Find the pH of 1.0E-8 M H2S04 (H2SO4 <==> 2H^+ + SO4^2-)

Question

I have got answer to be 7.70, can you confirm the answer for me please?
If it is not good, i'll be very grateful if you could leave some guides.

Answer 1

It isn't right and your gut should tell you that. How can you have an acid solution with a pH greater than 7. Remember 7 is neutral so a pH of 7.70 means your acid solution is basic and that can't be. Where are you in chemistry? If it is not beginning chemistry there are three things to do here. First, the ionization of H2SO4 occurs in two steps. The first step is 100% dissociation in which
H2SO4 ==> H^+ + HSO4^-.
Next, the HSO4^- dissociates but it is not 100%; i.e., the Ka is about 0.012. You should confirm that number from your Ka tables in your text.
You can take both of these into account as follows:
........H2SO4 ==> H^+ + HSO4^-
I.......1E-8......0......0
C......-1E-8.....1E-8...1E-8
E........0.......1E-8...1E-8

Then k2 kicks in.
...........HSO4^- ==> H^+ + SO4^2-
I.........1E-8........1E08....0
C..........-x.........x.......x
E........1E-8 - x....1E-8+x...x

k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
0.012 = (1E-8+x)(x)/(1E-8-x)
If you are in beginning chemistry your prof PROBABLY doesn't expect you to do this so I've gone ahead and solved the quadratic above and obtained x = 1e-8 and that added to 1E-8 = 2E-8M and that is exactly the same number you would obtain if you worked the problem straight. Starting with 1E-8M H2SO4 you get 2H^+/mol H2SO4 and that gives you 2E-8M. That takes care of the first two reasons and my guess is the prof expects you to simply multiply 2*1E-8 = 2E-8M.

The next step is to realize that at this low concentration you can't ignore the H^+ that comes from the ionization of water. So this is what you do.
...........H2O ==> H^+ + OH^-
I.........liquid...0......0
C.........liquid...x......x
E.........liquid...x......x
-----------------------------
then add the 2E-8M to this to give you
add...............2E-8..........
new E............2E-8+x.....x

Kw H2O = 1E-14 = (H^+)(OH^-)
1E-14 = (2E-14+x)(x)
Solve that quadratic for x, add x to 2E-8 and convert that H^+ to pH. I ran through it very fast and came up with 6.95 and we know it should be less than 7. Hope this helps.