Write the equilibrium equation (Keq), the solubility expression (Ksp) and calculate the solubility of PbSO4 in moles/ L. The Ksp value of lead sulphate is 1.3 * 10^-8. :)))
To write the equilibrium equation for the solubility of PbSO4 and the associated equilibrium constant (Keq), we need to know the dissociation reaction. In this case, PbSO4 dissociates into lead ions (Pb2+) and sulfate ions (SO42-).
The equilibrium equation for the dissociation of PbSO4 is as follows:
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq)
The equilibrium constant Keq can be expressed using the concentrations of the ions at equilibrium. In this case, the concentrations of Pb2+ and SO42- will be represented as [Pb2+] and [SO42-], respectively.
Keq = [Pb2+] * [SO42-]
Now, let's move on to the solubility expression (Ksp) and the calculation of the solubility of PbSO4.
The solubility product expression (Ksp) is the equilibrium expression for the dissolution of a solid compound in water. For PbSO4, the Ksp expression is as follows:
Ksp = [Pb2+][SO42-]
Given that the Ksp value of lead sulfate (PbSO4) is 1.3 * 10^-8, we can equate it to the [Pb2+] and [SO42-] concentration at equilibrium:
1.3 * 10^-8 = [Pb2+][SO42-]
Since PbSO4 dissociates into 1 Pb2+ ion and 1 SO42- ion, the solubility of PbSO4 in moles/L is equal to the concentration of Pb2+ or SO42- at equilibrium.
Therefore, the solubility of PbSO4 in moles/L (M) is the square root of the Ksp value:
Solubility of PbSO4 = (Ksp)^0.5 = (1.3 * 10^-8)^0.5
Using a calculator, we find that the solubility of PbSO4 is approximately 3.6 * 10^-4 M (moles/L).