Question

You have a solution that contains 0.1 M Ag+ , 0.1 M Cu2+ and 0.1 M Ba2+. You would like to separate these ions by precipitation. For quality purposes, the relative error must be below 0.1%, i.e., the precipitation of each ion must be complete (>99.9%) and there must not be more than 0.1% (by weight) of one of the other ions in the precipitate. Explain your answer: nature of precipitate, pH, concentrations etc. Be quantitative in your answer. Note: at the end, you want one vial with >99.9% of the silver, one with >99.9% of the copper, and one with >99.9% of the barium.

Please show the math for each cation

Answer 1

The following scheme of separating desired ions should be helpful to get the ions present in a solution containing Ag^+ , Cu^2+, & Ba^2+ -

Treat the given solution with excess of 0.2M HCl solution

Result - Ag^+ would get precipitated while the other two ions will remain in solution as CuCl2 , and Ba Cl2

Note that excess of HCl solution would ensure complete precipitation of Ag Cl which has to be filtered, washed bly with water & dried carefully.

The concentration of 0.2M HCl ( pH = 0.69 ) to be used is based upon the stoichiometry of the reactions

Ag^+ ( aq ) + Cl^- ( aq ) ------------------> AgCl ( precipitate )

1mole / L----------- 1 mole / L ---------------------- complete precipitation

Cu^2+ + 2Cl^- (aq ) ---------------> Cu Cl2 (aq)

1 mole / L ---- 2 moles / Litre ---------------------- 1 mole / L

Ba ^2+ ( aq ) + 2Cl^- (aq ) --------------> Ba Cl2 (aq)

1 mole / l --------- 2 mole / litre ------------------ 1mole / litre

Step - 2 / The filtrate now contains Ba Cl2 & CuSO4 with Ba^2+ and Cu^2+ ions.

Treat the filtrate with 0.1M - H2SO4 solution (excess).[ pH = 0.69 ] This will separate Ba^2+ ions as BaSO4 . Filter the solution to separate Ba^2+ . Wash this precipitated BaSO4 with distilled water weakly acidulated with a few drops of H2SO4, Dry the precipitate , thereby separating Ba^2+ as BaSO4.

Preserve the filtrate

Reactions - BaCl2 (aq ) + H2SO4 (aq ) ---------------------> BaSO4 ( precipitate )

CuCl2 (aq ) + H2SO4 (aq) ------------------ --> CuSO4 ( aq )

Step 3 / The above filtrate ( from step 2 ) contains CuSO4 only. This solution on careful evaporation and subsequent crystallization would yield CuSO4..H2 O. This hydrated copper sulphate on careful dehydration would result to give CuSO4 (anhydrous ) only containing Cu^2+ ion which was present in the original solution.

Alternatively:

The filtrate can be treated with dilute NaOH solution ( excess ) gives a precipitate of Cu (OH)2 which on filtration , drying and subsequent heating yields CuO containing the Cu^2+ ions present in the original mixture of solutions.

CuSO4 + 2NaOH ------------------> Cu( OH )2 + Na2SO4

Cu(OH)2 ----------------heat ------------------------------> CuO + H2O

* Note that using excess of the precipitating reagent shifts the reaction to RHS causing maximum yield of the products or complete separation of the desired ions as a precipitate.

* Maths / Calculations involved -

The stoichiometry of precipitation reactions (as shown above ) & K_sp values of the precipitate forms the basis of their separation . K_sp AgCl = 1.755 x 10^-10  , K_sp Ba SO4 = !.1 x 10^-10  )., K_sp  Cu(OH)2 = 1.6 x !0^-19  The given concentrations of each cations in the solution mixture is 0.1M ,hence the concentration of respective anions should be so adjusted that the product of ionic concentrations exceeds their solubility product. Further , since their is a possibility of backward reaction an excess of reagents providing anion has to be used.. This would shift the precipitation reaction to RHS ensuring complete precipitation (ie 99.9 % ) of cation along with the desired concentrations of the precipitating reagents as per above scheme of separation.

However, the % error with respect to purity of the cation separated can be estimated through gravimetric analysis of the precipitates separated.