Ok so anyway I got CaCO3 <===> Ca + CO3
Ksp = 4.96x 10^-9 = x^2
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Sq root = 7.043x10^-5 I am lost on where togo from here IF I am even doing it right
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
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.
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 !!
Thanks again !!
3 answers