H2SO4 is a diprotic acid. The first H is ionized completely but the second one has a k2 = 0.0120. We need to go through two ICE charts.
For the first H^+ it is as follows:
.......................H2SO4 ==> H^+ + HSO4^-
I...................0.00267...........0...........0
C.................-0.00267......0.00267....0.00267
E........................0..........0.00267.....0.00267 so the first H^+ contributes 0.00267 M. The second H is done this way.
....................HSO4^- ==> H^+ + SO4^2-
I..................0.00267........0.00267.....0
C..................-x..................+x.............x
E............0.00267 - x.......0.00267+x...x
Then k2 = (H^+)(SO4^2-)/(HSO4^-)
Substitute the E line into k2 expression and solve for x (you will need to solve the quadratic equation), evaluate 0.00267 + x which will give you the total (H^+), then convert that to pH. Post your work if you get stuck.
Calculate the pH of a 2.67×10-3 M solution of H2SO4.
(Ka = 0.0120 for HSO4-)
1 answer