Question

In: Chemistry

2(a). Describe how adding a solution of KOH to the original sulfide precipitate results in the...

2(a). Describe how adding a solution of KOH to the original sulfide precipitate results in the separation of Sn4+ ion from the other Group II cations.

2(b). Why wouldn’t a weak base, such as aqueous NH3 work just as well as aqueous KOH in (a)?

Solutions

Expert Solution

2(a).

Sn4+ hydroxide, which is Sn(OH)4, is extremely insoluble, just as the sulfide is (Ksp = 1.0 x10-57 and 1.0 x10-70, respectively). However, the hydroxides of the group II cations are relatively soluble, much more soluble than the corresponding sulfides. Thus, adding KOH to the solution will dissolve these other metal ions back into solution, leaving only the Sn4+ precipitate, which you could then just filter or centrifuge out.

(b)

Ammonia can't be used because it forms a complex with all of the cations mentioned. Thus, it would dissolve everything, and you would have no precipitate. It has nothing to do with the fact that NH3 is a weaker base, it has to do with the fact that ammonia forms these complexes, in other words, it is a good chelating agent and isn't very selective.

Even can't use other bases, since they aren't more basic than the conjugate base of hydrogen sulfide. Hydrogen sulfide is a very weak acid, thus the conjugate base (S2-) is a pretty strong base. So, you have to use a base that is more basic than the sulfide, which leaves you with very few choices, especially in aqueous solution.


Related Solutions

2(a). Describe how adding a solution of KOH to the original sulfide precipitate results in the...
2(a). Describe how adding a solution of KOH to the original sulfide precipitate results in the separation of Sn4+ ion from the other Group II cations. 2(b). Why wouldn’t a weak base, such as aqueous NH3 work just as well as aqueous KOH in (a)?
will a precipitate form when Solutions of Ba(NO3)2 and KOH are mixed
will a precipitate form when Solutions of Ba(NO3)2 and KOH are mixed
1. Calculate [OH−] for a solution formed by adding 5.00 mL of 0.120 M KOH to...
1. Calculate [OH−] for a solution formed by adding 5.00 mL of 0.120 M KOH to 20.0 mL of 7.1×10−2 M Ca(OH)2. 2. Calculate pH for a solution formed by adding 5.00 mL of 0.120 M KOH to 20.0 mL of 7.1×10−2 M Ca(OH)2. 3. Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine...
When ammonia is added to a solution that has Cu(OH)2(s) precipitate in it, the precipitate dissolves....
When ammonia is added to a solution that has Cu(OH)2(s) precipitate in it, the precipitate dissolves. Use Le Châtelier’s principle to explain this.
What is the pH of the solution that results from adding 27.7 mL of 0.12 M...
What is the pH of the solution that results from adding 27.7 mL of 0.12 M HCl to 25.9 mL of 0.41 M NH3 ? (Ka for ammonium ion is 5.6 x 10-10 .) pH =
What is the pH of the solution that results from adding 26.2 mL of 0.14 M...
What is the pH of the solution that results from adding 26.2 mL of 0.14 M to 25.9 mL of 0.41 MNH3? (Ka for ammonium ion is 5.6*10^-10). pH-------------------?
When chlorine is bubbled through an ethanol solution of [NEt4]I a bright yellow precipitate results. Elemental...
When chlorine is bubbled through an ethanol solution of [NEt4]I a bright yellow precipitate results. Elemental analysis of the product reveals that it contains 3.51% nitrogen. Use this information to elucidate the product of this reaction. Write a balanced equation for the reaction and use VSEPR theory to predict the structure of the solid. What is the oxidation state of iodine at the end of the reaction?
1.) Why must the solution used to wash the precipitate be slightly acidified? 2.) If a...
1.) Why must the solution used to wash the precipitate be slightly acidified? 2.) If a student did not test the precipitate for presence of copper and some was included with the AgCl, will the final result (Ag %) be higher or lower than the true value? Why? 3.) If the precipitate was not dried completely before final weighing, will the final result (Ag %) be higher or lower than the true value? Why? 4.)Assuming the final concentration of chloride...
When a solution of Pb(NO3)2 and NaF are mixed, a precipitate is formed. The net ionic...
When a solution of Pb(NO3)2 and NaF are mixed, a precipitate is formed. The net ionic equation is: Pb+2(aq) + 2F-(aq) --> PbF2(s) and the Ksp for this compound is 3.3 x 10-8. What is the concentration of the ions remaining in solution? To solve, first do the stoichiometry of the reaction to determine the limiting reactant. Then, any excess reactant is treated as a common ion when the net ionic reaction is reversed and treated as a Ksp problem....
#3 R&D Exp. 26 Should Cu(OH)2 precipitate in 3 M NH3 if the original [Cu2+] is...
#3 R&D Exp. 26 Should Cu(OH)2 precipitate in 3 M NH3 if the original [Cu2+] is 0.1 M ( the Kb for ammonia is 1.6x10-5)? Please show work and explain a little of how you got this result
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT