Question

Determine the molar solubility of baf2 in pure water. ksp for baf2 = 2.45 x 10-5. 1.83 x 10-2 m 1.23 x 10-5 m 2.90 x 10-2 m 4.95 x 10-3 m 6.13 x 10-6 m

Answer 1

The molar solubility of BaF2 in pure water is approximately 6.13 x 10⁻⁶ M.

Step-by-step explanation:

The molar solubility of BaF2 in pure water can be calculated using the solubility product constant (Ksp) of BaF2. The equation for the dissolution of BaF2 is BaF2 → Ba²+ + 2F¯. The Ksp expression for BaF2 is Ksp = [Ba²+][F¯]². Since there is a 1:2 stoichiometric relationship between BaF2 and fluoride ions in solution, the molar solubility of BaF2 can be represented as x, which gives the concentrations of Ba²+ and F¯ as x and 2x, respectively. Substituting these values into the Ksp expression, we get:

Ksp = (x)(2x)² = 4x³ = 2.45 x 10⁻⁵

Solving for x, we find that the molar solubility of BaF2 in pure water is approximately 6.13 x 10⁻⁶ M.

Answer 2

The Molar solubility of baf2 in pure water is 1.83 x 10⁽₋₂⁾ if ksp for baf2 = 2.45 x 10⁻⁵.
solubility product equilibrium reaction from the balance equation of reaction is:
k₍sp₎ = [Ba⁺²] [F⁻]²
using mole ratios from one to another, [Ba⁺²] = x and [F⁻]² = 2x
k₍sp₎ = [Ba⁺²] [F⁻]²
k₍sp₎ = [x][2x]²
ksp = 2.45 x 10⁻⁵ then,
2.45 x 10⁻⁵ = [x][2x]²
4x³ = 2.45 x 10⁻⁵
x = ∛(2.45 x 10⁻⁵)/4 = 1.83 x 10⁻²m
so, x is molar solubility which is 1.83 x 10⁻²m