Answer
$ K_{sp} (SrF_2) = (4.37 \times 10^{-9})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2F^{-}(aq)$
$ K_{sp} = [Sr^{2+}]^ 1[F^{-}]^ 2$
2. Determine the ionic concentrations:
$[Sr^{2+}] = 1.03 \times 10^{-3}M$
$[F^{-}] = [Sr^{2+}] * 2 = 2.06 \times 10^{-3}M$
3. Calculate the $K_{sp}$:
$ K_{sp} = (1.03 \times 10^{-3})^ 1 \times (2.06 \times 10^{-3})^ 2$
$ K_{sp} = (1.03 \times 10^{-3}) \times (4.24 \times 10^{-6})$
$ K_{sp} = (4.37 \times 10^{-9})$
