Solveeit Logo
Login

Question

Chemistry Question on Equilibrium

The solubility of a saturated solution of calcium fluoride is 2×10−42 \times 10^{-4} moles per litre. Its solubility product is

A

12×10−212 \times 10^{-2}

B

14×10−414 \times 10^{-4}

C

22×10−1122 \times 10^{-11}

D

32×10−1232 \times 10^{-12}

Answer

32×10−1232 \times 10^{-12}

Explanation

Solution

2×10−4M   CaF2⇌2×10−4M   Ca2++2××10−4M   2F−_{2 \times 10^{-4}M}^{\, \, \, CaF_2} \rightleftharpoons _{2 \times 10^{-4}M}^{\, \, \, Ca^{2+}} + _{2 \times \times 10^{-4}M}^{\, \, \, 2F^-} Ksp  of  CaF2=[Ca2+][F−]2K_{sp} \, \, of \, \, CaF_2 = [Ca^{2+}] [F^-]^2 =[2×10−4][4×10−4]2=32×10−12(mol/L)2= [2 \times 10^{-4} ] [4 \times 10^{-4} ]^2 = 32 \times 10^{-12} (mol/L)^2