The Ksp values of BaSO4 and SrSO4 are 1.1x10^-10 and 3.8x10^-7 respectively. Calculate the concentrations of Ba2+, Sr2+, and SO4 2- in a solution that is saturated with BaSO4 and SrSO4.
I would do this. Since the SrSO4 is relatively more soluble than BaSO4 (Ksp about 3,500 times) I would calculate the solubility of SrSO4 in a saturated solution ignoring the solubility of BaSO4 and figure the small solubility of BaSO4 makes very little difference.
So Ksp = (Sr^2+)(SO4^2-) = 3.8E-7 and solve for (Sr^2+) = (SO4^2-) = approx (but you need to do it more accurately) 6E-4. Then that SO4^2- acts as a common ion to make the more insoluble BaSO4 even more insoluble; i.e.,
..........BaO4 ==> Ba^2+ + SO4^2-
I.........solid....0.......6E-4
C.........solid....x........x
E.........solid....x.......6E-4+x
Then plug the E line into Ksp for BaSO4 and solve for (Ba^2+) and use the quadratic formula to solve it; i.e. don't call 6E-4+x = 6E-4 because x will make a small difference. When you finish you will have x = (Ba^2+), you have SO4^2- = 6E-4 or whatever it really is +x, and Sr you have from the first calculation.
Hello,Sorry, May I know what if the Ksp of both substances no different much??
I think that simplifies the problem somewhat but the process is the same.
After thinking about this for 5-6 hours I don't think it simplifies the problem and I don't think the process is the same.
Well, if the solution is saturated with BaSO4 and SrSO4, that means it has reached the maximum concentration of those compounds that it can hold. In other words, it's like a crowded elevator - any more Ba2+ or Sr2+ ions trying to get in would be told, "Sorry, this floor is already full."
So the concentrations of Ba2+ and Sr2+ in the solution would be equal to the solubility of their respective salts. For BaSO4, that's 1.1x10^-10, and for SrSO4, it's 3.8x10^-7. As for SO4 2-, since both BaSO4 and SrSO4 are fully ionized, the concentration of SO4 2- would be determined by a common ion effect.
As for me knowing the concentrations of Ba2+ and Sr2+ specifically, that's like asking a clown bot to solve advanced chemistry equations. All I can tell you is that the concentrations will be really low, just like my chances of getting into a serious scientific discussion.
To calculate the concentrations of Ba2+, Sr2+, and SO4 2- in a saturated solution of BaSO4 and SrSO4, you need to consider the solubility products (Ksp) of both compounds.
The Ksp expression for BaSO4 is:
BaSO4 ↔ Ba2+ + SO4 2-
Ksp = [Ba2+][SO4 2-]
Given that the Ksp value for BaSO4 is 1.1x10^-10, we can assume that at equilibrium, the concentrations of [Ba2+] and [SO4 2-] will be equal.
Let's assume that the concentration of Ba2+ and SO4 2- in the saturated solution is "x" M each.
Therefore:
[Ba2+] = x M
[SO4 2-] = x M
Substituting these values into the Ksp expression for BaSO4, we get:
Ksp = (x)(x) = x^2
Since the Ksp value for BaSO4 is 1.1x10^-10, we can set up the following equation:
x^2 = 1.1x10^-10
Solving this equation, we find that x ≈ 1.05x10^-5 M. Therefore, the concentrations of Ba2+ and SO4 2- in the saturated solution of BaSO4 are approximately 1.05x10^-5 M each.
Similarly, you can use the same approach to calculate the concentrations of Sr2+ and SO4 2- in the saturated solution of SrSO4.
The Ksp expression for SrSO4 is:
SrSO4 ↔ Sr2+ + SO4 2-
Ksp = [Sr2+][SO4 2-]
Given that the Ksp value for SrSO4 is 3.8x10^-7, we can assume that at equilibrium, the concentrations of [Sr2+] and [SO4 2-] will be equal.
Let's assume that the concentration of Sr2+ and SO4 2- in the saturated solution is "y" M each.
Therefore:
[Sr2+] = y M
[SO4 2-] = y M
Substituting these values into the Ksp expression for SrSO4, we get:
Ksp = (y)(y) = y^2
Since the Ksp value for SrSO4 is 3.8x10^-7, we can set up the following equation:
y^2 = 3.8x10^-7
Solving this equation, we find that y ≈ 6.2x10^-4 M. Therefore, the concentrations of Sr2+ and SO4 2- in the saturated solution of SrSO4 are approximately 6.2x10^-4 M each.