Qualitative analysis of inorganic compound

1. Lab Report
Qualitative analysis of inorganic compound.
Course Title: Chemistry for biologist I
Course code: CHE 108

Prepared by:
Md. Shabab Mehebub


Department of
Genetic engineering and Biotechnology
East West University, Dhaka

2. Name of the Experiment: Qualitative analysis of inorganic
compound.
Principal: Qualitative analysis deals with the identification of elements and compounds
with the detection of individual components in their mixtures. In inorganic compound analysis is
usually to identify a single substance, or to detect the cations and anions in a mixture. Inorganic
compounds reaction with reagent and gives precipitate. We can easily identify them by
precipitation.
Materials:
Equipment:
1. Test tube
2. Spatula
3. Dropper
Reagent:
1. Dilute HNO3
2. AgNO3 solution
3. KI solution
4. Ammonium oxalate solution
5. 0.1 N Ba(OH)2
6. 5% HCl
7. NH4OH
Supplied Compound:
8. Lead sulphate
9. Cu(OH)2
10. Calcium chloride
11. Sodium carbonate
Procedure:
Test for cations:
Calcium: To a solution of the substance being examined adding a few drops of a solution of
Ammonium oxalate [ (NH4)2C2O4 ]
(NH4)2C2O4 + CaCl2= CaC2O4 + 2NH4Cl
Copper: The solution of substance, add dil. Ammonia.
Cu(OH)2(s) + 4NH3(aq.) + 2H2O(l) [Cu(NH3)4(H2O2)]
2+
(aq.) + 2OH
-
.

3. Lead: Use 0.5 ml of the prescribed solution. Add 10 ml of water and 0.2 ml of M potassium
iodide.
2KI + PbSO4 = PbI2 +K2SO4 .
Test for anions:
Carbonate: 1 ml dilute HCl had taken in test tube. Dilute HCl acid gave vigorous
effervescence with carbonates, evolving carbon dioxide.
CO3
2-
(s) + 2H+(aq) = H2O(l) + CO2(g) .
Chloride: Dissolved in 2 ml of water a quantity of the substance had been examined
equivalent to about 2 mg of chloride ion or used 2 ml of the prescribed solution. Acidify with dil.
Nitric acid add 0.5 ml of silver nitrate solution.
AgCl(s) + 2NH3(aq) Ag(NH3)
2+
(aq) + H
+
(aq)
Ag(NH3)
2+
(aq) + H
+
(aq) + Cl
-
(aq) AgCl(s) + 2NH4
+
(aq)
Sulphate: Used 5 ml of the prescribed solution. Added 1 ml of dil. HCl acid and 1 ml of
barium chloride solution.
Ba
2+
(aq) + SO4
2-
(aq) BaSO4(s)
Observation:
Cations:
Calcium: A white precipitate had obtained. That had only sparingly soluble in dilute acetic
acid but soluble in hydrochloride acid.
Copper: A greenish blue precipitate is formed which dissolved in excess of reagent forming
deep blunish solution.
Lead: A yellow precipitate had been formed. It had been heated to boil for one or two minute,
and allowed to cool. The precipitate was formed as glistening, yellow plates.
Anions:
Carbonate: The colorless, odorless carbon dioxide had been identified by bubbling it through
a saturated solution Of barium hydroxide, with which it had formed a white precipitate.
CO2(g) + Ba
2+
(aq) + 2OH
-
(aq) BaCO3(s) + H2O(l) .
Chloride: A curdy white precipitate had formed and soluble in NH4OH.
Sulphate: A white precipitate had formed.

4. Result:
Cations:
Calcium: It was Ca
+
ion because a white precipitate had obtained and sparingly soluble in
dilute acetic acid but soluble in HCl.
Copper: It was Cu
+
ion because the mixture was given greenish blue precipitate and added
excess of reagent formed deep bluish solution.
Lead: A yellow precipitate had been formed that’s why it is Pb
+
ion.
Anions:
Carbonate: It was CO3
-
ion because carbon dioxide bubbling and had formed a white
precipitate with barium hydroxide.
Chloride: It was Cl
-
ion because a curdy white precipitate had formed and soluble in NH4OH.
Sulphate: It was SO4
-
ion because a white precipitate had formed and we know that SO4
-
ion gives
white precipitate in confirmation test.
Discussion:
Precaution:
1. I had worn Lab coat.
2. Volume had carefully measured.
3. The observation had recorded.
4. All apparatus had handled carefully.
5. Test tubes were cleaned by water during the experiment time.
Calcium: At first I had taken a cleaned test tube. Then took some substance and solute it into
the distill water. I had added a few drop of Ammonium oxalate into the solution. Then a white
precipitate had obtained. It was solute in HCl. That’s why we can say, that was about Calcium
and there was calcium ion.
Copper: A cleaned test tube had taken by me. Then I had taken substance and solute it into the
distill water. By adding dil. Ammonia I had seen a greenish blue precipitate had formed. When I
add excess of reagent it had formed deep bluish solution. Copper gives blue precipitates in
confirmation test. That’s why we can say; in this solution it was copper.
Lead: Firstly I had taken a cleaned test tube. Secondly, I had taken 0.5ml of the solution by
used of pipette. I had added 10 ml of water and 0.2 ml M Potassium iodide by used of pipette. A
yellow precipitates had formed. It had been heated by the Bunsen burner and then cooled. The
precipitate had formed glistening, yellow plates. Lead gives yellow precipitates in confirmation
test. That’s why we can say, in this solution it was lead.

5. Carbonate: A cleaned test tube had taken by me. Then I had taken 1 ml dil. HCl. After that I
added carbonate. The colorless carbon dioxide had been identified by bubbling. When added
Ba(OH)2 it reaction with CO2 and together with formed a white precipitate of barium carbonate.
. Carbonate gives the bubble, which is CO2 in confirmation test. That’s why we can say, in this
solution it was Carbonate ion.
Chloride: A cleaned test tube had taken by me. Firstly, I had taken 2 ml of water an added the
given substance. Secondly I had added dil. Nitric acid for acidify the solution. Thirdly I had
added 0.5 ml of silver nitrate and shook it well. Then a curdy white precipitate had seen in test
tube. Lastly NH4OH added and the precipitate solute. It indicated that chlorine in the inorganic
compound. That’s why we can say, in this solution it was contained chloride.
Sulphate: A cleaned test tube had taken by me. Firstly I had taken 5 ml of prescribed solution.
Secondly I had added 1 ml of dil. HCl acid and then also added BaCl2 solution. A white
precipitate had formed. That’s why we can say, in this solution it was Sulphate ion.

Experimental discuss: In that process of Qualitative analysis of in organic compound,
we can found the unknown in organic chemical substance because they reaction with reagent,
make precipitate and change their color in the solution.