Answer
Molar solubility: $5.54 \times 10^{-6} M$
Mass in a saturated solution in 15.0 L: $9.63 \times 10^{-3}g$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ FeCO_3(s) \lt -- \gt 1Fe^{2+}(aq) + 1CO_3^{2-}(aq)$
$3.07 \times 10^{-11} = [Fe^{2+}]^ 1[CO_3^{2-}]^ 1$
2. Considering a pure solution: $[Fe^{2+}] = 1S$ and $[CO_3^{2-}] = 1S$
$3.07 \times 10^{-11}= ( 1S)^ 1 \times ( 1S)^ 1$
$3.07 \times 10^{-11} = S^ 2$
$ \sqrt [ 2] {3.07 \times 10^{-11}} = S$
$5.54 \times 10^{-6} = S$
- This is the molar solubility value for this salt.
3. Find the number of moles:
$Concentration(M) = \frac{n(mol)}{V(L)}$
$5.54 \times 10^{-6} = \frac{n(mol)}{15}$
$5.54 \times 10^{-6} * 15 = n(mol)$
$8.31 \times 10^{-5} moles = n(mol)$
4. Determine the molar mass of this compound ($FeCO_3$):
55.85* 1 + 12.01* 1 + 16* 3 = 115.86g/mol
5. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 115.86 * 8.31 \times 10^{-5} = mass(g)$
$9.63 \times 10^{-3} = mass(g)$
