What is the molar solubility of BaF2 in a solution containing 0.0750 M LiF (Ksp = 1.7 x 10^-6) (a) 2.3 x 10^-5 M (b) 3.0 x 10^-4 M (c) 1.2 x 10^-2 M (d) 1.3 x 10^-3 M

Verified step by step guidance

1. Write the dissolution equation for BaF2: BaF2(s) ⇌ Ba2+(aq) + 2F-(aq).
2. Set up the expression for the solubility product constant (Ksp) of BaF2: Ksp = [Ba2+][F-]^2.
3. Recognize that the fluoride ion concentration (F-) is affected by the presence of LiF, which also dissociates to give F-. Since LiF is a strong electrolyte, it fully dissociates to give [F-] = 0.0750 M initially.
4. Let the molar solubility of BaF2 be 's'. The total [F-] in solution will then be 0.0750 M + 2s, because for each mole of BaF2 that dissolves, 2 moles of F- are produced.
5. Substitute the expressions for [Ba2+] and [F-] back into the Ksp expression and solve for 's' (the molar solubility of BaF2).

Key Concepts

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at equilibrium, expressed in moles per liter (M). It is a crucial concept in solubility equilibria, as it helps determine how much of a compound can dissolve before reaching saturation.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. It is calculated from the concentrations of the ions in a saturated solution, and it provides insight into the extent to which a compound can dissolve in water. For BaF2, Ksp is given as 1.7 x 10^-6.

Common Ion Effect

The common ion effect describes the decrease in solubility of a salt when a common ion is added to the solution. In this case, the presence of LiF introduces F- ions, which shifts the equilibrium of BaF2 dissolution, thereby reducing its molar solubility. This principle is essential for calculating the new solubility in the presence of a common ion.