InterviewSolution
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InterviewSolution
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`2M` solution of `Na_(2)CO_(3)` is boiled in a closed container with excess of `CaF_(2)` .Very little amount of `CaCO_(3)` and `NaF` are formed.If the solubility product `(K_(sp))` of `CaCO_(3)` is `x` and molar solubility of `CaF_(1)` is `y`.Find the molar concentration of `F^(-)` in resulting solution after equilibrium is attained. |
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Answer» `{:(,Na_(2)CO_(3)+,CaF_(2)(s)hArr,2NaF(aq)+,CaCO_(3)(s)),(t=0,2,-,0,-),(t=eq,2-a,-,2a,-):}` where is very small For `CaCO_(3),K_(sp)=x=[Ca^(2+)][CO_(3)^(2-)]=[Ca^(2+)]xx2( :.CO_(3)^(2-)` mainly coming from `Na_(2)CO_(3)` `[Ca^(2+)]=(x)/(2)` For `CaF_(2),K_(sp)=4y^(3)=((x)/(2))[F^(-)]^(2)rArr [F^(-)]=sqrt((8y^(3))/(x))` |
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