Question 19.12: Separating Ions by Selective Precipitation Problem A solutio......

Plan Because both compounds have a 1/2 ratio of cation/anion (see the discussion on page 868), when we compare their K_{sp} values, we find that Mg(OH)_2 is about 1010 times more soluble than Cu(OH)_2; thus, Cu(OH)_2 precipitates first. We write the dissolution equations and ion-product expressions. We are given both cation concentrations, so we solve for the [OH^−] that gives a saturated solution of Mg(OH)_2, because this [OH^−] will precipitate the most Cu^{2+}. Then, we calculate the [Cu^{2+}] remaining to see if the separation was accomplished.

Solution Writing the equations and ion-product expressions:
          Mg(OH)_2(s)  \xrightleftharpoons[]{}  Mg^{2+}(aq)  +  2OH^−(aq)           K_{sp}  =  [Mg^{2+}][OH^−]^2  =  6.3×10^{−10}
          Cu(OH)_2(s)  \xrightleftharpoons[]{}  Cu^{2+}(aq)  +  2OH^−(aq)           K_{sp}  =  [Cu^{2+}][OH^−]^2  =  2.2×10^{−20}
Calculating the [OH^−] that gives a saturated Mg(OH)_2 solution:

          [OH^−]  =  \sqrt{\frac{K_{sp}}{[Mg^{2+}]}}  = \sqrt{\frac{6.3×10^{−10}}{0.20}}  =  5.6×10^{−5}  M
This is the maximum [OH^−] that will not precipitate Mg^{2+} ion.
Calculating the [Cu^{2+}] remaining in the solution with this [OH^−]:

          [Cu^{2+}]  =  \frac{K_{sp}}{[OH^−]^2}  =  \frac{2.2×10^{−20}}{(5.6×10^{−5})^2}  =  7.0×10^{−12}  M
Since the initial [Cu^{2+}] is 0.10 M, virtually all the Cu^{2+} ion is precipitated.

Check Rounding, we find that [OH^−] seems right: ∼  \sqrt{(6×10^{−10})/0.2}  =  5×10^{−5}. The [Cu^{2+}] remaining also seems correct: (200×10^{−22})/(5×10^{−5})^2  =  8×10^{−12}.