Analisis Kation Oleh : Heri Satria, M.Si

The general procedure for separating ions in qualitative analysis Figure 19.16 The general procedure for separating ions in qualitative analysis Add precipitating ion Add precipitating ion Centrifuge Centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A qualitative analysis scheme for separating cations into five ion groups Acidify to pH 0.5; add H2S Centrifuge Add NH3/NH4+ buffer(pH 8) Centrifuge Add (NH4)2HPO4 Centrifuge Add 6M HCl Centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

A qualitative analysis scheme for separating cations into five ion groups Figure 19.17 Acidify to pH 0.5; add H2S Centrifuge Add NH3/NH4+ buffer(pH 8) Centrifuge Add (NH4)2HPO4 Centrifuge Add 6M HCl Centrifuge Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Using pH and complexation to Separate Ions For Qualitative Analysis

Hot water AgCl(s); Hg2Cl2(s) Pb2+ ppt

Hot water AgCl(s); Hg2Cl2(s) Pb2+ NH3 HgNH2Cl(s) ppt Ag(NH3)6+

Step 5 Dissolve in HCl and add KSCN Figure 19.18 A qualitative analysis scheme for Ag+,Al3+,Cu2+, and Fe3+ Step 1 Add NH3(aq) Centrifuge Centrifuge Step 2 Add HCl Step 3 Add NaOH Centrifuge Step 4 Add HCl, Na2HPO4 Step 5 Dissolve in HCl and add KSCN Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Background Valdosta State University

Background For this experiment, the group III ions are Fe3+, Ni2+, Mn2+, Al3+ and Zn2+. These ions initially precipitate as either metal sulfides (in an alkaline environment) or metal hydroxides. This requires the chemist to generate a small quantity of sulfide ion to precipitate the metals. A convenient source of S2- is thioacetamide, which decomposes when heated to give hydrogen sulfide (H2S) which yields S2- in chemical reactions. A reagent that is made and consumed in the same flask is said to be produced in situ. Valdosta State University

Background – Hydrogen Sulfide Thioacetamide H2S(aq) + 2 H2O(l) D 2 H3O+(aq) + S2-(aq) The addition of base to the second reaction consumes the hydronium ion and drives the reaction to the right, increasing the concentration of S2-(aq). Valdosta State University

Background – Group III Separation Scheme HNO3 Valdosta State University

Background – Group III Separation Scheme Group III unknown A – Preparation of Group III cations NH3, H2S, Heat The group III ions are initially separated from the bulk solution by precipitation as either insoluble metal sulfides or hydroxides. NiS, FeS, MnS, Fe(OH)3, Al(OH)3, ZnS Group IV ions Ni2+(aq) + S2(aq) D NiS(s) (black) Fe2+(aq) + S2(aq) D FeS(s) (black) Zn2+(aq) + S2(aq) D ZnS(s) (white) Mn2+(aq) + S2(aq) D MnS(s) (pink) Al3+(aq) + 3 OH(aq) D Al(OH)3(s) (white, gel) Valdosta State University

Background – Group III Separation Scheme Group III unknown A – Preparation of Group III cations NH3, H2S, Heat Since iron has two common oxidation states, its chemistry in this step is more complex. If iron(III) is present it is reduced to iron(II) and elemental sulfur in produced. NiS, FeS, MnS, Fe(OH)3, Al(OH)3, ZnS Group IV ions 2 Fe3+(aq) + H2S (aq)  2 Fe2+(aq) + S(s) + 2 H+(aq) Valdosta State University

Background – Group III Separation Scheme Group III unknown A – Preparation of Group III cations NH3, H2S, Heat Alternately, the iron(III) can combine with the hydroxide ion and precipitate as iron(III) hydroxide. NiS, FeS, MnS, Fe(OH)3, Al(OH)3, ZnS Group IV ions Fe3+(aq) + 3 OH-(aq) D Fe(OH)3 (rust color) Valdosta State University

Background – Group III Separation Scheme B1 – Separation of Group III cations NiS, FeS, MnS, Fe(OH)3, Al(OH)3 Following the precipitation, the metal ions are combined with acid to form the free (and soluble) metal ions. HCl, HNO3, Heat Waste Ni2+, Fe3+, Mn2+, Zn2+, Al3+ 3NiS(s) +8H+(aq) + 2NO3(aq)  3Ni2+(aq) + 2NO(g) + 3S(s) + 4H2O(l) FeS(s) + 2 H+ (aq)  Fe2+(aq) + H2S(aq) 3Fe2+(aq) + 4H+(aq) + NO3(aq)  3Fe3+(aq) + NO(g) + 2H2O(l) MnS(s) + 2 H+(aq)  Mn2+(aq) + H2S(aq) ZnS(s) + 2 H+(aq)  Zn2+(aq) + H2S(aq) Al(OH)3(s) + 3 H+(aq)  Al3+(aq) + H2O(l) Valdosta State University

Background – Group III Separation Scheme B2 – Separation of Group III cations Ni2+, Fe3+, Mn2+, Zn2+, Al3+ Aluminum and zinc ions are amphoteric. This means that at high acid or base concentrations, these metals form soluble complexes, but precipitate at moderate pH. Iron, manganese and nickel form insoluble hydroxides at high pH. 6 M NaOH Fe(OH)3, Ni(OH)2, Mn(OH)2 Al(OH)4-, Zn(OH)42- Fe3+(aq) + 3 OH(aq) D Fe(OH)3(s) (rust-color) Ni2+(aq) + 2 OH(aq) D Ni(OH)2(s) (green) Mn2+(aq) + 2 OH(aq) D Mn(OH)2(s) (light brown) Valdosta State University

Background – Group III Separation Scheme B2 – Separation of Group III cations Ni2+, Fe3+, Mn2+, Zn2+, Al3+ Aluminum and zinc ions are amphoteric. This means that at high acid or base concentrations, these metals form soluble complexes, but precipitate at moderate pH. Iron, manganese and nickel form insoluble hydroxides at high pH. 6 M NaOH Fe(OH)3, Ni(OH)2, Mn(OH)2 Al(OH)4-, Zn(OH)42- Al3+(aq) + 3 OH(aq) D Al(OH)3(s) (white, gelatinous) Zn2+(aq) + 2 OH(aq) D Zn(OH)2(s) (white) Valdosta State University

Background – Group III Separation Scheme B2 – Separation of Group III cations Ni2+, Fe3+, Mn2+, Zn2+, Al3+ 6 M NaOH Fe(OH)3, Ni(OH)2, Mn(OH)2 Al(OH)4-, Zn(OH)42- Excess Acid Al(OH)3(s) + 3H+(aq) D Al3+ + 3 H2O(aq) Zn(OH)2(s) + 2H+(aq) D Zn2+ + 2 H2O(aq) Valdosta State University

Background – Group III Separation Scheme B2 – Separation of Group III cations Ni2+, Fe3+, Mn2+, Zn2+, Al3+ 6 M NaOH Fe(OH)3, Ni(OH)2, Mn(OH)2 Al(OH)4-, Zn(OH)42- Excess Base Al(OH)3(s) + OH(aq) D Al(OH)4-(aq) Zn(OH)2(s) + 2OH(aq) D Zn(OH)42-(aq) Valdosta State University

Background – Group III Separation Scheme C1 – Test for Mn2+, Fe3+, Ni2+ Fe(OH)3, Ni(OH)2, Mn(OH)2 The precipitate is redissolved by adding acid to the precipitate. The addition of nitric acid neutralizes the sodium hydroxide and regenerates the free cations. There is no easy method which will allow Mn2+, Fe3+ and Ni2+ to be separated; therefore, the sample is divided. HNO3, KNO2 Fe3+, Ni2+, Mn2+ Divide Sample Valdosta State University

Background – Group III Separation Scheme C2 – Test for Mn2+ Fe3+, Ni2+, Mn2+ If sodium bismuthate is added to a solution containing manganese(II), a redox reaction occurs resulting in the formation of the purple permanganate ion. Divide Sample NaBiO3 MnO4- purple 14H+(aq) + 2Mn2+(aq) + 5BiO3-(s)  2 MnO4-(aq) + 5Bi3+(aq) + 7H2O(l) Valdosta State University

Background – Group III Separation Scheme D1 – Separation of Fe3+ and Ni2+ Fe3+, Ni2+, Mn2+ The nickel and iron ions can be separated by the addition of ammonia. The increased pH causes the formation of the insoluble iron(III) hydroxide. The nickel ion combines with ammonia to form a soluble complex ion, hexaamminenickel(II). Divide Sample Conc. NH3 Fe(OH)3 Ni(NH3)62+ Fe3+(aq) + 3NH3(aq) + 3H2O(l)  3NH4+(aq) + Fe(OH)3(s) (brown) Ni2+(aq) + 6NH3(aq) D Ni(NH3)62+(aq) (blue) Valdosta State University

Background – Group III Separation Scheme D2 – Test for Fe3+ Fe3+, Ni2+, Mn2+ The presence of the iron(III) ion is confirmed by the addition of ammonium thiocyanate. If iron(III) is present, a blood red solution forms. Divide Sample Conc. NH3 Fe(OH)3 Ni(NH3)62+ HCl / NH4SCN Fe(SCN)63- blood red Fe3+(aq) + 6SCN-(aq) D Fe(SCN)63-(aq) blood red Valdosta State University

Background – Group III Separation Scheme E – Test for Ni2+ Fe3+, Ni2+, Mn2+ The presence of the nickel ion is confirmed by the addition of dimethylglyoxime. Dimethylglyoxime combines with the nickel ion to form a complex which forms a strawberry red precipiate. Divide Sample Conc. NH3 Fe(OH)3 Ni(NH3)62+ HCl / NH4SCN H2DMG Fe(SCN)63- blood red Ni(DMG)2 strawberry red ppt. Ni(NH3)62+(aq) + 2 HC4H7N2O2(aq)  4NH3(aq) + 2NH4+(aq) + Ni(C4H7N2O2)2(s) (red) Valdosta State University

Background – Group III Separation Scheme F1 – Separation of Al3+ and Zn2+ Al(OH)4-, Zn(OH)42- Careful control of pH allows for the separation of aluminum and zinc ions. The solution is made very slightly basic. At these conditions, the aluminum ion precipitates as aluminum hydroxide. The zinc ion remains in solution. HNO3 NH3 Al(OH)3 Zn(NH3)42+ Al3+(aq) + 3 NH3(aq) + 3 H2O(l) D 3 NH4+(aq) + Al(OH)3(s) Zn2+(aq) + 4 NH3(aq) D Zn(NH3)42+(aq) Valdosta State University

Background – Group III Separation Scheme F2 – Test for Al3+ A successful test for aluminum requires that the previous reactions and their pH control were properly performed. If not, false positive tests result. The test for aluminum requires the free aluminum ion to react with ammonia in the presence of a reagent called aluminon and form a red precipitate. Be careful, if there is iron or zinc left in the sample, a red precipitate will form resulting in a false positive. Al(OH)4-, Zn(OH)42- NH3 HNO3 Al(OH)3 Zn(NH3)42+ NH3, aluminon Al(OH)3 aluminon cherry red ppt. Al3+(aq) + 3 NH3(aq) + 3 H2O + aluminon(aq) D 3 NH4+(aq) + Al(OH)3aluminon(s) (red) Valdosta State University

Background – Group III Separation Scheme F2 – Test for Al3+ Al(OH)4-, Zn(OH)42- To confirm that the red precipitate is the aluminum complex, ammonium carbonate is added. If the red color does not fade, aluminum is present. NH3 HNO3 Al(OH)3 Zn(NH3)42+ NH3, aluminon Al(OH)3 aluminon cherry red ppt. Valdosta State University

Background – Group III Separation Scheme G – Test for Zn2+ Al(OH)4-, Zn(OH)42- To test for the zinc ion, a solution of potassium hexacyanoferrate(II) is added to the test solution. If zinc is present a white precipitate forms. The exact color of the precipitate can vary depending on the presence of other ions. If iron is present the color can change to yellow, green or blue. NH3 HNO3 Al(OH)3 Zn(NH3)42+ NH3, aluminon K4Fe(CN)6 Al(OH)3 aluminon cherry red ppt. K2Zn3[Fe(CN)6]2 white ppt. 3 Zn2+(aq) + 2 K+(aq) + 2 Fe(CN)64 (aq) D K2Zn3[Fe(CN)6]2 (s) Valdosta State University

Background – Group III Separation Scheme Zn2+ Al3+ Ni2+ Fe3+ Mn2+ Valdosta State University

Background Valdosta State University

Background – Group IV Separation Scheme Group IV Unknown A – Flame test for Na+ and K+ Insoluble salts of sodium and potassium are not known. One method of determining the presence of these ions is the flame test. Flame Test K+ lavender flame Flame Test Na+ orange-yellow flame Na K Valdosta State University

Background – Group IV Separation Scheme B – Test for Ba2+ HC2H3O2, K2CrO4 The formation of a yellow precipitate on the addition of potassium chromate indicates the presence of the barium ion. BaCrO4 yellow ppt. 6M HCl Ba2+(aq) + K2CrO4(aq)  BaCrO4(s) + 2K+(aq) Ba2+ Flame Test apple - green 6M H2SO4 BaSO4 white ppt. Valdosta State University

Background – Group IV Separation Scheme B – Test for Ba2+ HC2H3O2, K2CrO4 A flame test (apple green) is used to confirm the presence of the ion. BaCrO4 yellow ppt. 6M HCl Ba2+ Flame Test apple - green 6M H2SO4 BaSO4 white ppt. Valdosta State University

Background – Group IV Separation Scheme B – Test for Ba2+ HC2H3O2, K2CrO4 The final test for barium is the formation of a white precipitate on the addition of a sulfate to the solution. BaCrO4 yellow ppt. 6M HCl Ba2+(aq) + H2SO4(aq)  BaSO4(s) + 2H+(aq) Ba2+ Flame Test apple - green 6M H2SO4 BaSO4 white ppt. Valdosta State University

Background – Group IV Separation Scheme C – Test for Ca2+ Ca2+, Mg2+ The calcium ion is separated from magnesium by precipitating calcium oxalate. (NH4)2C2O4 Ca2+(aq) + (NH4)2C2O4(aq)  CaC2O4(s) + 2NH4+(aq) CaC2O4 Mg2+ 6M HCl Flame test red-orange Valdosta State University

Background – Group IV Separation Scheme C – Test for Ca2+ Ca2+, Mg2+ A flame test (red-orange flame) is used to confirm the presence of the calcium ion. (NH4)2C2O4 CaC2O4 Mg2+ 6M HCl Flame test red-orange Valdosta State University

Background – Group IV Separation Scheme C – Test for Mg2+ Ca2+, Mg2+ K2C2O4 The magnesium ion is precipitated as a white solid (ammonium phosphate) in an alkaline solution. CaC2O4 Mg2+ Mg2+(aq) + NH3(aq) + HPO42-(aq)  MgNH4PO4(s) 6M HCl NH3(aq) Na2HPO4(aq) Flame test red-orange MgNH4PO4–6H2O white ppt. Valdosta State University

Sample Problem 19.12 Separating Ions by Selective Precipitation PROBLEM: A solution consists of 0.20M MgCl2 and 0.10M CuCl2. Calculate the [OH-] that would separate the metal ions as their hydroxides. Ksp of Mg(OH)2= is 6.3x10-10; Ksp of Cu(OH)2 is 2.2x10-20. PLAN: Both precipitates are of the same ion ratio, 1:2, so we can compare their Ksp values to determine which has the greater solubility. It is obvious that Cu(OH)2 will precipitate first so we calculate the [OH-] needed for a saturated solution of Mg(OH)2. This should ensure that we do not precipitate Mg(OH)2. Then we can check how much Cu2+ remains in solution. SOLUTION: Mg(OH)2(s) Mg2+(aq) + 2OH-(aq) Ksp = 6.3x10-10 Cu(OH)2(s) Cu2+(aq) + 2OH-(aq) Ksp = 2.2x10-20 [OH-] needed for a saturated Mg(OH)2 solution = = 5.6x10-5M