Chemistry

what is the molar solubility of lead sulfate in 1.0 X 10^-3 M Na2SO4?

The Solubility Product Constant PbSO4

Ksp= 1.8 x 10^-8

The Ksp expression is:
Na2SO4 = [2Na][SO4^-2]

[2Na]=2x
[SO4^-2]=x

Ksp= 1.8 x 10^-8 = (x)(2x)
Ksp= 1.8 x 10^-8 /4x^3

1.8 x 10^-8/4 = x^3
4.5 x 10^-9 = x^3

is this right so far... what do i do next please

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asked by LT
  1. No, you have made some wrong assumptions. First, PbSO4 is the insoluble material. Na2SO4 is soluble in water. Second, this is a common ion problem in which the presence of the sulfate ion from Na2SO4 will decrease the solubility of PbSO4.

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  2. So what do i do then to solve for the ksp

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    posted by LT
  3. The problem doesn't ask for you to solve for Ksp. It asks for the solubility of PbSO4 in 1 x 10^-3 M Na2SO4.

    PbSO4 ==> Pb^+2 + SO4^=

    Ksp = (Pb^+2)(SO4^=) = 1.8 x 10^-8
    If solubility of PbSO4 = S (or x if you wish), then (Pb^+) = S and (SO4^=) = S

    The Na2SO4 in solution is
    Na2SO4 ==> 2Na^+ + SO4^=
    Since it is 1 x 10^-3 M in Na2SO4, then (SO4^=) = 1 x 10^-3.

    Now set those up in the Ksp expression and solve for S. Remember that the total sulfate is that from the PbSO4 + that form the Na2SO4.

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  4. So im solving for two expressions... is that it.

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    posted by LT
  5. ok I got x^2 = 1.8 x 10^-8 = 1.34 x 10^-4 for PbSO4 is this right

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    posted by LT
  6. You are solving for the only unknown in the equation, which is S.
    Ksp = (Pb^+2)(SO4^-2) = 1.8 x 10^-8
    (S)(S + 1E-3)= 1.8 x 10^-8
    S =

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  7. i subtracted 1E-3 from the 1.8E-8 and it got me a -9.999E-4 then i took the sqrt cause (X)(X) is x^2 and then the calculator told me an error

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    posted by LT
  8. Ksp = (Pb^+2)(SO4^=) = 1,8 x 10^-8
    (Pb^+2) = x
    (SO4^=) = x+0.001
    (x)(x+0.001) = 1.8 x 10^-8
    You can see that this is a quadratic equation and we could solve it as a quadratic. The exact solution is (and I'm switching to y so as not to confuse the x with the times sign),
    y^2 + 0.001y = 1.8 x 10^-8

    There are two ways to do this.
    1. Simplify the equation so there is no quadratic. You do that by making the assumption that y + 0.001 = 0.001; that is, that y is such a small number that a small number + 0.001 is still 0.001. If we do that then we have
    y = 1.8 x 10^-8/0.001 = 1.8 x 10^-5 M. We need to test to see if our assumption is ok. 0.001 + 1.8 x 10^-5 = 0.00102 and there is just a small difference so we let it go and work it the easy way.
    2. we can solve the quadratic by equating
    y^2 + 0.001 -1.8 x 10^-8 = 0
    I solved this by the quadratic formula and obtained 1.77 x 10^-5 so you see we don't make a bad assumption by solving it the easy way.

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