Anachem-Lab-M5 Activity 2 LAB PDF

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M5 Activity 2 ANSWER SHEETThis is a group activity which covers Group 1Cations – Grp III Cations**A. Guide Questions for Group I Cations What metals are members of Group I Cations?** Pb+2, Hg 2 +2, Ag+ How can they be separated from other group cations?Group I cations form insoluble chlorides and ar...

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M5 Activity 2 ANSWER SHEET This is a group activity which covers Group 1Cations – Grp III Cations A. Guide Questions for Group I Cations 1. What metals are members of Group I Cations? Pb+2, Hg2+2, Ag+ How can they be separated from other group cations? Group I cations form insoluble chlorides and are separated from the solution using 2-3M HCl as a group precipitant. 2.

How is lead ion separated and identified? Lead (II) chloride can be separated from the other two chlorides based on its increased solubility at higher temperatures. This means that the separation of lead (II) chloride from the precipitate can be done by leaching with hot water. The presence of lead ion can be identified by the formation of a yellow precipitate of PbCrO4 when we add aqueous K2CrO4. 3. What is the purpose of adding ammonia to the precipitate? Ammonia will aid in separation of AgCl and Hg2Cl2

B. Guide Questions for Group II Cations 1. What metals are members of Group II Cations? Hg+2,Bi+3, Cu+2, Cd+2, As+3, Sb+3, Sn+4 2. For each of the following pairs, identify a reagent that will dissolve one substance but not the other:

HgS and As2S3

Reagent: (NH4)2 - As2S3 dissolves in (NH4)2, but HgS does not dissolve in (NH4)2 hence it will form a dark precipitate.

HgS and Bi2S3

Reagent: HNO3 - HNO3 dissolves Bi2S3 and forms bismuth nitrate, whereas HgS does not dissolve in HNO3.

As2S3 and Sb2S3

Reagent : HCl - In concentrated HCl, Sb2S3 dissolves and forms soluble chlorides, whereas As2S3 does not dissolve.

PbSO4 and CuSO4 Reagent: HCl - CuSO4 dissolves in dilute HCl, whereas PbSO4 does not, which then results in a white sulfate precipitate.

3. Explain the theoretical aspect of the separation of Group IIA from Group IIB. Theoretically, Group IIA is distinguished from Group IIB because Group II-A comprises cations whose sulfide compounds are insoluble in KOH, whereas Group II-sulfide B's compounds dissolve in KOH solution.

C. Guide Questions for Group III Cations: 1. Account for the separation of aluminum ion from manganese ion, ferric ion, and cobalt ion. Write the complete reaction involved. - In an NH3-NH4Cl buffered solution containing ammonium sulfide (NH4OH + H2S), group III cations precipitate as sulfides or hydroxides. The action of a strong base (KOH) on a solution of cations divides the cations into two subgroups. Group IIIA hydroxides (Cr+2 , Zn+2, Al+3) dissolve in x's KOH to create the soluble group. The insoluble group is formed from the hydroxides of Group IIIB (Mn+2 , Fe+3, Ni+2 , Co+2 ). Cr+3 Cr+6O4-2, Fe+2 Fe+3, Mn+2 Mn+4, and Co+2 Co+3 are all oxidized when H2O2 is added. Higher valence levels of hydroxides are more insoluble than lower valence states. The use of peroxides facilitates the separation of group III subgroups.

Al+3(aq) + 3OH-(aq) → Al (OH)3(aq) Mn+2 (aq)+ 2OH- (aq) → Mn (OH)2(s) Fe+3(aq) + 3OH-(aq) → Fe(OH)3(s) CO+2(aq) + 2OH-(aq) → CO(OH)2(s) As a result, the formed centrifugate contains Al+3 ion, which is now isolated from the Fe3+, Co2+ ,and Mn2+ ions that are present in the precipitate. 2. What is the property of Al(OH)3 which, when reacting with organic dyes, serves as a confirmatory test? - Al(OH)3 has the property of forming a complex with alizarin dye which serves as the confirmatory test for the aluminum ion.

3. How is manganous ion identified? What is the reaction involved? - Manganous ion will be identified by the formation of purple MnO4. Mn(OH)4 dissolves in HCl, and Mn+4 is reduced back to Mn+2 in an acidic medium. Manganous ions in acid solution are oxidized to permanganate by sodium bismuthate.

4. Enumerate the ions that form colored complexes during the confirmatory test. Identify their characteristic colors. Chemical Formula of the Complex Ion

Characteristic Color

1. Aluminum ion (Al3+) Al3+(aq) +3NH3(aq) +3H2O(l) ⇄ Al(OH)3 + 3NH4+(aq)

alizarin dye (yellow solution) and reddishpurple supernatant

2. Ferric ion (Fe3+) Fe3+ + SCN- ⇄ FeSCN2+

dark-reddish brown complex

3. Manganese ion (Mn2+) 2Mn2+(aq) + 5NaBiO3 ⇄ 2MnO4+5Bi3+(aq) + 5Na+(aq) +7H2O

formation of purple complex

4. Cobalt ion (Co2+) Co2+ + 4SCN- ⇄ [Co(SCN)4]2-

intense blue colored complex

D. After studying the separation scheme for Group I to III cations (a) devise a flow chart or schematic diagram to separate and confirm the presence of the given ions in the solution below;

(b) write the chemical formula of the precipitates formed; and 1. PbCrO4 2. Sb2S3

3. 4. 5. 6. 7.

Fe(OH)3 MnS Al(OH)3 FeS FeSCN-

(c) provide a brief explanation of your schematic diagram. The schematic diagram shows the separation and confirmation of the presence of ions belonging to Group I to III. The first step is adding HCl, which is a precipitant agent for Group I cations. Pb2+ belongs to Group I, that’s why it undergoes precipitation and when a drop of K2CrO4 is added, a yellow precipitate of PbCrO4 is formed. The remaining ions are the decantates and when the precipitant H2S (in acidic solution) is added, the ion belonging to Group II, which is Sb3+, will precipitate and becomes Sb2S3 which can be confirmed when drop to a silver coin and a black deposit is observed. The remaining ions belonging to Group III can be precipitated by adding (NH4)2S and they will form the following precipitates, Fe(OH)3 , MnS and Al(OH)3. To confirm the presence of Al3+ ions, alizarin dye can be added to the soluble Al(OH)3 and a reddish supernatant will be observed. Next is to add HCl to the remaining two insoluble ions. If you add sodium bismuthate to MnS, it will produce a purple solution. On the other hand, adding SCN- to FeS will yield a deep red solution of FeSCN-....