Question

Determine the molar solubility of baf2 in pure water. ksp for baf2 = 2.45 × 10⁻5. (single choice)

a.0
b.1.83 × 10⁻2 m
c.1.23 × 10⁻5 m
d. 2.90 × 10⁻2 m
e.4..95 × 10⁻3 m
f..6.13 × 10⁻6 m

Answer 1

To determine the molar solubility of BaF2 in pure water, we use the solubility product constant (Ksp) of BaF2. The molar solubility is found by solving the Ksp expression for BaF2. The molar solubility of BaF2 in pure water is 1.83 × 10⁻2 M (Option B).

Step-by-step explanation:

To determine the molar solubility of BaF2 in pure water, we need to use the solubility product constant (Ksp) of BaF2. The Ksp for BaF2 is given as 2.45 × 10⁻5. In this case, BaF2 dissociates into Ba²⁺ and 2F⁻ ions. Let's assume that the molar solubility of BaF2 is x M. Therefore, the concentration of Ba²⁺ and F⁻ ions will also be x M.

Using the Ksp expression for BaF2, we have [Ba²⁺] * [F⁻]² = (x)(x)² = x³. The given Ksp for BaF2 is 2.45 × 10⁻5. Equating this to x³, we have x³ = 2.45 × 10⁻5. Taking the cube root of both sides, we get x = 1.83 × 10⁻2 M.

Therefore, the molar solubility of BaF2 in pure water is 1.83 × 10⁻2 M (Option B).