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Question

Solid BaF2 is added to a solution that has 0.3 mole of a sodium oxalate solution (1 litre) until equilibrium is reached. The Ksp of BaF2 and BaC2O4(s) is 106 and 2×107 respectively. Assume that the addition of BaF2 does not cause any change in the volume and causes no hydrolysis of any of the cations or anions. If the concentration of Ba2+ ions in the resulting solution at equilibrium is represented as y×10x, then (x+y) is:

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Solution

BaF2(s) to Ba2+(aq.)+2F(aq) Let S be the solubility of BaF2 , but Ba2+ reacts with C2O24 present in this solution & almost completely converts into BaC2O4 Ba2++C2O24toBaC2O4(s)
Let y mole per litre of Ba2+ be left after reaching equilibrium
So, Ba2+y+C2O240.3sBaC2O4(s) K=2×107 So, we get the following two equations,
y(0.3s)=2×107 & y(2s)2=106=(Ksp)BaF2
Solving these two equations, we get, s=0.25M
So , [C2O24)=0.30.25=0.05M [F]=2s=2×0.25=0.5M [Ba2+]=y=4×106M

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