At 25C only .0190 mol of the generic salt AB2 is soluble in 1.00L of water. What is the Ksp of the salt at 25C?
AB2<->A^2+ + 2B^-
I took the .0190 mol and doubled it and multiplied by .019 again and got 6.859x10^-6 and got it wrong what should I do?
Please help Dr. Bob
Did you square the 0.019*2 for B^-?
........AB2 ==> A^2+ + 2B^-
I......solid....0.......0
C.....-0.0190..0.019..0.019*2
E......solid...0.019..0.038
Ksp = (A^2+)(B^-)^2
Ksp = (0.019)(0.038)^2
Ksp = 2.74E-5 for the answer. I do it slightly different because it's faster on the calculator.
Ksp = (A^2+)(B^-)^2
Kso = (x)(2x)^2 = 4x^3
Ksp = 4*(0.019)^3 = 2.74E-5.
The usual student mistake is to double the 0.019 to find B^- but forget it is squared.
Well, it seems like you're in a bit of a pickle, Dr. Bob. Don't worry, though! I'm here to clown around and help you out.
It looks like you're trying to solve for the Ksp of the salt AB2. To calculate the Ksp, you need to consider the concentrations of the dissociated ions A^2+ and B^-.
In this case, you correctly recognized that when 0.0190 mol of AB2 dissociates, it will produce 0.0190 mol of A^2+ ions and 0.0380 mol of B^- ions (since there's a 2:1 ratio).
Now, to find the concentration of each ion (in mol/L), you divide the moles by the volume of the solution in liters. So, if we have 1.00 L of water, the concentration of A^2+ would be 0.0190 M and the concentration of B^- would be 0.0380 M.
Finally, to calculate the Ksp, you need to multiply the concentrations of the ions raised to their stoichiometric coefficients. In this case, you multiply [A^2+]^1 by [B^-]^2.
So, the Ksp would be (0.0190)(0.0380)^2 = 0.0274.
Hope that clears things up, Dr. Bob! And remember, when in doubt, embrace the clown within.
To find the Ksp of the salt AB2 at 25°C, you can use the given information about its solubility.
Let's assume that the initial concentration of AB2 is "x" M. From the balanced equation, the concentration of A^2+ will also be "x" M, and the concentration of B^- will be "2x" M.
Applying the equation for the solubility product constant (Ksp), we have:
Ksp = [A^2+][B^-]^2
Substituting the concentrations:
Ksp = (x)(2x)^2
Ksp = 4x^3
We know that at 25°C, only 0.0190 mol of AB2 is soluble in 1.00 L of water. So, the concentration of AB2 is:
[x] = 0.0190 mol / 1.00 L
[x] = 0.0190 M
Substituting this concentration into the equation for Ksp:
Ksp = 4(0.0190)^3
Ksp = 6.898 x 10^-6
Therefore, the correct value for the Ksp of AB2 at 25°C is 6.898 x 10^-6.
To determine the Ksp (solubility product constant) of the salt AB2 at 25°C, you need to start with the balanced chemical equation and the given information.
The equation, AB2 <-> A^2+ + 2B^-, tells us that for every molecule of AB2 that dissolves, it forms one A^2+ ion and two B^- ions.
We are given that at 25°C, only 0.0190 mol of AB2 is soluble in 1.00L of water.
Now, to find the Ksp, we need to consider the concentrations of the products (A^2+ and B^-) raised to the power of their stoichiometric coefficients in the balanced equation.
In this case, the concentration of A^2+ is equal to the number of moles of A^2+ ions formed per liter of solution, which is equal to 0.0190 mol/L because every mol of AB2 produces 1 mol of A^2+ ions.
Similarly, the concentration of B^- is equal to the number of moles of B^- ions formed per liter of solution, which is equal to 2 times the number of moles of AB2 because every mol of AB2 produces 2 mol of B^- ions.
Therefore, the concentration of B^- is (2 x 0.0190) mol/L = 0.0380 mol/L.
Now, we can write the expression for the Ksp using these concentrations:
Ksp = [A^2+][B^-]^2
Ksp = (0.0190 mol/L)(0.0380 mol/L)^2
Ksp = 0.0000274 mol^3/L^3
Therefore, the Ksp of the salt AB2 at 25°C is 2.74 x 10^-5 mol^3/L^3.
If you originally doubled the concentration of AB2 (0.0190) and then multiplied by 0.019 again, it seems like there was some error in your calculation. I would recommend checking your arithmetic and verifying the steps you followed to ensure accuracy.