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Author: Subject: Equilibrium concentration of sulphate ions

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Registered: 8-11-2005
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sad.gif posted on 12-11-2005 at 01:48
Equilibrium concentration of sulphate ions

Barium and strontium reacts with a 0.5 M solution of Na2SO4. Assume that 0,05 ml Na2SO4 has been used (a surplus) , that the total volume is 1 ml and pH=1. Calculate the equilibrium concentration of sulphate ions.

This is what I've found so far:

Na2SO4 (aq) + H2O (l) -> 2 Na+ (aq) + SO42- (aq)

SO42- (aq) + H2O (l) -> HSO4- + (aq) + OH- (aq)

Ks = [HSO4-] * [OH-] / [HSO42-]

Ks = 1,02 * 10^-2 M (I am supposed to use Ks for HSO4-, right?)

[NA2SO4] = 0,5 M * 0,05 ml / 1 ml = 25 * 10^-3

[HSO4-] + [SO42-] = [Na2SO4]

pH = 1
pH + pOH = 14 =>
pOH = 13 => [OH-] = 10^-13 M.

Now I'm lost. I just recently gained understanding of the whole idea of equilibrium calculations and this problem is quite a thing .
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National Hazard

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Registered: 27-3-2005
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[*] posted on 12-11-2005 at 02:25

I'm thinking that you would need to find the Ksp of barium sulfate and strontium sulfate first (in a table in a textbook, or on the internet).

They will react with the SO4 ions in solution to form the insoluble precipitate. From there, you can calculate the amount of SO4 ions using the equilbrium table (ICE table)

And if this is wrong, it's 5:30 am local time :)
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