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Chemistry Question on Equilibrium

If KspK_{sp} of CaF2CaF_2 at 25??25^?? is 1.7×10−101.7 \times 10^{-10}, the combination amongst the following which gives a precipitate of CaF2CaF_2 is

A

1×10−2M Ca2+1 \times 10^{-2} M \,Ca^{2+} and 1×10−3MF−1 \times 10^{-3} MF^-

B

1×10−4M Ca2+1 \times 10^{-4} M \,Ca^{2+} and 1×10−4MF−1 \times 10^{-4} MF^-

C

1×10−2M Ca2+1 \times 10^{-2} M \,Ca^{2+} and 1×10−5MF−1 \times 10^{-5} MF^-

D

1×10−3M Ca2+1 \times 10^{-3} M \,Ca^{2+} and 1×10−5MF−1 \times 10^{-5} MF^-

Answer

1×10−2M Ca2+1 \times 10^{-2} M \,Ca^{2+} and 1×10−3MF−1 \times 10^{-3} MF^-

Explanation

Solution

When ionic product i.e. the product of the concentration of ions in the solution exceeds the value of solubility product, formation of precitpiate occurs. CaF2<=>Ca2++2F− {CaF2<=>Ca^{2+} + 2F^-} Ionic product =[Ca2+][F−]2= [Ca^{2+}] [F^-]^2 when, [Ca2+]=1×10−2 M[Ca^{2+}] = 1 \times 10 ^{-2} \,M [F−]2=(1×10−3)2 M[F^-]^2 = (1 \times 10^{-3})^2\,M =1×10−6 M=1 \times 10^{-6}\,M ∴[Ca2+][F−]2=(1×10−2)(1×10−6)=1×10−8\therefore \left[Ca^{2+}\right] \left[F^{-}\right]^{2}= \left(1\times10^{-2}\right) \left(1\times 10^{-6}\right) = 1\times 10^{-8} In this case, Ionic product (1×108)>\left(1\times 10^{8}\right) > solubility product (1.7×10−10)\left(1.7\times 10^{-10}\right) ∴\therefore Hence (a) is correct option.