Determine the pH of a .50 M solution of H2S04 and also determine the concentration of the sulfate ion (SO42-)

Answer 1

I don't know where you are in chemistry but this problem is not as easy as it looks. The problem is that H2SO4 is a strong acid (100% ionized) for the first H^+ BUT the second one is not completely ionized (k2 = 1.2E-2). As a result, the first H that comes off reduces the ionization of the second one even further so the pH is essentially determined by the first ionization. ]
H2SO4 ==> H^+ + HSO4^-
HSO4^- ==> H^+ + SO4^2-
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
So you set up an ICE chart for k1 and k2.
..............H2SO4 ==> H^+ + HSO4^-
initial.......0.5M0......0......0
equilibrium.....0.......0.50....0.50*see below.

..............HSO4^- ==> H^+ + SO4^2-
initial......0.50.......0.......0
change.........-x.......x.......x
equil........0.50-x......x.......x

k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Now substitute.
(H^+) 0.50 from first equn + x from second.
(SO4^2-) = x from second equn
(HSO4^-) = 0.50-x frm second equn.
0.012 = (0.50+x)(x)/(0.5-x) and solve for x. You will need to solve the quadratic OR you can do it by successive approximations.
The (H^+) then is 0.5+x and x is the sulfate.

Answer 2

In the above I used k1 after saying H2SO4 had no k1. It doesn't. I should have said the first ionization which I referred to k1.