Answer
The $K_{sp}$ for $SrF_2$ is $2.7 \times 10^{-9}$ at $25 ^{\circ}C$
Work Step by Step
1. Calculate the molar mass:
87.62* 1 + 19* 2 = 125.62g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.011}{ 125.62}$
$n(moles) = 8.757\times 10^{- 5}$
3. Find the concentration in mol/L:
- 100ml = 0.100L
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 8.757\times 10^{- 5}}{ 0.1} $
$C(mol/L) = 8.757\times 10^{- 4}$
4. Write the $K_{sp}$ expression:
$ SrF_2(s) \lt -- \gt 1Sr^{2+}(aq) + 2F^{-}(aq)$
$ K_{sp} = [Sr^{2+}]^ 1[F^{-}]^ 2$
5. Determine the ions concentrations:
$[Sr^{2+}] = [SrF_2] * 1 = [8.757 \times 10^{-4}] * 1 = 8.757 \times 10^{-4}$
$[F^{-}] = [SrF_2] * 2 = 1.750 \times 10^{-3}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (8.757 \times 10^{-4})^ 1 \times (1.750 \times 10^{-3})^ 2$
$ K_{sp} = (8.757 \times 10^{-4}) \times (3.067 \times 10^{-6})$
$ K_{sp} = (2.686 \times 10^{-9})$
