What is Qualitative analysis?

Qualitative analyses are mainly used for the determination of anions and cations present in the given sample. This analysis is a type of analytical chemistry that is used to determine the elemental composition present in the inorganic compounds via different reagents.

Qualitative analysis of Anions

Before the qualitative analysis of anions, some preliminary tests should be done.

Smell and color

The physical examination of the given inorganic salt includes the review of smell, color, and density. The anions like sulfide and acetate could be determined with the help of their characteristic smell.

Dry heating test

In this test, a pinch of salt is heated in the dry test tube. On heating, these salts could undergo decomposition, and specific gases evolve. By noting the smell of gas, the anions could be identified.

Determination of Anions

The anions could be determined by carrying out the tests below.

Reaction with dilute Sulfuric Acid (H₂SO₄)

A pinch of salt is reacted with dil. H₂SO₄. The type of anions can be identified by observing the type of gas evolved.

Sulfide: Sulphides on reaction with dil. H₂SO₄ gives hydrogen sulfide (H₂S) gas.

Carbonate: Carbonate when reacted with dil. H₂SO₄, carbon dioxide (CO₂) gas evolved with brisk effervescence.

Sulfite: Sulfite on reaction with dil. H₂SO₄, sulfur dioxide (SO₂) gas is liberated.

Nitrite: Nitrites on reaction with dil. H₂SO₄ gives nitrogen monoxide (NO) gas which will react with oxygen present in the air to form nitrogen dioxide (NO₂). It is identified by the liberation of brown gas.

Reaction with Conc. H₂SO₄

When there is no effect on reaction with dil. H₂SO₄, then heat a pinch of salt with conc. H₂SO₄ and determine the gas evolved.

Bromides: Bromide will result in the formation of bromine gas on reaction with con. H₂SO₄.

Chlorides: Chloride gas is evolved on the treatment of chloride with con. H₂SO₄.

Iodides: Iodide on reaction with conc. H₂SO₄, produces the vapors of iodine.

Nitrates: Nitrate on reaction with con. H₂SO₄ forms NO₂ gas.

Oxalate: Oxalate anions on reaction with conc. H₂SO₄ produces a mixture of carbon monoxide and carbon dioxide.

Acetate: Acetate anions on reaction with acetic acid forms the vapors of acetic acid.

Independent groups ( PO₄^3- and SO₄^2-) test

When the unknown sample does not react with the dilute and concentrated sulphuric acid, the sample is tested for sulfate and phosphate by doing their individual tests.

Potassium permanganate (KmnO₄) test

This test could be carried out with the help of KMnO₄& sulfuric acid as reagents. This test is useful for the detection of anions including chloride, bromide, iodide, and oxalate.

Wet tests for the analysis of Anions

The confirmation reactions of anions could be described below.

Confirmation tests for Carbonate ions

Reaction with dilute Hydrochloric Acid (HCl)

Carbonate ion when reacted with dilute HCl gives carbon dioxide which reacts with lime water to produce a white precipitate of calcium carbonate that turns the lime water milky. This test is carried out with water if the carbonate is soluble. If it is insoluble carbonate, then the test is carried out in solid-state. The reaction can be represented as follows:

$C a C O_{3} + 2 H C l \to C a C l_{2} + C O_{2} + H_{2} O C a {(O H)}_{2} + C O_{2} \to C a C O_{3} + H_{2} O c a l c i u m w h i t e p p t c a r b o n a t e$

Magnesium sulphate test

The soluble carbonates can be characterized by using this test. Soluble carbonates on reaction with magnesium sulfate produce the white precipitate of magnesium carbonate. The reaction could be represented as follows:

$N a_{2} C O_{3} + M g S O_{4} \to M g C O_{3} + N a_{2} S O_{4}$

Confirmation tests for sulfite ion

Barium chloride (BaCl₂) test

Sulfite ions reacts with $BaC l_{2}$ to produce the white precipitate (barium sulfite). The formed barium sulfite is dissolved in dilute hydrochloric acid to produce sulfur dioxide gas.

$N a_{2} S O_{3} + B a C l_{2} \to 2 N a C l + B a S O_{3} (s) w h i t e p p t B a S O_{3} + 2 H C l \to B a C l_{2} + S O_{2} (g) + H_{2} O$

Potassium Dichromate test

Sulfites react with potassium dichromate to form a green color solution. The reaction can be represented as follows:

$K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} + 4 H_{2} S O_{4} \to 3 N a_{2} S O_{4} + K_{2} S O_{4} + C r_{2} {(S O_{4})}_{3} + 4 H_{2} O$

Confirmation test for sulphide

Sodium Nitroprusside test

When sulfide reacts with sodium nitroprusside solution, violet or purple color solution is formed. This is because of the formation of $N a_{4} [F e {(C N)}_{5} N O S]$ .

$N a_{2} S + N a_{2} [F e {(C N)}_{5} N O] \to N a_{4} [F e {(C N)}_{5} N O S]$

Lead acetate test

Sulfides on treatment with lead acetate produces black precipitate due to the formation of lead sulfide.

$N a_{2} S + {(C H_{3} C O O)}_{2} P b \to P b S + 2 C H_{3} C O O N a B l a c k p p t$

Cadmium carbonate (CdCO₃) test

Sulfides when treated with the suspensions of $CdC O_{3}$ , produce cadmium sulfide which is formed as a yellow precipitate.

$N a_{2} S + CdC O_{3} \to CdS + N a_{2} C O_{3}$

Confirmation tests for Nitrite

Starch-Iodide test

Nitrite on treatement with KI in the presence of dilute. $H_{2} S O_{4}$ results in the formation of iodine. When iodine is treated with starch, it results in the formation of a blue-black complex. The reaction could be represented as follows:

$2 K I + 2 H_{2} S O_{4} + 2 H N O_{2} \to 2 K H S O_{4} + I_{2} + 2 N O + 2 H_{2} O I_{2} + S t a r c h \to B l u e - b l a c k c o m p l e x$

Diphenlyl Amine test

Diphenylamine, when reacted with nitrites, gets oxidized to form blue coloration.

Ferrous sulphate test

Nitrites, on reaction with ferrous sulfate, forms black or brown color solution due to the formation of $F e S O_{4} . N O$ .

$K N O_{2} + C H_{3} C O O H \to C H_{3} C O O K + H N O_{2} 3 H N O_{2} \to H N O_{3} + 2 N O + H_{2} O F e S O_{4} + N O \to F e S O_{4} . N O$

Confirmation tests for Nitrate

Diphenylamine test

Nitrate on treatment with copper chips undergoes reduction to form reddish-brown NO₂ gas.

Brown-ring test

In this test, the iron sulfate is mixed with the solution of nitrate and conc. sulphuric acid is slowly added to this mixture. As a result, a brown ring is produced at the junctions of the solutions. It shows the presence of nitrate.

$K N O_{3} + H_{2} S O_{4} \to K H S O_{4} + H N O_{3} 6 F e S O_{4} + 3 H_{2} S O_{4} + 2 H N O_{3} \to 3 F e_{2} {(S O_{4})}_{3} + 4 H_{2} O + 2 N O F e S O_{4} + N O + 5 H_{2} O \to [F e (N O) {(H_{2} O)}_{5} S O_{4}] B r o w n r i n g$

Confirmation test for chloride

Chromyl Chloride test

Chloride on reaction with conc.sulphuric acid & potassium dichromate produces red fumes of chromyl chloride that will react with sodium hydroxide to produce sodium chromate (yellow solution). This solution when treated with dilute CH₃COOH and lead acetate to produce lead chromate (yellow precipitate).

$4 N a C l + K_{2} C r_{2} O_{7} + 3 H_{2} S O_{4} \to 2 N a_{2} S O_{4} + K_{2} S O_{4} + 2 C r O_{2} C l_{2} + 3 H_{2} O C r O_{2} C l_{2} + 4 N a O H \to N a_{2} C r O_{4} + 2 H_{2} O + 2 N a C l N a_{2} C r O_{4} + {(C H_{3} C O O)}_{2} P b \to P b C r O_{4} + 2 C H_{3} C O O N a Y e l l o w p p t$

Confirmation tests for Bromide

Silver Nitrate test

The reaction of bromide with silver nitrate produces yellow precipitate (silver bromide) that is sparingly soluble in ammonium hydroxide (NH₄OH).

$K B r + A g N O_{3} \to K N O_{3} + A g B r$

Manganese dioxide test

Bromide salt on reaction with concentrated sulphuric acid and manganese dioxide liberates bromine.

$2 K B r + M n O_{2} + 3 H_{2} S O_{4} \to 2 K H S O_{4} + M n S O_{4} + 2 H_{2} O + B r_{2}$

Confirmation test for iodide

Silver nitrate test

In this test, a small unknown sample is taken in the test tube and silver nitrate solution is added to it. The formation of yellow precipitate shows the presence of iodide which is insoluble in NH₄OH.

$K I + A g N O_{3} \to K N O_{3} + A g I$

Manganese dioxide test

A small amount of the unknown sample is taken in the test tube. Concentrated sulphuric acid & manganese dioxide is added to the test tube. The liberation of iodine gas shows the presence of iodide.

$2 K I + M n O_{2} + 3 H_{2} S O_{4} \to 2 K H S O_{4} + M n S O_{4} + 2 H_{2} O + I_{2}$

Confirmation test for acetate

Oxalic acid test

A small amount of sample is taken in the test tube, oxalic acid is added to it. It forms acetic acid that is characterized by a vinegar-like smell.

$C O O H - C O O H + 2 C H_{3} C O O N a \to C O O N a - C O O N a + 2 C H_{3} C O O H$

Confirmation test for oxalate

Calcium Chloride test

A small amount of sample is taken in the test tube and calcium chloride is added to it. It forms calcium oxalate (white precipitate).

${(N H_{4})}_{2} C_{2} O_{4} + C a C l_{2} \to C a C_{2} O_{4} + 2 N H_{4} C l$

Confirmation test for sulfate

Barium Chloride test

When sulfate is reacted with barium chloride, white precipitate of barium sulfate is produced.

$N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l$

Confirmation test for phosphate

Magnesia mixture test

When phosphate is reacted with magnesia mixture, magnesium ammonium phosphate is produced. It is formed as a white precipitate.

$N a_{2} H P O_{4} + M g C l_{2} + N H_{4} O H \to M g (N H_{4}) P O_{4} + 2 N a C l + H_{2} O$

Context and Applications

This topic is important for both undergraduate and postgraduate courses, particularly for Bachelors and Masters in Chemistry.

Practice Problems

Question 1: A brown ring is produced in the test for __________.

Iron
Bromide
Nitite
Nitrate

Answer: Option 4 is correct.

Explanation: In the case of the nitrate test, a brown ring is formed at the junction of salt solution and the ferric sulfate solution with concentrated sulfuric acid.

Question 2: Among the given compounds, the compound which is insoluble in CH₃COOH is _______.

Calcium oxalate
Calcium hydroxide
Calcium carbonate
Calcium oxide

Answer: Option 1 is correct.

Explanation: Calcium oxalate could not be dissolved in weak acid like acetic acid. It can only be dissolved in strong acids.

Question 3: The result of chromyl chloride test is _____________.

A colourless solution will be formed
A dark green solution will be formed
A yellow precipitate will be formed
A yellow solution will be formed

Answer: Option 3 is correct.

Explanation: The chromyl chloride vapor on react with NaOH, acetic acid & lead acetate will form lead chromate (yellow precipitate).

Question 4: The presence of effervescence is because of ______.

Presence of carbonate
Presence of chloride
Absence of bicarbonate
Absence of carbonates

Answer: Option 1 is correct.

Explanation: Carbonate on reaction with dil. H₂SO₄, carbon dioxide (CO₂) gas evolved with brisk effervescence.

Question 5: Unknown sample on heated with dil. H₂SO₄ & the lead acetate paper turns black with the liberation of gas. The sample contains _______.

Sulphide
Sulphite
Thiosulphate
Sulphate

Answer: Option 1 is correct.

Explanation: Sulphide when reacted with dil. H₂SO₄, the lead acetate paper turns black with the liberation of gas. Hence, the mixture has a sulfide ion.

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Science Chemistry

Inorganic Chemistry

Qualitative Inorganic Analysis