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?
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.
To determine whether a precipitate will form when the 50% saturated solution of BaCO3 is mixed with 20.00 mL of 0.0100 M Na2CO3, you need to consider the solubility product constant (Ksp) of BaCO3 and the product of the concentrations of Ba2+ and CO3^2- ions in the solution.
First, let's clarify what "50% of the saturated solution" means. The solubility of BaCO3 is given as 7.1 x 10^-5 M. When we say the solution is 50% saturated, it means that the concentration of Ba2+ ions is half of the solubility, which is 0.5 * 7.1 x 10^-5 M = 3.55 x 10^-5 M.
Now, let's calculate the concentrations of Ba2+ and CO3^2- ions in the final solution after mixing.
BaCO3 dissociates in water as follows: BaCO3(s) ⇌ Ba2+(aq) + CO3^2-(aq)
Based on your calculations, the concentration of Ba2+ ions ([Ba2+]) is 0.0491 M, and the concentration of CO3^2- ions ([CO3^2-]) is 0.0670 M.
Since the stoichiometric ratio between Ba2+ and CO3^2- ions is 1:1, both the concentration values are already in the correct ratio. Therefore, you don't need to add them together.
Now, compare the Qsp value (the reaction quotient) to the Ksp value to determine if a precipitate will form. Qsp is calculated by multiplying the concentrations of the products (Ba2+ and CO3^2-) at a given time.
Qsp = [Ba2+][CO3^2-] = (0.0491 M)(0.0670 M) = 0.0032937
The Ksp value for BaCO3 is given as 5.1 x 10^-9.
Since Qsp (0.0032937) is greater than Ksp (5.1 x 10^-9), the solution is already supersaturated. Therefore, when the 20.00 mL of the 50% saturated solution is mixed with 20.00 mL of 0.0100 M Na2CO3, a precipitate of BaCO3 will likely form due to the excess of Ba2+ and CO3^2- ions.
Note: The 50% saturated solution implies that only half of the maximum amount of BaCO3 that can dissolve in the solvent (water) is present in the solution.