Question

A solution is saturated with respect to ${\text{SrC}}{{\text{O}}_{\text{3}}}{\text{ and Sr}}{{\text{F}}_{\text{2}}}$. The concentration of the carbonate ion was found to be $1.2 \times {10^{ - 3}}M$. The concentration of fluoride ion in the solution would be: (values of the solubility products of ${\text{SrC}}{{\text{O}}_{\text{3}}}$ and ${\text{Sr}}{{\text{F}}_{\text{2}}}$ are $7.0 \times {10^{ - 10}}{\text{ and }} 7.9 \times {10^{ - 10}}$ respectively)
A ) $1.3 \times {10^{ - 3}}M$
B ) $2.6 \times {10^{ - 2}}M$
C ) $3.7 \times {10^{ - 2}}M$
D ) $5.8 \times {10^{ - 7}}M$

Answer 1

In the solubility product expressions for strontium carbonate and strontium fluoride, the strontium cation is the common ion and cancels out when the solubility product expressions are divided. Substitute the value of the solubility product expressions of strontium carbonate and strontium fluoride, and the value of carbonate ion concentration to calculate the value of the fluoride ion concentration.

Complete step by step answer:
Strontium carbonate and strontium di-fluoride are sparingly soluble compounds. The solubility product constant for a sparingly soluble salt is the equilibrium constant for the dissolution of the sparingly soluble salt in water. It is equal to the product of the concentrations of cations and anions obtained by dissociation of the salt. These concentration terms are raised to appropriate stoichiometric coefficients.
Given the values of the solubility product constants for strontium carbonate and strontium di-fluoride as $7.0 \times {10^{ - 10}}{M^2}{\text{ and }}7.9 \times {10^{ - 10}}{M^3}$ respectively. You will find that these two values are comparable to each other, so you cannot neglect none of the values. Hence, during the calculation, you should consider both solubility equilibria (for strontium carbonate and strontium fluoride).
If one value was much smaller than another value, then you could have neglected the smaller value. This would have simplified your calculation.
Write the expressions for the dissociation of strontium carbonate and strontium fluoride.

{\text{ Sr}}{{\text{F}}_{\text{2}}}{\text{ }} \rightleftharpoons {\text{ S}}{{\text{r}}^{2 + }}{\text{ + }}2{\text{ }}{{\text{F}}^ - } \\\\$$ Write the expressions for the solubility product constants of strontium carbonate and strontium fluoride. $${K_{sp}}\left( {{\text{SrC}}{{\text{O}}_{\text{3}}}} \right) = \left[ {{\text{S}}{{\text{r}}^{2 + }}} \right] \times \left[ {{\text{CO}}_3^{2 - }} \right] \\\ {K_{sp}}\left( {{\text{Sr}}{{\text{F}}_{\text{2}}}} \right) = \left[ {{\text{S}}{{\text{r}}^{2 + }}} \right] \times {\left[ {{{\text{F}}^ - }} \right]^2} \\\\$$ Divide the expression for the solubility product constant of strontium carbonate with that of strontium fluoride: $$\dfrac{{{K_{sp}}\left( {{\text{SrC}}{{\text{O}}_{\text{3}}}} \right)}}{{{K_{sp}}\left( {{\text{Sr}}{{\text{F}}_{\text{2}}}} \right)}} = \dfrac{{\left[ {{\text{S}}{{\text{r}}^{2 + }}} \right] \times \left[ {{\text{CO}}_3^{2 - }} \right]}}{{\left[ {{\text{S}}{{\text{r}}^{2 + }}} \right] \times {{\left[ {{{\text{F}}^ - }} \right]}^2}}} \\\ \dfrac{{{K_{sp}}\left( {{\text{SrC}}{{\text{O}}_{\text{3}}}} \right)}}{{{K_{sp}}\left( {{\text{Sr}}{{\text{F}}_{\text{2}}}} \right)}} = \dfrac{{\left[ {{\text{CO}}_3^{2 - }} \right]}}{{{{\left[ {{{\text{F}}^ - }} \right]}^2}}} \\\ \left[ {{{\text{F}}^ - }} \right]^2 = \sqrt {\dfrac{{{K_{sp}}\left( {{\text{Sr}}{{\text{F}}_{\text{2}}}} \right)}}{{{K_{sp}}\left( {{\text{SrC}}{{\text{O}}_{\text{3}}}} \right)}} \times \left[ {{\text{CO}}_3^{2 - }} \right]} \\\ $$ Substitute values in the above expression: $$\left[ {{{\text{F}}^ - }} \right] = \sqrt {\dfrac{{{K_{sp}}\left( {{\text{Sr}}{{\text{F}}_{\text{2}}}} \right)}}{{{K_{sp}}\left( {{\text{SrC}}{{\text{O}}_{\text{3}}}} \right)}} \times \left[ {{\text{CO}}_3^{2 - }} \right]} \\\ = \sqrt {\dfrac{{7.9 \times {{10}^{ - 10}}}}{{7.0 \times {{10}^{ - 10}}}} \times 1.2 \times {{10}^{ - 3}}M} \\\ = \sqrt {1.35 \times {{10}^{ - 3}}M} \\\ = 3.7 \times {10^{ - 2}}M \\\\$$ The concentration of fluoride ion in the solution would be $$3.7 \times {10^{ - 2}}M$$ _**Hence, the correct option is the option C ) $$3.7 \times {10^{ - 2}}M$$.**_ **Note:** In the solubility product constant expression for the strontium difluoride, do not forget to raise the concentration of fluorine to the power 2. This is because, when one molecule of strontium fluoride dissociates, two fluoride ions are obtained. If the concentration of fluorine is not raised to the power 2, an error will be introduced in the calculation and the wrong value of fluoride ion concentration will be obtained.