You are given a solution that is 0.032 in Cu 2+ and 0.032 in Ca 2+ that you wish

Question

You are given a solution that is 0.032 in Cu2+ and 0.032 in Ca2+ that you wish to separate using selective ion precipitation with OH- ions. The Ksp value for Cu(OH)2 = 4.80 X 10-20 and the Ksp value for Ca(OH)2 = 5.02 X 10-6. The two metal hydroxides used to set the pH threshold for the separation is Ca(OH)2

b) What is the threshold pH, below which you get precipitation of the least soluble metal hydroxide while the most soluble one remains in solution?



c) Assume you are at the pH threshold that you calculated in part b. What is the molar concentration of Cu2+ cation that remains in solution at this pH?

Explanation / Answer

Selective precipitation of Cu2+ from Ca2+

b) [OH-] for Cu(OH)2 = sq.rt.(Ksp/[Cu2+])

                                  = sq.rt.(4.8 x 10^-20/0.032) = 1.225 x 10^-9 M

pOH = -log[OH-] = -log(1.225 x 10^-9) = 8.912

pH = 14 - pOH = 5.09

So below pH 5.1 we would get only Cu(OH)2 precipitation.

c) [OH-] concentration at which Ca(OH)2 precipitates = sq.rt.(Ksp/[Ca2+]

                                    = sq.rt.(5.02 x 10^-6/0.032) = 0.0125 M

so the amount of [Cu2+] remained in solution at this pH = 4.8 x 10^-20/0.0125^2 = 3.072 x 10^-16 M

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