Question

One of the main components of hard water is CaCO₃. When hard water evaporates, some of the CaCO₃ is left behind as a white mineral deposit. If a hard water solution is saturated with calcium carbonate, what volume of the solution has to evaporate to deposit 1.00

Answer 1

To deposit 1.00 gram of CaCO3 from a saturated solution, approximately 140845.07 liters of hard water would need to evaporate, based on the solubility of 7.1 mg/L. This calculation takes the solubility of calcium carbonate at face value without accounting for variations due to changing temperature and concentration.

Step-by-step explanation:

The volume of hard water that must evaporate to deposit 1.00 gram of CaCO3 depends on the solubility of calcium carbonate in water. As indicated by the solubility product, CaCO3 has a molar solubility of 7.1x10-5 M, which translates to 7.1 mg/L.

To calculate the volume of solution needed, divide the mass of CaCO3 deposited (1.00 g) by the solubility (7.1 mg/L).

1.00 g = 1000 mg, so:

1000 mg / (7.1 mg/L) = approximately 140845.07 L

This calculation assumes the solubility doesn't change as evaporation occurs, which is a simplification because typically solubility varies with temperature and concentration.