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Author: Subject: Equilibrium concentration of sulphate ions
Xque
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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 .
National Hazard

Posts: 411
Registered: 27-3-2005