Question

Calculate the molar solubility of barium fluoride, baf2, in each liquid or solution. for barium fluoride, ksp =2.45 x 10-5.

Answer 1

The molar solubility of barium fluoride is 0.018 M.

Step-by-step explanation:

The molar solubility is the maximum molar concentration of solute (in this case BaF₂) in any solvent. In this type of problems there are two important numbers, the "molar solubility" (represented by the letter "s") and the "Solubility Product Constant" (ksp).

- To start solving the problem, we have to write the dissociation reaction where the BaF₂ turns to ions Ba⁺² and F⁻.

- Then, it is necessary to multiply the molar solubility of each ion by its coefficient. After doing this, we can write the product of molar solubility of each ion which is equal to the Ksp. Here, it is very important rise to the power indicated by the coefficient of each ion. So, we get:

Ksp = [Ba⁺²] . [F⁻]²

After solving this, we find that the molar solubility is 0.018 M. Check the step by step resolution in the attach.

Remember that molar (M) means mol/L.

Answer 2

BaF₂ when it dissolves, dissociates as follows;
BaF₂ --> Ba²⁺ + 2F⁻
Molar solubility is the number of moles that can be dissolved in 1 L of solution. If molar solubility of BaF₂ is x, then molar solubility of Ba²⁺ is x and solubility of F⁻ is 2x.
ksp = [Ba²⁺][F⁻]²
ksp = (x)(2x)²
2.45 x 10⁻⁵ = 4x³
x³ = 0.6125 x 10⁻⁵
x = 0.0183 mol/L is molar solubility of BaF₂