Chemistry
posted by Ashley on .
Calculate the pH of a 5.90 103 M solution of H2SO4
I have tried doing this as pH=log[H+] and also by multiplying the Molarity by two because there are two hydrogens and then doing the log but both ways have gotten me the wrong answer is there something i am missing?
same thing happened with: Calculate the pH of a 0.44 M solution of NaHSO4
please help, so confused

First, do you have answers. The first H^+ from H2SO4 is 100% ionized, the second one is not; therefore, the H^+ is not twice M H2SO4 but it is more than one times that. Second thing is the NaHSO4. That is not the M HSO4^ because
HSO4^ ==> H^+ + SO4^2 and k2 = about 0.012. If you have answers I can see how you are expected to solve this. 
I worked this and I expect you are supposed to take all of this into consideration. Here is how you do it.
...........H2SO4 ==> H^+ + HSO4^
initial....0.0059....0.......0
change....0.0059..0.0059..0.0059
equil.......0......0.0059..0.0059
So the first ionization, which is 100%, produces 0.0059M H^+. Then the second H, which is not 100% ionized, kicks in.
............HSO4^ ==> H^+ + SO4^2
initial....0.0059......0......0
change......x.........x.......x
equil......0.0059x....x.......x
k2 = (H^+)(SO4^2)/(HSO4^)
You need to use the k2 in your text or notes for they change from text to text. My copy I use list it as 0.012.
For (H^+) you want to substitute 0.0059+x (that's 0.0059 from the first H and x from the HSO4^. For SO4^2 substitute x and for HSO4^ substitute 0.0059x. This is a quadratic and you solve for x. Then total H^+ = 0.0059+x and determine pH from that.
For the NaHSO4, follow the second part of the above.
...........HSO4^ ==> H^+ + SO4^2
initial...0.44.......0.......0
change......x.......x........x
equil....0.44x.......x.......x
k2 = (H^+)(SO4^2)/(HSO4^)
Substitute from the ICE chart and solve for x, then convert to pH.