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* 10^(-5)moles/L`

Step-by-step 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* 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* 10^(-9)=[s][s]`

`3.8* 10^(-9)=s^2`

`s=6.2* 10^(-5)moles/L`

Hence, the concentration in moles/liter of calcium ion in a saturated solution of calcium carbonate is
`6.2* 10^(-5)moles/L`