Asked by justin
Calculate the pH of a 5.20 10-3 M solution of H2SO4.
Answers
Answered by
DrBob222
How difficult is this course supposed to be? H2SO4 is a strong acid for the first ionization; therefore, that will be 0.00520M. To that, if you wish to be accurate, is added the H^+ contribution of the second ionization, which is relatively strong, but still a weak acid. k2 in my text is listed as 0.012 but you should confirm that from your text.
HSO4^- ==> H^+ + HSO4^-
k2 = (H^+)(SO4^-2)/(HSO4^-)
So the (H^+) is 0.00520 + x and the HSO4^- is 0.00520 - x. You can set that up and solve the quadratic to find x, then 0.0052 + x is H^+. Then convert that to pH.
HSO4^- ==> H^+ + HSO4^-
k2 = (H^+)(SO4^-2)/(HSO4^-)
So the (H^+) is 0.00520 + x and the HSO4^- is 0.00520 - x. You can set that up and solve the quadratic to find x, then 0.0052 + x is H^+. Then convert that to pH.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.