The correct option is D √12x3y
Given, Ksp(CaCO3)=y and Molar solubility of CaF2=x, [Na2CO3]=3M
Na2CO3(aq)+CaF2(s)⇌2NaF(aq)+CaCO3(s)t=0 3 − − −t=eq 3−a − 2a a
where a is very small
For CaCO3,
CaCO3(s)⇌Ca2+(aq)+CO2−3(aq)
Ksp=y=[Ca2+][CO2−3]=[Ca2+]×3 (∵CO2−3 mainly coming from Na2CO3
∴ [Ca2+]=y3.......(1)
For CaF2,
CaF2(s)⇌Ca2+(aq)+2F−(aq) C 0 0 C−x x 2x
Ksp=[Ca2+][F−]2
Ksp=x×(2x)2=4x3
[Ca2+][F−]2=4x3
(y3)[F−]2=4x3..........(from equation 1)
⇒[F−]=√12x3y