You would have done well to write the full question. I assume you are mixing 20.00 ml of a 50% saturated solution of BaCO3 (with the given Ksp) with 20.00 mL of 0.0100 M Na2CO3 although you don't say what the solutions are.
Next, I understand the first sentence that you don't know how to do this but the rest of your paragraph leaves me blank. I don't know where that 0.0670M and 0.0491 M came from or where they fit in. If the Na2CO3 assumption above is correct, this is what you do. If I've not interpreted the problem correctly please correct me and respond.
The 50% saturated solution is handled this way.
You are correct that if Ksp is 5.1E-9 then solubility is 7.1E-5M. So if it is only 50% saturated then (Ba^2+) = (CO3^2-) = 7.14E-5/2 = 3.57E-5M.
So you take 20 mL of this 3.57E-5M BaCO3 and it is diluted to 40 mL when you add 20 mL of the 0.01M Na2CO3.
Final Ba^+ = 3.57E-5M/2 = about 1.78E-5M
Final (Na2CO3) = 0.01 x 1/2 = 0.005M.
Then Qsp = (Ba^2+)(CO3^2-) =
(1.78E-5)(0.005) = about 9E-8 and that is larger than Ksp of 5.1E-9 so a ppt will form. Note that I ignored the CO3^2- from the BaCO3 solution; technically it should be added to the CO3 of the Na2CO3 but I didn't since the Qsp is larger than Ksp already. Increasing CO3 would have just made it even more obvious that it would ppt.
Hope I've not interpreted this wrong. Lotta work gone down the drain if I did.
Predict if a precipitate will form when 20.00 mL of the 50% saturated solution is mixed with 20.00 mL of 0.0100M of the following solutions.?
the saturated solution is : BaCO3, it has the Ksp of 5.1 x 10^-9
the solubility is 7.1 x 10^-5 M.
I don't know how to do this.
I am given the molarity of Ba and CO3, using the formula Mx -> <- M^2+ + x^2-
with [x^2-] = 0.0670 M and [M^2+] = 0.0491 M
so would I add those two molarities together to get the molarity of BaCO3?
if so, the molarity of BaCO3 is 0.1161M
so this would be: BaCO3 + Na2CO3 -> <- BaCO3 + Na2CO3
but what caused problems for me is what the question means by 50% of the saturated solution?
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