posted by Trista Nigh .
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 hard water solution is saturated with calcium carbonate, what volume of the solution has to evaporate to deposite 1.00x10^2 mg of CaCO3
CaCO3 ==> Ca^2+ + CO3^2-
Look up Ksp.
Ksp = (Ca^2+)(CO3^2-)
Set up an ICE chart and solve for solubility CaCO3. That value will be in moles/L. Multiply by molar mass to obtain grams/L. Then convert to volume necessary to hold 100 mg. Post your work if you get stuck.
I took the square root of (4.96x10^-9)= 7.04x10^-5 mol/L
then 100.09g/mol x 7.04 mol/L= 7.05x 10^-3 g/L
then .100 g x 1 L/ 7.05x10^-3= 7.05x10^-4 L, so that was my final answer, does this seem right?
Trista is almost correct. However she failed to realize that the problem called for 100 mg which is equal to 0.1 grams. This means that (7.05x10^-3)g/L / 0.1g = 0.0705/1L (Where the L is in the denominator).
So in order to finalize this problem all you have to do is take the inverse to get 1L/0.0705 = 14.2 L
The correct anwser.