Posted by May (PLEASE HELP) on Monday, October 11, 2010 at 5:22pm.
If only one H^+ is involved, you will have 0.065 moles/0.456L = ??M
I don't know if you are to calculate the second H^+ or not? I suspect so since the question asks for M.
The second ionization is not a strong acid; the Ka (k2) is 1.2 x 10^-2 in my text but you should look it up in your text and use that number.
The ionization is
HSO4^- ==> H^+ + SO4^-2
k2 = 1.2 x 10^-2 = (H^+)(SO4^-2)/(HSO4^-).
Then you make an ICE chart, substitute and solve.
If you call this second H^+ x, the SO4^-2 will be x and HSO4^- will be 0.1425-x. Then into k2 you substitute
0.1425+x for H^+, x for SO4^- and 0.1425-x for HSO4^-.
I went through it and got close to 0.01 for x but you need to confirm that. The total H^+ then would be 0.1425 + 0.01 and to two s.f. I would round to 0.15 M. Check my thinking. Check my work carefully.
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