What is the pH of a 0.051 M solution of Phosphorous acid (H2PHO3) in water? The value of Ka1 for H2PHO3 is 1.600 x 10-2 and the value of Ka2 is 7.000 x 10-7.
...........H3PO4 ==> H^+ + H2PO^-
initial....0.051......0......0
change.......-x.......x......x
equil......0.051-x....x.......x
k1 = 1.600E-2 = (H^+)(H2PO4^-)/(H3PO4)
Substitute from the ICE chart above into the ka1 expression and solve for x, then convert H^+ to pH.
Note that I ignored k2 since it it about 10,000 times less than k1.