3 A solution is 0.0010 M in each of the metal cations Fe and Hg 2. a. Enough thi

Question

3 A solution is 0.0010 M in each of the metal cations Fe and Hg 2. a. Enough thioacetamide and Ha are added to give the following equilibrium concentrations: IH,Sleg m 0.1 M and[H,0 M. Calculate the concentrations of Fe' and Hg", which remain in solution under these conditions. Under these conditions, h solution by precipitation? Explain. 4, what conclusion could be drawn if a Group unknown is blue in color? Explain. 5. What would happen in a general cation unknown if Group I metal cations were not removed before beginning Group II? 6. In adjusting the just prior to precipitation of the Group metal cations (Pb'2,Hg'a, Bi", and i), a precipitate will sometimes form upon addition of NH, (aq). Explain. 610 p ve Ana I trpe rimes 9 alysis

Explanation / Answer

3. Solution with 0.01 M Fe2+ and 0.01 M Hg2+

a. concentraton of cations after thioacetamide and HCl is added

a. Ksp of HgS is much lower than Ksp for FeS. therefore, HgS would be precipitated first from the solution while FeS remained in soluble state at this thioacetamide concentration,

[S^2-] needed to precipitate Fe2+ = Ksp/[Fe2+]

                                                      = 6 x 10^-2/0.01 = 6 M

since only 0.1 M S^2- is present, amount Fe2+ remained = 6 x 10^-2/(6 - 0.1) = 0.01 M

So practically all of Fe2+ remained unprecipitated at this stage.

[S^2-] needed to precipitate HgS = Ksp/[Hg2+]

                                                    = 2 x 10^-12/0.01 = 2 x 10^-10 M

Therefore, all of HgS would have been precipitated under these conditions.

b. Under these conditions all of Hg2+ has been precipitated but not all of Fe2+ has been precipitated from the solution.

4. Blue color of Group II solution indicates presence of Cu2+ ion in solution. Cu2+ in NH3 forms a strong blue color due to complex of ammonia formed in solution. this confirms Cu2+ is present.

5. It is important to remove all of Group I cations from solution prior to testing for other group cations in solution. This is done as Group I metal cations may interfere with the analysis and give false positive test in many cases with the tests performed for other metal ions in solution.

6. A precipitate (light color) of sulfide may appear prior to the analysis of Group II cations while adjusting the pH of the solution by addition of NH3.

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.