Answer to Question #251996 in General Chemistry for angle

Question

calculate the solubility of BaF-2 at 18°C

Answer 1

The solubility product constant for BaF₂ at 18°C is 1.7 × 10⁻⁶.

The solubility product constant K_sp of a compound is the product of molar equilibrium concentrations of the ions in the compound, each concentration raised to a power equal to its coefficient in the solubility equation Molar concentration of ions is related to molar solubility of ionic compound.

Molar solubility is the number of moles of compound dissolved per liter to produce a saturated solution.

Let assume the molar solubility of barium fluoride BaF₂ be x. Barium fluoride is dissolved in a liter of water to form x moles of Ba⁺ and 2x moles of F^-.

Summarize the data in the following table:

Substitute the equilibrium concentrations in the equilibrium equation:

$K_{sp} = [Ba^+][F^-]^2 \\ 1.7 \times 10^{-6} =(x)(2x)^2 \\ 1.7 \times 10^{-6} = 4x^3$

Solving for x

$x = (\frac{1.7 \times 10^{-6}}{4})^{1/3} \\ x= 7.5 \times 10^{-3} \;M$

Therefore, molar solubility of barium fluoride is $7.5 \times 10^{-3} \;mol/L.$