn't appear to me that the x can be ignored with respect to 2.0. That looks like about 15%. I would feel better if the quadratic equation were used.
0.316 = (H^+) with the assumption that
2.0-x = 2.0
But it really is 2.0-x and it doesn't appear to me that the x can be ignored with respect to 2.0. That looks like about 15%. I would feel better if the quadratic equation were used.