A solution is 1.5x10^-5 M in Ag+ and 2.5x10^-3 in Cu+ ions. you want to selectiv

Question

A solution is 1.5x10^-5 M in Ag+ and 2.5x10^-3 in Cu+ ions. you want to selectively remove the silver from the solution since it is more valuable and decide to use NaCl as your ion source. Answer the questions below to see if adding NaCl would be more efficient in removing silver from solution.

Ksp(AgCl)= 1.77x10^-10
Ksp (CuCl)= 1.72x10-7

A) what is the minimum chloride concentrationto start the precipitation of Ag+?

B) Calculate the chloride concentration required to precipitate Cu+?

C) How much Ag+ remains when Cu+ starts to saturate the solution?

D) determine the seperation efficiency (percentage of Ag+ seperated)

Explanation / Answer

A) for Ag+ precipitation

Ksp(AgCl)=[Ag+][Cl-]

Putting values of Ag+ concentration and Ksp

We get [Cl-]=1.18×10^-5

B) for precipitation of Cu+

[Cu+][Cl-]=Ksp(CuCl)

Putting values of Cu+ concentration and Ksp

We get [Cl-]=6.88×10^-5

C) when Cu+ start precipitating

[Cl-]=6.88×10^-5

Ksp=[Ag+][Cl-]

We get [Ag+]=2.57×10^-6

D) % Ag+ separated=(1.5×10^-5 -2.57×10^-6)/1.5×10^-5

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