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Teknologi dan Rekayasa

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Presentation on theme: "Teknologi dan Rekayasa"— Presentation transcript:

1 Teknologi dan Rekayasa
Kompetensi Keahlian Analisis Kimia

2 Qualitative Inorganic Analysis

3 Anions Analysis

4 Anions Analysis Prelimininary test can identificate some anions, e.g. carbonate, acetate, Sulphide, etc The soda extract solution (SES) should be made to identificate the other anions The soda extract is made by put some sample in saturated Na2CO3 solution, the solution then heated and filtered. The filtrate is soda extract solution (SES).

5 . The anions have been separated from the cations.
Soda Extract Solution . SES is the solution containing the anions in form of alkali salt. . The anions have been separated from the cations. . The cations are precipitated as carbonates and hydroxide.

6 Preliminary test for Identification of anions
Test of reducing agent 1 mL of SES acidified with a few drops of H2SO4 3M until free of carbonate. Add a bit of KMnO4. If the color of KMnO4 is disappear  there are reducing agentt like : SO32-, S2O32-, S2-, NO2-, SCN-, Br- dan I- If the color of KMnO4 is not disappear, heat for a moment. When the larutan become colorless  it could be C2O42-

7 Preliminary test for Identification of anions
Test of oxidator ions 1 mL of SES acidified with a few drops of H2SO4 3M until free of carbonate. Add a drop of diphenilamine solution  blue color indicate the existence of oxidizing agent oxidizing ions : MnO4-, Cr2O72-, etc

8 Separation and Identification of Sulphates, Sulphites, Oxalates, Chromates, Flourides
Extract Soda Acidity 1 mL of ES with 1 M HCl solution, free carbonate (Add Cd(NO3)2 solution to discharge sulphides ions)* Filtrate Residue : CdS, S Identification : CdS  yellow residue S  white residue from Na2S2O3 Add BaCl2 solution Residue : BaSO4 Identification : BaSO4  white residue There is SO42- present Filtrate Add Br2 solution shake well until the filtrate is not decolourized Residue : BaSO4 Identification : BaSO4  white residue There is SO32- present Filtrate

9 Residue : H2CrO4/H2Cr2O7, HF and H2C2O4
Separation and Identification of Sulphades, Sulphites, Oxalates, Chromates, Flourides Filtrate Add NaOAc solution (centrifuge) Residue : H2CrO4/H2Cr2O7, HF and H2C2O4 Identification : BaCrO4  yellow residue (there is H2CrO4 or H2Cr2O7 present) BaF2  white (if there have a lot of HF and H2C2O4 ) Filtrate : Identification : Add CaCl2  white residue of CaC2O4 and white residue of CaF2 (centrifuge) Place the residue in 2 M HOAc solution and add 1 drops of 0,2 M KMnO4 shake and warming (If KMnO4 is bleached, add KMnO4 solution in excess until the colour is constant). If all of the residue is dissolved  just H2C2O4 present If there is still the residue  H2C2O4 and HF present If adding 1 drops of 0,2 M KMnO4 is not decolourized  CaF2 residue and just HF present

10 Separation and Identification of Halides, (Chloride, Iodide, Bromide)
Acidity 1 mL of ES with 6M HNO3 solution, free carbonate. Than Add AgNO3 solution, centrifuge. Residue may contain : AgCl (white), AgSCN (white), AgBr (white-yellow), AgI (yellow) If there is SCN-, AgSCN must be removed first as follows : Add 1mL of HNO3 concentrate and steam solution until almost dry (repeat once more until oxidation AgSCN has finished) Wash the residue has been free AgSCN with 1 mL of H2O and 5 drops of 6 M HNO3 until free of Ag+ Add 10 drops of 1 M (NH4)2CO3(centrifuge) Filtrate : Ag(NH3)2+, Cl- Identification : At a part of filtrate add 1 drops 1 M KBr  yellowish white residu of AgBr , Cl- present Acidity with 6 M HNO3  white residue of AgCl, Cl- present Residue may contain : AgBr, Agl

11 Residue may contain : AgBr, Agl
Separation and Identification of Sulphades, Sulphites, Oxalates, Chromates, Flourides Residue may contain : AgBr, Agl Add a few of Zn solid and 1 mL 2M H2SO4 (by reduce coming about HBr and HI) centrifuge Identification filtrate about Iodide : To the filtrate add 1drop of 0,1 M FeCl3 (1:1) and place 1 drop of solution upon starch paper  blue spot of I- If Iodide is present must be remove first as follows : A part of the filtrate adding 0,1 FeCl3 (1:1) and boil gently until all iodide are volatilized (test with starch paper) 3. For the filtrate of free from Iodide add 3 drops of 0,1 M KMnO4 and 3 drops of 3 M H2SO4 then shake well. KMnO4 excess must be remove with 10% H2O2 Add CHCl3  CHCl3 layer coloured brown  Bromide present


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