Answer
$ K_{sp} = (2.36 \times 10^{-4})$
Work Step by Step
1. Calculate the molar mass $(CaSO4)$:
40.08* 1 + 32.07* 1 + 16* 4 = 136.15g/mol
2. Calculate the number of moles $(CaSO4)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{0.209}{ 136.15}$
$n(moles) = 1.54\times 10^{- 3}$
3. Find the concentration in mol/L $(CaSO4)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 1.54\times 10^{- 3}}{ 0.1} $
$C(mol/L) = 0.0154$
4. Write the $K_{sp}$ expression:
$ CaSO_4(s) \lt -- \gt 1Ca^{2+}(aq) + 1{SO_4}^{2-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[{SO_4}^{2-}]^ 1$
5. Determine the ion concentrations:
$[Ca^{2+}] = [CaSO_4] * 1 = [0.0154] * 1 = 0.0154$
$[{SO_4}^{2-}] = [CaSO_4] * 1 = 0.0154$
6. Calculate the $K_{sp}$:
$ K_{sp} = (0.0154)^ 1 \times (0.0154)^ 1$
$ K_{sp} = (0.0154) \times (0.0154)$
$ K_{sp} = (2.36 \times 10^{-4})$
