Posted by LT on .
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

Chemistry 
DrBob222,
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.

Chemistry 
LT,
So what do i do then to solve for the ksp

Chemistry 
DrBob222,
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. 
Chemistry 
LT,
So im solving for two expressions... is that it.

Chemistry 
LT,
ok I got x^2 = 1.8 x 10^8 = 1.34 x 10^4 for PbSO4 is this right

Chemistry 
DrBob222,
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 + 1E3)= 1.8 x 10^8
S = 
Chemistry 
LT,
i subtracted 1E3 from the 1.8E8 and it got me a 9.999E4 then i took the sqrt cause (X)(X) is x^2 and then the calculator told me an error

Chemistry 
DrBob222,
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.