Question

An aqueous solution of sodium fluoride is slowly added to a water sample that contains barium ion (2.75×10-2M ) and calcium ion (6.70×10-2M ). What is the remaining concentration of the first ion to precipitate when the second ion begins to precipitate?

Answer 1

Answer : The remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, $1.07\times 10^{-6}M$

Explanation :

The dissociation of barium fluoride is written as:

$BaF_2\rightleftharpoons Ba^{2+}+2F^-$

The expression for solubility constant for this reaction will be,

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

As we know that at room temperature $25^oC$ the $K_{sp}$ of barium fluoride is, $1\times 10^{-6}$.

Now put all the given values in this expression, we get:

$1\times 10^{-6}=(2.75\times 10^{-2})\times [F^-]^2$

$[F^-]=6.03\times 10^{-3}M=0.00603M$

The barium fluoride precipitate when fluoride ion is equal to 0.00603 M.

The dissociation of calcium fluoride is written as:

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

The expression for solubility constant for this reaction will be,

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

As we know that at room temperature $25^oC$ the $K_{sp}$ of calcium fluoride is, $3.90\times 10^{-11}$.

Now put all the given values in this expression, we get:

$3.90\times 10^{-11}=(6.70\times 10^{-2})\times [F^-]^2$

$[F^-]=2.41\times 10^{-5}M=0.0000241M$

The calcium fluoride precipitate when fluoride ion is equal to 0.0000241 M.

Since, the fluoride ion concentration in calcium fluoride is less then the fluoride ion concentration in barium fluoride. That means, calcium fluoride will precipitate first.

Thus, the concentration $Ca^{2+}$ ion remaining at $F^-$ concentration (0.00603 M) is calculated as:

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

Now put all the given values in this expression, we get:

$3.90\times 10^{-11}=[Ca^{2+}]\times (0.00603)^2$

$[Ca^{2+}]=1.07\times 10^{-6}M$

Therefore, the remaining concentration of the first ion to precipitate when the second ion begins to precipitate is, $1.07\times 10^{-6}M$