Question #ad761

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, #k_(sp)#, which is listed as being equal to #2.8 * 10^(-9)#, 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 #Ca^(2+)# and #CO_3^(2-)# ions

#CaCO_(3(s)) rightleftharpoons Ca_((aq))^(2+) + CO_(3(aq))^(2-)#

By definition, #K_(sp)# will be equal to

#K_(sp) = [Ca^(2+)] * [CO_3^(2-)]#.

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

#K_(sp) = 2.8 * 10^(-9) = x * x = x^2 => x = 5.3 * 10^(-5)#

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

#5.3 * 10^(-5)cancel("moles")/"L" * "100.0869 g"/(1cancel("mole")) ~= "0.0053 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

#100cancel("mL") * "0.0053 g"/(1000cancel("mL")) = "0.00053 g "# #CaCO_3#

The molarity of the #Ca^(2+)# and #CO_3^(2-)# ions would be

#[Ca^(2+)] = [CO_3^(2-)] = (5.3 * 10^(-6)"moles")/(100 * 10^(-3)"L") = 5.3 * 10^(-6)"M"#

Far from the 0.5 M of your target solution.

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