Solid CaF2 and solid SrF2 are both in equilibrium with pure water. K_sp(CaF2) = 3.9 x 10^-11, K_sp(SrF2) = 2.9 x 10^-9. Ignore hydrolysis of F-. Select the correct statement(s).

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia073

Solid $CaF X_{2}$ and solid $SrF X_{2}$ are both in equilibrium with pure water. $K_{s p} (CaF X_{2}) = 3.9 \times 1 0^{- 11}$ , $K_{s p} (SrF X_{2}) = 2.9 \times 1 0^{- 9}$ . Ignore hydrolysis of $F X^{-}$ . Select the correct statement(s).

Charge balance:

[F X^{-}] = 2 ([Ca X^{2 +}] + [Sr X^{2 +}]) = 2 (K_{1} + K_{2}) / [F X^{-}]^{2}

, so

[F X^{-}]^{3} = 2 (K_{1} + K_{2})

and

[F X^{-}] = 1.8 \times 1 0^{- 3} M

[Sr X^{2 +}] / [Ca X^{2 +}] = K_{2} / K_{1} = 74

(rounded), and

[Ca X^{2 +}] = K_{1} / [F X^{-}]^{2} = 1.2 \times 1 0^{- 5} M

. The last value comes from an incorrect formula.