Respuesta :
Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.
Explanation:
The equation for the reaction will be as follows:
[tex]BaF_2\rightarrow Ba^{2+}+2F^[/tex]
By Stoichiometry,
1 mole of [tex]BaF_2[/tex] gives 2 moles of [tex]F^-[/tex] and 1 mole of [tex]Ba^{2+}[/tex]
Thus if solubility of [tex]BaF_2[/tex] is s moles/liter, solubility of [tex]Ba^{2+}[/tex] is s moles/liter and solubility of [tex]F^-[/tex] is 2s moles/liter
Therefore,
[tex]K_sp=[Ba^{2+}][F^{-}]^2[/tex]
[tex]2.45\times 10^{-5}=[s][2s]^2[/tex]
[tex]4s^3=2.45\times 10^{-5}[/tex]
[tex]s^3=6.12\times 10^{-6}moles/liter[/tex]
[tex]s=0.0183moles/liter[/tex]
Thus the molar solubility of barium fluoride is 0.0183 moles/liter.
The molar solubility of barium fluoride is 0.0183 mole/liter.
What is molar solubility?
It is defined as the no. of moles of solute dissolved per liter of solution before of the saturation.
Chemical equation for barium fluoride is shown as:
BaF₂ → Ba⁺ + 2F⁻
Molar solubility for the BaF₂ i.e. [tex]${{\rm{K}}_{{\rm{sp}}}}$[/tex] = 2.45 × 10⁻⁵ (given)
And it is calculates as [tex]${{\rm{K}}_{{\rm{sp}}}}$[/tex] = [Ba⁺] [F⁻]²
Let, the solubility of Ba⁺ is x mol/liter and solubility of F⁻ is 2x mole/liter according to the stoichiometry. Now putting all the values in the above [tex]${{\rm{K}}_{{\rm{sp}}}}$[/tex] equation we get,
2.45 × 10⁻⁵ = [x] [2x]²
4x³ = 2.45 × 10⁻⁵
x = 0.0183 mole/liter
Hence, the molar solubility of barium fluoride is 0.0183 mole/liter.
To learn more about molar solubility, visit the below link:
https://brainly.com/question/4170005
