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July 25, 2014

July 25, 2014

Posted by **Ashley** on Monday, March 5, 2012 at 9:54pm.

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

- Chemistry -
**DrBob222**, Monday, March 5, 2012 at 10:06pmFirst, 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.

- Chemistry -
**DrBob222**, Monday, March 5, 2012 at 10:26pmI 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......0.0059-x....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 0.0059-x. 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....0.44-x.......x.......x

k2 = (H^+)(SO4^2-)/(HSO4^-)

Substitute from the ICE chart and solve for x, then convert to pH.

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