In a qualitative analysis procedure, a chemist adds HCI to an unknown group of i

Question

In a qualitative analysis procedure, a chemist adds HCI to an unknown group of ions and then saturates the solution with H2S so that the solution is 0.4 M in HCl and 0.10 M in H2S. (a) What is the [HS 40 2E-8 of the solution? (Note: The answer should be rounded to one significant figure. Be careful, 3.5 rounds up to 4, but 4.5 rounds down to 4!) (b) If o.01 M of each of the following ions is in the solution, which wll form a precipitate? (Select all that apply.) Mn2+ Ni2 + Fe2+ 0 Ag Cu2+ Pb2+ Hint: First, calculate [HS'] produced by the dissociation of H2S from the given concentrations of HCl and H2S. Second, calculate [OH] from the given concentration of HCI. Ther

Explanation / Answer

Ag+ Hg2+ Pb2+ Cu2+  

Group I (Ag+, Pb2+, Hg2+) cations produce insoluble chlorides so they can be precipitated with dilute HCl, while all other cations remain in solution.

Group II (Cu2+, Bi3+, Cd2+, Hg2+, As3+, Sb3+, Sn4+) cations produce very insoluble sulfides (Ksp values less than 10-30) so they can be precipitated by low amounts of sulfide ion; this can be achieved by adding an acidic solution of H2S.

Group III (Al3+, Cr3+, Fe3+, Zn2+, Ni2+, Co2+, Mn2+) cations produce slightly soluble sulfides (Ksp values more than 10-20) so they can be precipitated by relatively high amounts of sulfide ion; this can be achieved by adding a basic solution of H2S.

Group IV (Mg2+, Ca2+, Sr2+, Ba2+) cations, as well as all of the above groups, produce insoluble carbonates so they can be precipitated by the addition of carbonate once the ions of the first three groups have been removed.

Group V (Na+, K+, NH4 +) cations do not precipitate with any of the above reagents

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