Answer
$ K_{sp} (CaSO_4) = (2.2 \times 10^{-4})$
Work Step by Step
1. Calculate the molar mass:
40.08* 1 + 32.07* 1 + 16* 4 = 136.15g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 2.03}{ 136.15}$
$n(moles) = 0.01491$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.01491}{ 1} $
$C(mol/L) = 0.01491$
4. Write the $K_{sp}$ expression:
$ CaSO_4(s) \lt -- \gt 1Ca^{2+}(aq) + 1S{O_4}^{2-}(aq)$
$ K_{sp} = [Ca^{2+}]^ 1[S{O_4}^{2-}]^ 1$
5. Determine the ions concentrations:
$[Ca^{2+}] = [CaSO_4] * 1 = [0.01491] * 1 = 0.01491$
$[S{O_4}^{2-}] = [CaSO_4] * 1 = 0.01491$
6. Calculate the $K_{sp}$:
$ K_{sp} = (0.01491)^ 1 \times (0.01491)^ 1$
$ K_{sp} = (0.01491) \times (0.01491)$
$ K_{sp} = (2.222 \times 10^{-4})$
