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.
Calculate the pH of a 5.90 10-3 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
2 answers
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......0.0059-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.
...........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.