1. Separation and Identification of the Group II Hydroxides in an Unknown
Qual II
Separation and Identification of the Group II Hydroxides in an Unknown

2. Goals
To use the qual scheme provided to ____________________________ the ions Zn+2, Al+3, Cu+2, Ni+2, Co+2, Fe+3, and Mn+2 from each other so their presence can be verified

3. The standard qual.
The standard qual analysis scheme for cations begins by separating ions into groups (cation groups) based on __________ of their _________, _________, __________, and _______________.
Each group, which consists of a small number of cations, is then analyzed further.
Usually, additional reactions are carried out that _________ ions in the group from each other, and then _______________ tests are performed to _________________________________________.

4. Cation groups.
The initial separation into cation groups is accomplished in the following order:
Cation group I: Ag+, Hg22+, Pb2+ are ________ as their __________.
Cation group II: Cu2+, Hg2+, Pb2+, Sn4+, Sn2+, Bi3+, Sb3+, Cd2+, As3+ are __________ as ________ from their acidic solution.
All sulfides are black, except CdS (orange), As2S2 (yellow), SnS2 (yellow), Sb2S3 (orange).
!!! Abbreviated scheme: only Bi3+ and Sn4+ and no sulfur containing reagent that has a strong odor.

5. Cation groups cont’d.
Cation group III: Co2+, Ni2+, Mn2+, Fe2+, Fe3+, Al3+, Cr3+, Zn2+.
All ions in this group, except ______ and ____ are precipitated as s__________ from a slightly basic solution.
Under these conditions ____ and ____ precipitate as ______________.
!!! Abbreviated group: Al3+, Zn2+, Co2+, Ni2+, Fe3+, Mn2+ and Cu2+ (group II) are precipitated as ____________, except zinc and aluminum that form colored amphoteric solids

6. Groups cont’d.
Cation group IV: Ba2+, Sr2+, Ca2+ are precipitated as _________ from a slightly basic solution (all white precipitates).
Cation group V “soluble group”: Mg2+, NH4+, Na+, K+
All these separations are based on differences in _________ and formation of complex ions.

7. Qual approach
Sometimes the _______ of one precipitate obscures the ________ of the other.
For example, PbS is black and Sb2S3 is orange.
So if you have both present, you see ______, but no ____________.
To resolve such difficulties, the qual analysis of a sample is usually approached ___________.
First, reactions are carried out to ________ ions from each other.
Then, the tests are performed to confirm either the presence or absence of each ion.

8. LeChatelier’s principle in qual.
The separations demonstrate the concept of chemical equilibrium and Le’Chatelier’s principle, which states that if a stress is placed on a system, then the system shifts to minimize the effect of that stress.
The idea of shifting equilibrium one way or another is frequently used in qualitative analysis experiments.
Qualitative analysis separations usually involve one of three types of equilibrium:

9. Solubility Equilibrium:
AgCl (s) ↔ Ag+ (aq) + Cl- (aq)
The solubility of AgCl will be decreased by addition of Cl- in form of _____.
Cl- from the _____ is the same as Cl- from the _____
The AgCl solubility is shifted to the ____when ____ is added

10. Acid-base equilibrium
NH3(aq) + H+(aq) ↔ NH4+(aq)
Adding acid (H+) shifts equilibrium to the ______

11. Complexation Equilibrium
Ag+ (aq) + 2 NH3 (aq) → Ag(NH3)2+
If the amount of ammonia is decreased the equilibrium is shifted to the ______

12. All three equilibria:
All five substances are present: AgCl, Ag+, Cl-, H+, NH3
A single change effects all three equilibria:
If more acid is added (H+), the reaction NH3(aq) + H+(aq) ↔ NH4+(aq) is shifted to the _____, which decreases the concentration of _______.
This will decrease the amount of _____ available for reaction Ag+ (aq) + 2 NH3 (aq) → Ag(NH3)2+
Thus, will also increase the concentration of _____
The increase in -___ will shift the equilibrium in reaction AgCl (s) ↔ Ag+ (aq) + Cl- (aq) to the ____ forming more _______

13. To obtain correct results
___________________________ to insure that your sample is not contaminated with other ions
_________________________ to insure that one substance does not mask the test for another
For ppt.s, __________ are a critical step
Use proper ________________________________
Insure that the reagents are _______________ so that the reaction occurs throughout the test tube
Do not __________________ until you are finished!!!!
Remember: you have 2 weeks for this one, if you need it

14. S.II
Use half of unknown
Mix it up and get some of the _______, if any appears to be present
Add 3M NH3 (NH4OH) until you do not notice more solid (precipitate) forming
This step separates the ions that you are interested in from other _______ that may be present
Mn n OH-  M(OH)n ppt
(Mn represents any metal ion)

15. S.II.A Isolation of Zn & Al
S.II + 6d 3M NaOH
Zn(OH)2 + 2OH-  [Zn(OH)4]2-
Al(OH)3 + OH-  [Al(OH)4]1-
Both products are ions that allow these metals to be dissolved due to the _____________

16. T.II.A.1-3 Testing for Al S.II.A. supernate + 5d 6M HCl
[Zn(OH)4]2- + 4H+  Zn2+ + 4H2O
[Al(OH)4]1- + 4H+  Al3+ + 4H2O
T.II.A.1 + 1mL 6M NH3 (NH4OH)
Al3+ + 3NH4OH  Al(OH)3 + 3NH4+
gelatinous

17. T.II.A. 4&5 Testing for Zn
T.II.A.3 supernatant + 1d 6M HNO3 + 2d 0.05M Co(NO3)2 then E
Only add Na2S2O4 if there is a blue tint present
Blue is from ____________
2[Cu(NH3)4]2++S2O42-+2H2O2Cu1++2SO32-+4NH4++4NH4OH
2Cu+1+S2O42-+2H2O  2SO Cu0 + 4H+
[Zn(NH3)4]2++H++H2OH2ZnO2(stoich?)
H2ZnO2+Co(NO3)2 CoZnO2 +2HNO3
lime green

18. S.II.B. Isolation of Cu & Ni
S.II.A + 1mL 6M NH3
Cu(OH)2 + 4H2O  [Cu(H2O)4]2+ + 2OH-
Ni(OH)2 + 6H2O  [Ni(H2O)6]2+ + 2OH-
[Cu(H2O)4]2++4NH3  [Cu(NH3)4]2+ deep blue
[Ni(H2O)6]2++ 6NH3  [Ni(NH3)6] blue
Both products are ions that allow these metals to be ___________ due to the formation of the _______________________

19. T.II.B Testing for Ni &Cu
S.II.B.3 supernatant + 30d DMG
DMG is

20. Ni2+ + 2DMG-  Another coordination complex
The H-bonds help to stabilize this complex
This complex is red and will ppt out
The resulting supernatant will be blue (or purple) if Cu is present

21. S.II.C Isolation of Cobalt
S.II.B.3 ppt + 5d 6M HCl
Co(OH)2 + 6H2O  [Co(H2O)6]2++ 2OH-
[Co(H2O)6]2++ 2OH- Co(OH)2(s)+6H2O
Add H+ to get rid of this possibility (react with OH-)
Co(OH)2 + 4Cl- CoCl42- +2OH-
Co2+ + 6NH3  [Co(NH3)6]2+

22. T.II.C S.II.C.3 supernatant + 10d NH4SCN Co2+ + 4SCN-  [Co(SCN)4]2-
excess unstable in H2O
in methanol stable in alcohol

23. S.II.D Separation of Mn from Fe
S.II.C.3 ppt + 4d H2O + 3d H2O2 + 
Mn(OH)2+ in air  MnO2 + 2H2O
Redox: Mn2+  Mn4+
MnO2 + H2O2  MnO(s) + H2O + O2
Redox: Mn+4  Mn2+
2Fe(OH)3 + 3H2SO4  Fe2(SO4)3 + 12H2O
Fe2(SO4)3 is water soluble

24. T.II.D Testing for Iron S.II.D.3 supernatant + 10d KSCN
Fe3+ + 6SCN-  [Fe(SCN)6]3-

25. T.II.E Testing for Manganese
S.II.D.3 +20d 2M H2SO4 + 3d H2O2
MnO2 + H2O2  MnO(s) + H2O + O2
MnO + 2H+  Mn2+ + H2O
+ 10d H2O + NaBiO3 + 2d 3M HNO3
2e-+ 6H+ + BiO3-  Bi3+ + 3H2O
Do this so we have a non-interfering species to accept e-s
4H2O + Mn2+ MnO4-+ 8H++5e-