2) Use the Kspa expressions for CuS and ZnS to calculate the pH where you might be able to precipitate as much Cu2 as possible w

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2) Use the Kspa expressions for CuS and ZnS to calculate the pH where you might be able to precipitate as much Cu2 as possible w

hile leaving the Zn2 in solution, and find what concentration of copper would be left. Assume the initial concentration of both ions is 0.075M.

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1 answer:

Phantasy [73] · Answer 1 · 2023-04-10T17:00:05+03:00

The pH where you might be able to precipitate as much Cu2 as possible will be any value that is less than 5.85.

<h3>How to illustrate the information?</h3>

The solubility product of salt is defined as the product of molar concentrations of ions in a saturated solution of it.

An ion will be precipitated if the ionic product exceeds the solubility product of the cation. In the case of Cu and Zn, the solubility product is greater for Zinc than Copper.

Also Cu will be precipitated in an acidic medium while Zn is precipitated in an alkaline medium. We have for CuS, K= S2 so S = 2.4495*10-8. For ZnS, S = 0.1414. So the lower the pH higher will be the amount of CuS precipitated.

If we are starting with 0.075M CuS and ZnS solution it will have 0.075 moles each of Cu2+, Zn2+, and S2. Since the anion concentration can be equated to POH, POH for Cus and Zns are 14.096 and 5.85.

The maximum precipitation will take place is any value which is less than 5.85.

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