Question

A solution has a Ca2+ concentration of 0.049 M and an F- concentration is 0.147 M at equilibrium. The value of ksp for CaF2 at 25°C is 4.0 x 10-11. Will this solution form a precipitate?

Answer 1

Yes

Step-by-step explanation:

Answer 2

The solution will precipitate.

Step-by-step explanation:

For a solid to precipitate, its ionic product should be more than the solubility product.

`CaF_2` will form its respective ions in the solution as:

`CaF_2_((s))\rightleftharpoons Ca^(2+)+2F^-`

Ionic product =
`[Ca^(2+)][F^-]^2`

Given that:

`[Ca^(2+)]=0.049\ M`

`[F^-]=0.147\ M`

So,

Ionic product = 0.049 × (0.147)² = 0.00106

Given =
`K_(sp)=4.0* 10^(-11)`

Ionic product > Solubility product

The solution will precipitate.