Question #ad761

1 Answer
Apr 22, 2015

Michael and Doreen are correct, calcium carbonate is considered insoluble in water, which is another way of saying that very, very little amounts will actually dissolve.

You can use calcium carbonate's solubility product constant, ksp, which is listed as being equal to 2.8109, to determine how much calcium carbonate would dissolve in a liter of water.

The very small amount of calcium carbonate that does dissolve will dissociate into Ca2+ and CO23 ions

CaCO3(s)Ca2+(aq)+CO23(aq)

By definition, Ksp will be equal to

Ksp=[Ca2+][CO23].

The 1:1 mole ratio between the ions will ensure that the two concentrations are equal, which implies

Ksp=2.8109=xx=x2x=5.3105

This is calcium carbonate's molar solubility in water at a temperature of 25C. To determine the mass that contains this many moles, use the compound's molar mass

5.3105molesL100.0869 g1mole0.0053 g/L

This means that you cannot add more than 0.0053 g of calcium carbonate to a liter of water without making a saturated solution.

So, at best, your 100-mL volume of water can only hold

100mL0.0053 g1000mL=0.00053 g CaCO3

The molarity of the Ca2+ and CO23 ions would be

[Ca2+]=[CO23]=5.3106moles100103L=5.3106M

Far from the 0.5 M of your target solution.

That's the best you can hope for in that much volume of water.