Question

The Ksp for CaCO3 is 3.8 x 10-9. Determine the concentration in moles/liter of calcium ion in a saturated solution of calcium carbonate. a. 3.8 x 10-9 b. 1.4 x 10-17 C. 6.2 x 10-5 d. 6.1 x 10-9

Answer 1

Answer: c. $6.2\times 10^{-5}moles/L$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the calcium carbonate is given as:

$CaCO_3\leftrightharpoons Ca^{2+}+CO_3^{2-}$

We are given:

$K_{sp}=3.8\times 10^{-9}$

By stoichiometry of the reaction:

1 mole of $CaCO_3$ gives 1 mole of $Ca^{2+}$ and 1 mole of $CO_3^{2-}$.

Expression for the equilibrium constant of $CaCO_3$ will be:

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

$3.8\times 10^{-9}=[s][s]$

$3.8\times 10^{-9}=s^2$

$s=6.2\times 10^{-5}moles/L$

Hence, the concentration in moles/liter of calcium ion in a saturated solution of calcium carbonate is $6.2\times 10^{-5}moles/L$