Calculate the solubility of Ag2CrO4 in
a) 0.05M KClO4
b) 0.005M AgNO3
a) I have figured out: Ag2CrO4=2Ag- + CrO42-
Ksp=1.12*10^-12.
Ksp=[2Ag]^2*[CrO4]
Ksp=(2x)^2(x) = 1.12*10^-12=4x^3= x=2.65*10^-5 (?)
b)Ag2CrO4=2Ag- + CrO42-
Ksp=1.12*10^-12.
Ksp=(2[Ag-]+0.005M)^2*[CrO42-]
1.12*10^-12=(2X+.005)^2*(x)
?
What you have done is almost correct but you've made two mistakes to get there for both parts.
The first big problem is Ksp. Let me do it correctly, then point out the error.
..........Ag2CrO4 --> 2Ag^+ + CrO4^2-
I........solid.........0.......0
C........solid........2x.......x
E........solid........2x.......x
Note that Ksp is this.
Ksp = 1.12E-12 = (Ag^)^2(CrO4^2-)
It is NOT (2Ag^+)^2(CrO4^2-)
But when you put in (2x)^2 for (2Ag^+)^2, a second error, you have made it exactly right. so 4x^3 = 1.12E-12. I didn't get your answer; probably you punched the wrong button on your calculator.
4x^3 = 1.12E-12
x^3 = 1.12E-12/4 = 2.80E-13
x = 6.5E-5
Note that you could have written that as Ksp = (2*CrO4^2-)^2(CrO4^2-) and technically that is what you did when you wrote (2x)^2(x) = Ksp.
For the second part, it works this way.
Refer to the first part for Ag2CrO4 => 2Ag^+ + CrO4^2-
Then ...AgNO3 --> Ag^+ + NO3
I.......0.005......0......0
C......-0.005...0.005....0.005
E........0......0.005....0.005
Ksp = (Ag^+)^2(CrO4^2-)
Ksp = (2x+0.005)(x)
Generally we way 2x + 0.005 is almost equal to 0.005 since 2x is so small).
1.12E-12 = (0.005)^2*x
x = 4.48E-8 M. Then you check it to see if your assumption above is right.
2x+0.005 = (2*4.48E-8)+0.005 = 8.96E-8 + 0.005 = 0.005 so our assumption above is correct.
You are amazing. I can't thank you enough!!
a) To find the solubility of Ag2CrO4 in 0.05M KClO4, we can use the solubility product constant (Ksp) expression and substitute the given values.
Ag2CrO4 = 2Ag+ + CrO42-
Ksp = [2Ag+]^2 * [CrO42-]
Ksp = (2x)^2 * x
Ksp = 4x^3
Given that Ksp = 1.12 * 10^-12, we can set up the equation:
1.12 * 10^-12 = 4x^3
Solving for x, the solubility of Ag2CrO4, we get:
x ≈ 2.65 * 10^-5
Therefore, the solubility of Ag2CrO4 in 0.05M KClO4 is approximately 2.65 * 10^-5 M.
b) To find the solubility of Ag2CrO4 in 0.005M AgNO3, we can again use the solubility product constant (Ksp) expression and substitute the given values.
Ag2CrO4 = 2Ag+ + CrO42-
Ksp = [2Ag+]^2 * [CrO42-]
Since the concentration of AgNO3 is given as 0.005M, let's denote it as [Ag+] = 0.005M.
Substituting this value in the Ksp expression, we have:
Ksp = (2 * 0.005)^2 * [CrO42-]
Ksp = 0.02^2 * [CrO42-]
Given that Ksp = 1.12 * 10^-12, we can set up the equation:
1.12 * 10^-12 = 0.02^2 * [CrO42-]
Solving for [CrO42-], we get:
[CrO42-] ≈ 7 * 10^-13
Therefore, the solubility of Ag2CrO4 in 0.005M AgNO3 is approximately 7 * 10^-13 M.
To calculate the solubility of Ag2CrO4 in each case, we need to use the solubility product constant (Ksp) expression for Ag2CrO4, which is given as Ksp = [Ag+]^2 * [CrO42-], where [Ag+] represents the concentration of Ag+ ions and [CrO42-] represents the concentration of CrO42- ions.
a) In the presence of 0.05M KClO4:
Ag2CrO4 dissociates into 2 Ag+ ions and 1 CrO42- ion.
So, we have Ksp = (2[Ag+])^2 * [CrO42-].
Given that Ksp for Ag2CrO4 = 1.12 * 10^-12, we can substitute the values into the expression:
1.12 * 10^-12 = (2x)^2 * (x),
where x represents the concentration of Ag+ and CrO42- ions.
By solving the above equation, we get the value of x as 2.65 * 10^-5 M, which represents the solubility of Ag2CrO4 in 0.05M KClO4.
b) In the presence of 0.005M AgNO3:
Ag2CrO4 dissociates into 2 Ag+ ions and 1 CrO42- ion.
So, we have Ksp = (2[Ag+] + 0.005)^2 * [CrO42-] (concentration of Ag+ ions is 0.005M higher due to AgNO3).
Given that Ksp for Ag2CrO4 = 1.12 * 10^-12, we can substitute the values into the expression:
1.12 * 10^-12 = (2x + 0.005)^2 * (x),
where x represents the concentration of Ag+ and CrO42- ions.
To find the solubility of Ag2CrO4 in 0.005M AgNO3, we need to solve this equation and find the value of x.