Question

Chem 2 solubility question. Please explain how you do this.

Answer 1

Solution.

The solubility product constant expression is

$[Ca^{2+}][CO_3^{2-}]=K_{sp};$

The concentration of calcium ions is

$[Ca^{2+}]=0.0540M;$

Therefore, the concentration of carbonate-ions is

$[CO_3^{2-}]=\frac{K_{sp}}{[Ca^{2+}]}=\frac{8.70\cdot 10^{-9}}{0.054}=1.611\cdot 10^{-7}M;$

As

$c=\frac{m}{MV}$,

the mass is

$m = cMV;$

The molar mass of calcium carbonate is 100.087 g/mol;

$m=1.611\cdot 10^{-7}\cdot 100.087\cdot 2.9=4.676\cdot 10^{-5}g.$