Answer
$5.225 \times 10^{-6} g$
Work Step by Step
Since $Ca(NO_3)_2$ is an ionic soluble salt, the concentration of calcium ions will be equal to $0.050$ M.
1. Write the $K_{sp}$ expression:
$ CaCO_3(s) \lt -- \gt 1Ca^{2+}(aq) + 1CO_3^{2-}(aq)$
$8.8 \times 10^{-9} = [Ca^{2+}]^ 1[CO_3^{2-}]^ 1$
$8.8 \times 10^{-9} = (0.05 + S)^ 1( 1S)^ 1$
2. Find the molar solubility.
Since 'S' has a very small value, we can approximate: $[Ca^{2+}] = 0.05$
$8.7 \times 10^{-9}= (0.05)^ 1 \times ( 1S)^ 1$
$ \frac{8.7 \times 10^{-9}}{0.05} = ( 1S)^ 1$
$1.74 \times 10^{-7} = S$
3. Find the number of moles:
** $3 \times 10^2$ mL = $300mL$ = $300 \times 10^{-3}$ L = 0.300 L
$Concentration(M) = \frac{n(mol)}{V(L)}$
$1.74 \times 10^{-7} = \frac{n(mol)}{0.3}$
$1.74 \times 10^{-7} * 0.3 = n(mol)$
$5.22 \times 10^{-8} moles = n(mol)$
4. Determine the molar mass of this compound ($CaCO_3$):
40.08* 1 + 12.01* 1 + 16* 3 ) = 100.09g/mol
5. Calculate the mass
$mm(g/mol) = \frac{mass(g)}{n(mol)}$
$mm(g/mol) * n(mol) = mass(g)$
$ 100.09 * 5.22 \times 10^{-8} = mass(g)$
$5.225 \times 10^{-6} = mass(g)$
