Question

The Ksp of CaF2 is 4.0 x 10-11. What is the solubility of CaF2 in 0.10 M Ca(NO3)2? Give your answer using scientific notation and to 2 significant figures (i.e., one decimal place).

Answer 1

Answer:

$1.0\times 10^{-5} M$ is the solubility of $CaF_2$ in 0.10 M $Ca(NO_3)_2$.

Explanation:

Concentration of calcium nitrate = $[Ca(NO_3)_2]=0.10 M$

$Ca(NO_3)_2\rightarrow Ca^{2+}+2NO_3^{-}$

$[Ca(NO_3)_2]=0.10 M=[Ca^{2+}]$

Solubility product of the calcium fluoride = $K_{sp}=4.0\times 10^{-11}$

Solubility of calcium fluoride in calcium nitrate solution :

$CaF_2\rightleftharpoons Ca^{2+}+2F^-$

            (0.10+S)   (2S)

The expression of solubility is given by :

$K_{sp}=[Ca^{2+}][F^-]^2$

$4.0\times 10^{-11}=(S+0.10)(2S)^2$

Solving for S:

$S=1.0\times 10^{-5} M$

$1.0\times 10^{-5} M$ is the solubility of $CaF_2$ in 0.10 M $Ca(NO_3)_2$.

Answer 2

The solubility of $CaF_2$ in 0.10 M  $Ca(NO_3)_2$  is equal to $3.7 \times 10^{-6}\;M$

Given the following data:

• Ksp of $CaF_2= 4.0 \times 10^{-11}.$

• Molarity of $Ca(NO_3)_2$ = 0.10 M.

How to determine the solubility.

First of al, we would write the properly balanced chemical equation for this chemical reaction:

                                     $CaF_2(s)\rightleftharpoons Ca^{2+} (aq)+ 2F^{-}(aq)$

Initial cond.                       c                    0             0

At equib.                          c                     x              2x

Mathematically, the Ksp for this chemical reaction is given by:

$K_{sp}=[Ca^{2+}]{[F^{-}]^2$

From the ICE table, we have:

$4.0 \times 10^{-11}.=x[2x]^2\\\\4.0 \times 10^{-11}.=3x^2\\\\x=\sqrt{\frac{4.0 \times 10^{-11}}{3} } \\\\x=3.7 \times 10^{-6}\;M$

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