Question

Barium fluoride, , is partially soluble with Ksp of 1.7e-06. What is the solubility of barium fluoride in pure water

Answer 1

`7.5x10^(-3)M`

Step-by-step explanation:

Hello,

In this case, since the dissociation barium fluoride is represented at equilibrium by:

`BaF_2(s)\rightleftharpoons Ba^(2+)(aq)+2F^-(aq)`

Hence, the equilibrium expression is:

`Ksp=[Ba^(2+)][F^-]^2`

Whereas the molar solubility is represented as the reaction extent
`x`:

`Ksp=[x][2x]^2`

In such a way, we can solve for
`x`:

`1.7x10^(-6)=4x^3\\\\x=\sqrt[3]{(1.7x10^(-6))/(4) } \\\\x=7.5x10^(-3)M`

Which as said before, is the molar solubility.

Best regards.