Question

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the concentration of Ca 2 + ( aq ) when Ag 2 SO 4 ( s ) begins to precipitate?

Answer 1

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M $Ca^{2+}(aq)$ and 0.0390 M $Ag^{+}(aq)$. What will be the concentration of $Ca^{2+}$(aq) when $Ag_{2}SO_{4}(s)$ begins to precipitate? What percentage of the $Ca^{2+}(aq)$ can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      $Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}$

$[Ag^{+}]$ = 0.0390 M

When $Ag_{2}SO_{4}$ precipitates then expression for $K_{sp}$ will be as follows.

         $K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]$

        $1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]$

       $[SO^{2-}_{4}]$ = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

         $CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}$

      $K_{sp} = [Ca^{2+}][SO^{2-}_{4}]$

     $4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788$

           $[Ca^{2+}]$ = 0.00625 M

Now, we will calculate the percentage of $Ca^{2+}$ remaining in the solution as follows.

               $\frac{0.00625}{0.05} \times 100$

                 = 12.5%

And, the percentage of $Ca^{2+}$ that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of $Ca^{2+}(aq)$ when $Ag_{2}SO_{4}(s)$ begins to precipitate.