Question

The solubility product of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ is $7.4 \times {10^{ - 11}}$
.Calculate the solubility of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ in 0.1 M NaF solution.

A.$7.4 \times {10^{ - 9}}$

B.$3.7 \times {10^{ - 9}}$

C. $3.7 \times {10^{ - 11}}$

D. $7.4 \times {10^{ - 11}}$

Answer 1

We know that, the product of concentration of the ions of the salt in its saturated solution at a given temperature raised to the power the number of ions produced by the dissociation of one mol of the salt is the solubility product.

Complete step by step answer: First, we write the solubility equilibrium in case of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$.

$MgF_2\rightleftharpoons Mg^{+2}+2F^{-}$ …… (1)

Let’s take the solubility of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ be${\rm{S}}\,{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}$. So, the concentration of magnesium ion $\left( {{\rm{M}}{{\rm{g}}^{2 + }}} \right)$ will be ${\rm{S}}\,{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}$ and concentration of fluoride ion $\left( {{{\rm{F}}^ - }} \right)$ is ${\rm{2S}}\,\,{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}$

Now, we write the dissociation of NaF.

${\rm{NaF}} \to {\rm{N}}{{\rm{a}}^ + } + {{\rm{F}}^ - }$

Given that NaF is of 0.1 M. So, concentration of sodium ion $\left( {{\rm{N}}{{\rm{a}}^ + }} \right)$ is 0.1 M and fluoride ion $\left( {{{\rm{F}}^ - }} \right)$ is 0.1 M.

Now, we write the solubility product expression of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ using equation (1).

${K_{sp}} = \left[ {{\rm{M}}{{\rm{g}}^{2 + }}} \right]{\left[ {{{\rm{F}}^ - }} \right]^2}$
…… (2)

The solubility product is given as $7.4 \times {10^{ - 11}}$. The concentration of ${\rm{M}}{{\rm{g}}^{2 + }}$ is S$\,{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}$. Now, we need the fluoride ion concentration.

As ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$is soluble in NaF, the fluoride concentration is equal to the summation of $\left[ {{{\rm{F}}^ - }} \right]$ from ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ and $\left[ {{{\rm{F}}^ - }} \right]$ from NaF.

$\left[ {{{\rm{F}}^ - }} \right] = \left[ {{{\rm{F}}^ - }} \right]\,\,{\rm{from}}\;\,{\rm{Mg}}{{\rm{F}}_{\rm{2}}} + \left[ {{{\rm{F}}^ - }} \right]\,\,{\rm{from}}\,\;{\rm{NaF}}$

The fluoride ion concentration from ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ is 2S$\,{\rm{mol}}\,\,{{\rm{L}}^{ - 1}}$ and fluoride ion concentration from NaF is 0.1 M. So, the total fluoride concentration is,

$\left[ {{{\rm{F}}^ - }} \right] = 2S + 0.1$

As value of ${K_{sp}}$ is very small, ${\rm{2S}} + 0.{\rm{1}} \approx 0.1$

So, the fluoride concentration is 0.1.

Now, we have to substitute the value of ${K_{sp}}$, $\left[ {{{\rm{F}}^ - }} \right]$ and $\left[ {{\rm{M}}{{\rm{g}}^{2 + }}} \right]$ in equation (2).

${K_{sp}} = \left[ {{\rm{M}}{{\rm{g}}^{2 + }}} \right]{\left[ {{{\rm{F}}^ - }} \right]^2}$

$7.4 \times {10^{ - 11}} = S \times {\left( {0.1} \right)^2}$

$S = \dfrac{{7.4 \times {{10}^{ - 11}}}}{{0.01}}$

$S = 7.4 \times {10^{ - 9}}\,$

Hence, the solubility of ${\rm{Mg}}{{\rm{F}}_{\rm{2}}}$ in 0.1 M NaF solution is $7.4 \times {10^{ - 9}}\,{\rm{M}}$.

Hence, the correct answer is A.

Note: Remember that solubility product is the product of ionic concentration in the saturated solution whereas ionic product is the product of ionic concentration at any concentration of the solution. Solubility products have a fixed value for a salt at a constant temperature but the value of ionic product changes with the change of concentration of the ion.