Can anyone tell me how to solve this problem:

The concentration of H2SO4 in a solution is 3.60x10^-6 M. Assume the acid dissociates completely in solution.

What is the [H3O+] in the solution?

What is the [OH-] in the solution?

What is the pH of the solution?

What is the pOH of the solution?
1. That's just the molarity times 2.

2. The antilog of (14+log(#1)). There's an easier way, but I forget it.

3. -log(#1)

4. -log(#2)

I'm pretty sure about all of that except #2. Check your book, and look for something about the number 1 x 10^14, that's ringing a bell.
it is 2.
like all other math/science qustions.
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edit: was talkin to hrdcorelaxplaya
Last edited by kenan6346 at Apr 13, 2008,

Since H2SO4 is a strong acid, it dissociates fully.

H2SO4 + H2O -> HSO4- + H3O+

So, the concentration of H3O+ is the same as the concentration of H2SO4-, which is 3.60 x 10^-6. Next, you have to do

HSO4- + H20 -> SO4-2 + H3O+

You take the given Ka, and set it equal to x^2 / 3.60 x 10^-4. Add x to the H3O+ before, and you get the [H3O+]. 1 x 10^-14 / ans = [OH-].

Then, pH = -log[H3O+], and pOH = 14 - pH.
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