Question

Derive the formula for the solubility product of sparingly soluble salt CaF₂

Answer 1

Answer:

$K_{sp}= 4x^{3}$

Explanation:

We know the ionic equation for CaF₂ which is

$CaF_{2}\rightarrow Ca^{2+} +2F^{-1}$

The solubility product constant is holds for the solid substance which are used to dissolve in aqueous solution during equilibrium.It is represented by $K_{sp}$.

$K_{sp} = [Ca^{2+}][(F^{-1} )^{2}]$................(1)

Let the concentration of

$Ca_{+2} =x$ and $F^{-1} =2x$

used these concentration in (1) we get,

$K_{sp} = [x][4x ^{2}] = 4x^{3}$