Question

The K sp for barium fluoride, BaF2, is 2.45 × 10-5. What is the molar solubility of barium fluoride?

Answer 1

Answer: The molar solubility of barium fluoride is 0.0183 moles/liter.

Explanation:

The equation for the reaction will be as follows:

$BaF_2\rightarrow Ba^{2+}+2F^$

By Stoichiometry,

1 mole of $BaF_2$ gives 2 moles of $F^-$ and 1 mole of $Ba^{2+}$

Thus if solubility of $BaF_2$ is s moles/liter, solubility of $Ba^{2+}$ is s moles/liter and solubility of $F^-$ is 2s moles/liter

Therefore,  

$K_sp=[Ba^{2+}][F^{-}]^2$

$2.45\times 10^{-5}=[s][2s]^2$

$4s^3=2.45\times 10^{-5}$

$s^3=6.12\times 10^{-6}moles/liter$

$s=0.0183moles/liter$

Thus the molar solubility of barium fluoride is 0.0183 moles/liter.