Question

A saturated solution of barium fluoride, BaF2BaF2, was prepared by dissolving solid BaF2BaF2 in water. The concentration of Ba2+Ba2+ ion in the solution was found to be 7.52×10−3 MM . Calculate KspKspK_sp for BaF2BaF2.

Answer 1

Answer:
`1.70* 10^(-6)`

Step-by-step explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as

The equation for the ionization of the
`BaF_2` is given as:

`BaF_2\rightarrow Ba^(2+)+2F^-`

When the solubility of
`BaF_2` is S moles/liter, then the solubility of
`Ba^(2+)` will be S moles/liter and solubility of
`F^-` will be 2S moles/liter.

By stoichiometry of the reaction:

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

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

`K_(sp)=s* (2s)^2`

`K_(sp)=4* (7.52* 10^(-3))^3`

`K_(sp)=1.70* 10^(-6)`

Thus
`K_(sp)` for
`BaF_2` is
`1.70* 10^(-6)`