Analytical Chemistry PHCMp 101

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

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 (0.2-0.3 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 (0.2-0.3 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

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

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).

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 NH4 + + Cl-

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.

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

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.