0.01 mol of H2SO4 is present in its solution. Find the pH of the solution
I don't know how advanced this class is. If beginning your prof PROBABLY (I'm guessing) will want you to do this.
H2SO4 ==> 2H^+ + SO4^2-
So 0.01M H2SO4 gives 2 x 0.01 = 0.02M H^+ and you convert that to pH with pH = -log(H^+).
However, it turns out that although the first H comes off at 100%, the second H does not so you get all of the first H and part of the second (part meaning that it isn't another 0.01).
Here is how you do that part if you want to be more exact in working it.
......H2SO4 ==> H^+ + HSO4^-
I......0.01.....0.......0
C.....-0.01.....0.01...0.01
E.......0.......0.01...0.01
Then for the second H,
......HSO4^- ==> H^+ + SO4^2-
I.....0.01.....0.01.....0
C......-x........+x.....+x
E....0.01-x...0.01+x....+x
(k2 in my book is 0.012)
0.012 = (0.01+x)(x)/(0.01-x) and solve for x. Then total
(H^+) is (0.01+x). You must solve the quadratic. I ran through it and obtained 0.0145 for total and change that to pH. You should confirm that; I may have goofed on the calculator. You can see in the first calcn you get 0.02 but in the more exact you get 0.0145. I'll leave you with that. Since you don't show a k2 (0.012) for H2SO4, I'm guessing your prof wants the more simple calculation.