Question

A solution contains \[{\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 6}}}}\,{\text{M}}\]\[{\text{Sr(N}}{{\text{O}}_3}{)_2}\] and \[{\text{5}}{.0 \times 1}{{\text{0}}^{{\text{ - 7}}}}\,{\text{M}}\]\[{{\text{K}}_3}{\text{P}}{{\text{O}}_4}\]. Will \[{\text{S}}{{\text{r}}_3}{{\text{(P}}{{\text{O}}_4})_2}({\text{s}})\] precipitate?[\[{K_{sp}}\] for \[{\text{S}}{{\text{r}}_3}{{\text{(P}}{{\text{O}}_4})_2} = {\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 31}}}}\]].

Answer 1

Hint: Solubility product is used to determine the solubility of the compound. The higher the solubility product higher will be the solubility of the compound. It is the product of the concentration of the ions of the compound with each ion concentration raised to the power equal to its stoichiometric coefficient.


Complete solution:
Strontium nitrate is dissociated as follows:
\[Sr{(N{O_3})_2} \to S{r^{ + 2}}\, + \,2NO_3^ - \]
Here, we can see that the stoichiometric ratio between the strontium nitrate and the strontium is 1:1indicates the concentration of the strontium ion is the same as the concentration of strontium nitrate.
\[{\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 6}}}}\,{\text{M}}\,\,Sr{(N{O_3})_2} = {\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 6}}}}\,{\text{M}}\,\,S{r^{ + 2}}\]
Potassium phosphate is dissociated as follows:
\[{{\text{K}}_3}{\text{P}}{{\text{O}}_4} \to 3{{\text{K}}^ + }\, + PO_4^{3 - }\]
Here, we can see that the stoichiometric ratio between the potassium phosphate and the phosphate ion is 1:1 it indicates the concentration of the phosphate ion is the same as the concentration of potassium phosphate.
\[{\text{5}}{.0 \times 1}{{\text{0}}^{{\text{ - 7}}}}\,{\text{M}}\,\,{{\text{K}}_3}{\text{P}}{{\text{O}}_4} = {\text{5}}{.0 \times 1}{{\text{0}}^{{\text{ - 7}}}}\,{\text{M}}\,\,PO_4^ - \]
Now, here the solubility product of the strontium phosphate is given. To determine whether it is precipitate or not we have to know its reaction quotient first. If the reaction quotient is greater than the solubility product then salt gets precipitated.
The dissociation reaction of the strontium phosphate is as follows:
\[{\text{S}}{{\text{r}}_3}{{\text{(P}}{{\text{O}}_4})_2} \to 3S{r^{ + 2}}\, + \,2PO_4^{3 - }\]
The reaction quotient for the reaction is as follows:
\[{{\text{Q}}_c}{\text{ = }}\dfrac{{{{\left[ {S{r^{ + 2}}} \right]}^3}{{\left[ {PO_4^{3 - }} \right]}^2}}}{{\left[ {{\text{S}}{{\text{r}}_3}{{{\text{(P}}{{\text{O}}_4})}_2}} \right]}}\]
Here, we can see that in denominator solid strontium nitrate is present whose concentration is unity, therefore the equation for reaction quotient is as follows:
\[{{\text{Q}}_c}{\text{ = }}{\left[ {S{r^{ + 2}}} \right]^3}{\left[ {PO_4^{3 - }} \right]^2}\]
Now, substitute \[{\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 6}}}}\,{\text{M}}\] for the concentration of strontium ion and \[{\text{5}}{.0 \times 1}{{\text{0}}^{{\text{ - 7}}}}\,{\text{M}}\] for the concentration of phosphate ion.
\[{{\text{Q}}_c}{\text{ = }}{\left[ {S{r^{ + 2}}} \right]^3}{\left[ {PO_4^{3 - }} \right]^2}\]
\[{{\text{Q}}_c}{\text{ = }}{\left[ {{\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 6}}}}} \right]^3}{\left[ {{\text{5}}{.0 \times 1}{{\text{0}}^{{\text{ - 7}}}}} \right]^2}\]
\[{{\text{Q}}_c}{\text{ = 2}}{.5 \times 1}{{\text{0}}^{{\text{ - 31}}}}\]
Now, the solubility product for the strontium phosphate is given as \[{\text{1}}{.0 \times 1}{{\text{0}}^{{\text{ - 31}}}}\] and the reaction quotient obtained is \[{\text{2}}{.5 \times 1}{{\text{0}}^{{\text{ - 31}}}}\].
\[{{\text{Q}}_c}{\text{ > }}{{\text{K}}_{sp}}\]

Thus, we can see that the given salt satisfies the condition of the precipitation therefore, strontium salt will precipitate.


Note:the important relationships among the solubility product and the reaction quotient as follows:
\[{{\text{Q}}_c}{\text{ < }}{{\text{K}}_{sp}}\] (i)
If this is the condition compound is more soluble and there is no precipitation.
\[{{\text{Q}}_c}{\text{ > }}{{\text{K}}_{sp}}\] (ii)
If this is the condition then precipitation takes place.