the solubility product of PbSO4 at 298K is 1.6x10^-8 mol^2dm^-6. Calculate the solubility of PbSO4 in pure water
PbSO4 ==> Pb^2+ + SO4^2-
..x.........x......x
Substitute into Ksp expression and solve or x.
To calculate the solubility of PbSO4 in pure water, we can set up an equilibrium expression called the solubility product constant (Ksp), which relates to the concentration of the dissolved ions at equilibrium.
The solubility product constant expression for lead(II) sulfate (PbSO4) is as follows:
Ksp = [Pb2+][SO42-]
From the equation, we can see that one mole of PbSO4 dissociates into one mole of Pb2+ ions and one mole of SO42- ions. Therefore, at equilibrium, the concentration of Pb2+ ions and SO42- ions will be equal.
Let's assume that x represents the solubility of PbSO4, which will be the concentration of both Pb2+ and SO42- ions.
Therefore, [Pb2+] = x
And [SO42-] = x
Since Ksp is given as 1.6x10^-8 mol^2dm^-6, we can substitute the respective values into the Ksp expression:
Ksp = [Pb2+][SO42-]
1.6x10^-8 = x * x
Simplifying the equation, we have:
1.6x10^-8 = x^2
To solve for x, we take the square root of both sides of the equation:
√ (1.6x10^-8) = √ (x^2)
x = √ (1.6x10^-8)
Evaluating this expression using a calculator, we get:
x ≈ 1.26x10^-4 mol/dm^3
Therefore, the solubility of PbSO4 in pure water is approximately 1.26x10^-4 mol/dm^3.