Solved Examples on Principles Related to Practical Chemistry

Question 1: The gas liberated on heating a mixture of two salts with NaOH gives a reddish brown precipitate with an alkaline solution of K₂HgI₄. The aqueous solution of the mixture on treatment with BaCl₂ gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture with K₂Cr₂O₇ and conc. H₂SO₄ red vapours  of A are produced. The aqueous solution of the mixture gives a deep blue colouration B with potassium ferricyanide solution. Identify the radicals in the given mixture and write the balanced equations for the formation of A and B.

Solution: Let us summarise the given facts of the question.

The given reactions lead to following conclusions.

i) Heating of mixture with NaOH to give NH₃ gas (indicated by reddish brown ppt. with alkaline solution of K₂HgI₄) indicates the presence of   ion in the mixture

ii) Heating of mixture with K₂Cr₂O₇ and conc. H₂SO₄ to give red vapours (of chromyl chloride) indicates the presence of Cl^– ion in the mixture.

iii)  Reaction of aqueous solution of the mixture with barium chloride solution to give white ppt. (of BaSO₄) sparingly soluble in conc. HCl indicates the presence of  ions in the mixture.

iv)  Reaction of aqueous solution of the mixture with potassium ferricyanide solution to give deep blue colour indicates the presence of Fe²⁺ ions in the mixture.

 Hence the mixture contains the following four ions. NH₄⁺, Fe^2,  SO₄^2-and Cl^–.
Equations for the formation of A and B

Question2. : When an orange coloured crystalline compound (A) was heated with common salt and concentrated sulphuric acid an orange - yellow coloured gas (B) was evolved. The gas (B) when passed through caustic soda solution gave a yellow solution (C) which in turn gave the following reactions.

1. Addition of silver nitrate solution to (C) gave first a white precipitate which then turns red. Quantitatively, 0.155 g of the gas (B) required 2.0 m moles of AgNO₃ to produce the first trace of red colour.
2. Acidification of the solution (C) with dil. H₂SO₄ gave an orange solution which contained chromium in +6 oxidation state. The solution liberated iodine from aqueous potassium iodide, leaving a green solution containing chromium in +3 oxidation state. Quantitatively 0.155 g of the gas (B) liberated 1.5 mmole of iodine.

Deduce the formula of A, B and C and explain the reactions.

Solution: Let us summaries the given reactions.

The above set of reactions indicates that compound A is K₂Cr₂O₇, B is chromyl chloride gas and C is sodium chromate which explains all the given reactions as below.

The quantitative data is explained in the following manner. From the above reactions we observe that

155 g of CrO₂Cl₂  require 2 mole of AgNO₃
 

This also coincides with the given data.

Question: 3, A white amorphous powder (A) when heated, gives a colourless gas (B), which turns lime water milky (which dissolves on passing excess of gas (B) and the  residue (C) which is yellow while hot but white when cold. The residue (C) dissolves in dilute HCl and the resulting solution gives a white precipitate on addition of potassium ferrocyanide solution. (A) dissolves in dil. HCl with the evolution of a gas which is identical in all respects to gas (B).  The solution of (A) in dil. HCl gives a white ppt. (D) on addition of  NH₄Cl  excess of NH₄OH and on passing H₂S gas. Another portion of this solution gives initially a white ppt. (E) on addition of NaOH  solution which dissolves in excess of NaOH. The solution on passing again H₂S gives back the white ppt. of (E), the white ppt. on heating with dil H₂SO₄ give a gas used in II and IV group analysis. What are (A) to (E)? Give balanced chemical equations of the reactions.

Solution:

From observations for section (a), one may conclude that the colourless gas is CO₂ because it turns lime water milky, due to formation of insoluble CaCO₃.

CaCO₃  is soluble in excess of CO₂ due to formation of soluble calcium bicarbonate

The compound (C) is zinc oxide (ZnO) because it is yellow when hot and white when cold, hence the initial compound (A) is zinc carbonate (ZnCO₃).

From section (b), it is inferred that (C) is a salt of Zn(II) which dissolves in dil. HCl and white ppt. obtained after addition of K₄Fe(CN)₆ is due to zinc ferrocyanid.

ZnCO₃ because on treatment with dil. HCl it gives gas (B), i.e., CO₂, while Zn (II) goes in solution, i.e., ZnCl₂. On passing H₂S gas in presence of NH₄OH, it gives a white ppt. of ZnS(D). ZnS on heating with dil. H₂SO₄ evolves H₂S, which is used for the precipitation of sulphides of group II in acidic medium and of IV group in alkaline medium. ZnCl₂, reacts with NaOH to give a ppt. of  Zn(OH)₂ which dissolves in NaOH, as Zn(OH)₂ is amphoteric in nature. The solution Na₂ZnO₂ again gives ZnS on passing H₂S gas into it. Different chemical equations concerned are given below: