1. Analytical Chemistry PHCMp 101
Lab 3
Identification of
Group III Cations
Group III Cations

2. І ΙΙ ΙΙΙ ΙV V VІ Group reagent Ions Formulae of precipitates
Grp
Group reagent
Ions
Formulae of precipitates
Distinguishing features І
Cold dil HCl
Ag+, Pb++,Hg2++
AgCl, PbCl2, Hg2Cl2
Chlorides insoluble in cold dil HCl ΙΙ
H2S in presence of HCl ( M)
Subgroup IIA Cu++,Cd++,Hg++,Bi++,
Pb++
Subgroup IIB
Sn++, Sn4+, Sb3+, As3+, As5+
CuS, CdS, PbS, HgS, Bi2S3, SnS, SnS2, Sb2S3,
Sb2S5,
Precipitated as Sulfides in acidic medium ( M HCl)
ΙΙΙ
NH4OH in Presence of NH4Cl
Al3+, Cr3+, Fe3+
Al(OH)3, Cr(OH)3,
Fe(OH)3
Precipitated as Hydroxides by NH4OH in presence of excess NH4Cl ΙV
H2S in Presence of NH4OH
Ni++, Co++, Mn++, Zn++
NiS, CoS, MnS, ZnS
Precipitated as Sulfides by H2S from ammoniacal solution in presence of NH4Cl V
(NH)2CO3 in presence of NH4OH &NH4Cl
Ba++, Ca++, Sr++
BaCO3, SrCO3,
CaCO3
Precipitated as Carbonates in alkaline medium by NH4OH in presence of NH4Cl VІ
No particular reagent
Na+, K+, NH+4 , Mg++
Different precipitated forms
Ions not precipitated in previous groups

3. Group III Fe3+ / Al3+ / Cr3+ Ferric / Aluminum / Chromium
Group reagent: NH4OH, forming insoluble hydroxide salts.
Fe+3+ 3 NH4OH  Fe(OH)3(s) Reddish brown ppt
Al+3+ 3 NH4OH  Al(OH)3(s) White gelatinous ppt
Cr+3+ 3 NH4OH  Cr(OH)3(s) Grey green ppt

4. The concentration of the [OH-] should be kept low. WHY?
To be sure that only the hydroxides of group III are precipitated (they have the lowest Ksp among all other cations and so they need less OH- to be precipitated).

5. HOW is the concentration of [OH-] kept LOW?
by the addition of NH4Cl common ion effect  shifts reaction backwards  decreasing [OH-]  precipitation of group III cations only.
NH4OH NH4 + + OH-
NH4Cl NH Cl-

6. After adding excess conc. NH4OH, we boil the test tube. WHY?
Oxidation of Fe+2 if
present to Fe+3 by boiling
with few drops of conc.
HNO3 (Fe+2 will not be
precipitated under the
conditions of precipitation
of group III).
2. Cr(OH)3 is slightly
soluble in excess NH4OH
in the cold forming a
violet soluble amine
complex. So boiling is
essential to expel excess
ammonia thus
decomposing that
complex.

7. Indicating Fe+3 Indicating Cr+3 Indicating Al+3
2 ml sample+ xss NH4Cl + xss conc. NH4OH
Boiling
Reddish brown ppt
White gelatinous ppt
Grey green ppt
New portion from test solution
New portion from test solution
Xss 30% NaOH
1 ml H2O2
Boil
NH4SCN
K4Fe(CN)6
Colorless solution
Yellow solution
Blood red color
Fe(SCN)3
Prussian
(intense)blue
Fe4 [Fe(CN)6]3
CH3COOH
Pb(CH3COO)2
Dil.HCl
White gelatinous ppt
Al(OH)3
Indicating Fe+3
Yellow ppt PbCrO4
Indicating Cr+3
Indicating Al+3

8. In the test for iron using SCN-
The reaction must be made on cold as the red color is decomposed on hot.
Reducing agents e.g. SnCl2 reduces Fe3+ into Fe2+, so bleach the color.
F- & PO43- form stable complexes with Fe3+, so prevent the formation of the red color Fe [(SCN)6] 3- .
HgCl2 forms with SCN- stable non ionizable Hg(SCN)2, so prevent the color formation.