Answer:
ksp = 2.2 x ⁻⁴
Explanation:
The equilibrium here is:
PbCl₂ (s) ⇄ Pb²⁺ + 2 Cl⁻
we can recognize it as a product solubilty equilibrium once we are told that some undissolved PbCl₂ remained.
The equilibrium constant, Ksp is given by the equation
Ksp = [Pb²⁺][Cl⁻]²
where [Pb²⁺] and [Cl⁻]² are the concentrations (M) of Pb²⁺ and Cl⁻ in solution.
we have the mass of solid PbCl₂ placed in solution, so we can determine the number of moles it represents, and if we substract the moles of undissolved PbCl₂ we will know the moles of Pb²⁺ and Cl⁻ which went into solution.
From there we can calculate the molarity (M= moles/L solution) and finally plug the values into our expression for Ksp to answer this question.
molar mas PbCl₂ = 278.1 g/mol
1 milligram = 1 x 10⁻³ g
mol PbCl₂ initially = 231.8 x 10⁻³ g / 278.1 mol = 8.3 x 10⁻⁴ mol
Volume solution = 15 mL x 1L / 1000 mL = 0.015 L
mol undissolved PbCl₂ = 74 x 10⁻³ g / 278.1 g/mol = 2.7 x 10⁻⁴ mol
mol PbCl dissolved = 8.3 x 10⁻⁴ mol - 2.7 x 10⁻⁴ mol = 5.7 x 10⁻⁴ mol
Concentration of Pb²⁺ in solution = 5.7 x 10⁻⁴ mol / 0.015 L = 3.8 x 10⁻² M
Concentration of Cl⁻ in solution = 2 x 3.8 x 10⁻² M = 7.6 x 10⁻² M
(Note from the formula we we get 2 mol Cl⁻ per mol PbCl₂)
Plugging these values into the expression for Ksp we have
Ksp = 3.8 x 10⁻² x (7.6 x 10⁻²)² = 2.2 x 10⁻⁴
