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Question

3M solution of Na2CO3 is boiled in a closed container with excess of CaF2. Very little amount of CaCO3 and NaF are formed. If the solubility product Ksp of CaCO3 is y and molar solubility of CaF2 in water is x, find the molar concentration of F in resulting solution after equilibrium is attained.

A
8y3x
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B
8x3y
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C
3y3x
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D
12x3y
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Solution

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 3a 2a a
where a is very small
For CaCO3,
CaCO3(s)Ca2+(aq)+CO23(aq)
Ksp=y=[Ca2+][CO23]=[Ca2+]×3 (CO23 mainly coming from Na2CO3
[Ca2+]=y3.......(1)
For CaF2,
CaF2(s)Ca2+(aq)+2F(aq) C 0 0 Cx x 2x
Ksp=[Ca2+][F]2
Ksp=x×(2x)2=4x3
[Ca2+][F]2=4x3
(y3)[F]2=4x3..........(from equation 1)
[F]=12x3y

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