Question

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?

Answer 1

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

        $K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        $AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

     $9.84 \times 10^{-21} = s \times s$

             s = $9.92 \times 10^{-11}$

Also,     $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

                $2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

                            s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

         $K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

          $2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

       $2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

       $[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

                             = $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             $AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

           $K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

       $9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

                $[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.