Write the equilibrium equation (Keq), the solubility expression (Ksp) and calculate the solubility of PbSO4 in mole/ L. The Ksp value of PbSO4 is 1.3*10^-8
To write the equilibrium equation and solubility expression for PbSO4, we need to understand the dissociation of the compound.
The chemical formula for lead(II) sulfate is PbSO4. In water, it dissociates into its constituent ions, Pb²⁺ and SO₄²⁻.
The equilibrium equation is:
PbSO4 (s) ⇌ Pb²⁺ (aq) + SO₄²⁻ (aq)
Next, let's write the solubility expression.
The solubility expression represents the equilibrium constant for the dissolution of a sparingly soluble salt and is denoted as Ksp (solubility product constant). It is obtained by multiplying the concentrations of the dissolved ions, raised to their stoichiometric coefficients.
The solubility expression for PbSO4 is:
Ksp = [Pb²⁺] * [SO₄²⁻]
Now, to calculate the solubility of PbSO4 in moles per liter (Molarity or M), we need to determine the concentration of either Pb²⁺ or SO₄²⁻ when the compound is at equilibrium. Since PbSO4 is a 1:1 electrolyte, meaning it dissociates into equal amounts of Pb²⁺ and SO₄²⁻ ions, we only need to consider one of them. Let's use Pb²⁺ as our example.
To calculate the solubility, S, we assume that x moles of PbSO4 dissolve in water, which means x moles of Pb²⁺ and SO₄²⁻ ions are formed. Therefore, the concentration of Pb²⁺ at equilibrium is x moles/L.
Since the Ksp value is given as 1.3 × 10⁻⁸, we can substitute these values into the solubility expression.
1.3 × 10⁻⁸ = x * x
Solving for x gives the solubility of PbSO4:
x² = 1.3 × 10⁻⁸
x = √(1.3 × 10⁻⁸)
Calculating this gives us the approximate solubility of PbSO4 in moles per liter.
.............PbSO4 ==> Pb^2+ + [SO4]^2-
I...............solid............0............0
C..............solid............x.............x
E...............solid............x.............x
Ksp = (Pb^2+)(SO4^2-)^2- = 1.3E-8
Plug in the E line into the Ksp expression and solve for x = solubility in mols/L for PbSO4.