Solid BaF_2 is added to a solution containing 0.1 mole of sodium oxalate solution (1 litre) until equilibrium is reached.If the Ksp of BaF_2 and BaC_2O_4(s) is 10^(-6) & 10^(-7) respectively.Assume addition of BaF_2 does not cause any change in volume and no hydrolysis of any of the cations or anions. (Given :sqrt116=10.77) If concentration of Ba^(2+) ions in resulting solution at equilibrium is represented as 2.7xx10^(-4), then x is :

Solid BaF_2 is added to a solution containing 0.1 mole of sodium oxala

1.	Solid BaF_2 is added to a solution containing 0.1 mole of sodium oxalate solution (1 litre) until equilibrium is reached.If the Ksp of BaF_2 and BaC_2O_4(s) is 10^(-6) & 10^(-7) respectively.Assume addition of BaF_2 does not cause any change in volume and no hydrolysis of any of the cations or anions. (Given :sqrt116=10.77) If concentration of Ba^(2+) ions in resulting solution at equilibrium is represented as 2.7xx10^(-4), then x is :
Answer» Solution :`BaF_2(s)toBa^(2+)(AQ.)+2F^(-)(aq)` Let 'S' be the solubility of `BaF_2` , but `Ba^(2+)` REACTS with `C_2O_4^(2-)` present in this solution & almost completely convert into `BaC_2O_4` `Ba^(2+)+ C_2O_4^(2-)toBaC_2O_4(s)` Let y mole PER it. of `Ba^(2+)` is left after reaching equilibrium So, `underset(y)(Ba^(2+))+underset(0.1-s)(C_2O_4^(2-))toBaC_2O_4(s) " " Keq=10^(-7)` So, we get the following two equations, `y(0.1-s)=10^(-7) " &" y(2S)^2=10^(-6)=(K_(sp))BaF_2` Solving there two equations, we get, S=0.096 M So , `[C_2O_4^(2-))=0.1-0.096=4xx10^(-3) M` `[F^(-)]=2s=2xx0.096=0.192 M` `[Ba^(2+)]=y=2.7xx10^(-5)M`