If water is saturated with CaCO3, how much of it has to evaporate to deposit .250g of CaCO3.

Ksp CaCO3 = 4.96 x 10^-7

3 answers

Ok so anyway I got CaCO3 <===> Ca + CO3
Ksp = 4.96x 10^-9 = x^2
\/
Sq root = 7.043x10^-5 I am lost on where togo from here IF I am even doing it right
There is a discrepancy in your Ksp. You have E-7 in the problem but you substituted E-9. Check to make sure that is ok.
When you solve for x, that gives you the solubility CaCO3 in moles/L. I obtained about 7E-5 M and that x molar mass CaCO3 = about 7E-3 or 0.007 g CaCO3/L
Then I realized 1 L evaporated would only give me 0.007 g which isn't enough. Therefore, I took 100 L which gives 0.7 g. Now, what volume will contain 0.25g? That will be 100 L x 0.25/0.7 = about 36 L (approximate--you need to do it more accurately). So if you evaporated 36 L of 100L to leave 64L you would ppt 250 mg CaCO3. Your problem didn't specify how much of the saturated solution you had to work with.
Thanks Dr Bob The Ksp = 4.96 x 10^-9, so the second one I was using was the correct one, Reaching the M wasn't the hard part But I am still not sure of the equation. I will ask my teacher today in class. Somehow I came up with 52.133L but I did not factor the H2O in there anywhere , not sure if I even had to. And If I had to they gave me no saturation level to work with.
Thanks again !!