Chemistry

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)
?

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  1. 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.

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  2. You are amazing. I can't thank you enough!!

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