Calculate the solubility, in molL―1, of lead sulfate, PbSO4, in a 0.0185 M solution of sodium sulfate, Na2SO4 .

Ksp (PbSO4) = 1.6 x 10―8

I get 8 x 10^-9...but that is not right! Thanks!

You need to write a balanced equation for this reaction and write an ice table. since you know ksp.. you can solve for the x value. and plug it back into your ice table. easy stuff.

If I am reading it correctly...

1.6E-8=(2x)^2 (.0185+x)
ignoring the +x (it is small)
x=1.6E-8/(4*.0185)

To calculate the solubility of lead sulfate (PbSO4) in a 0.0185 M solution of sodium sulfate (Na2SO4), we need to use the concept of the common ion effect.

The common ion effect states that if a compound is already dissolved in a solution containing one of its ions, the solubility of that compound in another solution containing the same ion will be reduced.

In this case, both lead sulfate (PbSO4) and sodium sulfate (Na2SO4) contain the sulfate ion (SO4^2-). Since we have a solution of sodium sulfate, we can assume that the sulfate ion concentration will be equal to the concentration of sodium sulfate, which is 0.0185 M.

Given that the solubility product constant (Ksp) for lead sulfate (PbSO4) is 1.6 x 10^-8, we can use the equation:

Ksp = [Pb^2+][SO4^2-]

However, since we have the concentration of the sulfate ion ([SO4^2-]), we can rewrite the equation as:

Ksp = [Pb^2+][(0.0185 M)]

Now, we can solve for the concentration of the lead ion ([Pb^2+]):

[Pb^2+] = Ksp / (0.0185 M)

Plugging in the given values:

[Pb^2+] = (1.6 x 10^-8) / (0.0185)

[Pb^2+] ≈ 8.6486 x 10^-7 mol/L

Therefore, the solubility of lead sulfate in a 0.0185 M solution of sodium sulfate is approximately 8.6486 x 10^-7 mol/L.