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Topic: Problem of the week - 06/02/2012  (Read 14377 times)

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Offline XGen

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Re: Problem of the week - 06/02/2012
« Reply #15 on: February 10, 2012, 08:51:50 PM »
Let me see if I understand it correctly.

Taking the 0.000164492 mol of H+ and dividing it by 0.05 L gives 0.00328984 M of [H+] and 0.00328984 M for [HSO4-].
At pH 2, Ka2 = [10^-2][SO4 2-]/[HSO4-], and the sulfate and hydrogen sulfate concentrations are the same, leaving Ka2 = 10^-2.
Then, letting x = the amount of hydrogen sulfate dissociated, 10^-2 = [0.00328984 + x][ x]/[0.00328984 - x]. Solving, x = 0.00213308 M. Then, 0.00328984 + 0.00213308 = 0.00542292 M, or 2.266, which rounds to 2.27.

It seems that I had made a computation error before. :3

Offline Borek

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Re: Problem of the week - 06/02/2012
« Reply #16 on: February 11, 2012, 05:04:13 AM »
Taking the 0.000164492 mol of H+ and dividing it by 0.05 L gives 0.00328984 M of [H+] and 0.00328984 M for [HSO4-].

Not exactly. You don't know these concentrations. All you can say at this moment is that analytical concentration of sulfuric acid is 0.000164492/2/0.05L. It doesn't say anything about equilibrium concentrations of H+ and HSO4-.

Quote
At pH 2, Ka2 = [10^-2][SO4 2-]/[HSO4-], and the sulfate and hydrogen sulfate concentrations are the same, leaving Ka2 = 10^-2.

Right.

Quote
Then, letting x = the amount of hydrogen sulfate dissociated, 10^-2 = [0.00328984 + x][ x]/[0.00328984 - x]. Solving, x = 0.00213308 M. Then, 0.00328984 + 0.00213308 = 0.00542292 M, or 2.266, which rounds to 2.27.

Right.
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