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?

Respuesta :

Answer:

The solution will precipitate.

Explanation:

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

[tex]CaF_2[/tex] will form its respective ions in the solution as:

[tex]CaF_2_{(s)}\rightleftharpoons Ca^{2+}+2F^-[/tex]

Ionic product = [tex][Ca^{2+}][F^-]^2[/tex]

Given that:

[tex][Ca^{2+}]=0.049\ M[/tex]

[tex][F^-]=0.147\ M[/tex]

So,

Ionic product = 0.049 × (0.147)² = 0.00106

Given = [tex]K_{sp}=4.0\times 10^{-11}[/tex]

Ionic product > Solubility product

The solution will precipitate.

Answer:

Yes

Explanation:

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