Problem 13: Write down the products of the following reactions.

(a) CuSO₄ solution is treated with KI solution.

(b) AgNO₃ solution is added to Na₂S₂O₃ solution.

Solution:

(a) Free iodine is liberated along with the formation of a white precipitate of cupric iodide.

            CuSO₄ + 2Kl ————→ Cul₂ + K₂SO₄

(b) A white precipitate of Ag₂S₂O₃ is obtained which turns yellow, brown and finally black on keeping.

            2AgNO₃ + Na₂S₂O₃ ————→ Ag₂S₂O₃ + 2NaNO₃

            Ag₂S₂O₃ + H₂O ————→ Ag₂S + H₂SO₄

                                                         ^{black ppt.}

Problem 14: Explain the following

(a) Zinc readily liberates H₂ form cold dil.H₂SO₄ but not from cold conc. H₂SO₄.

(b) Blue colour of the CuSO₄ solution is discharged slowly when an iron rod is dipped into it.

Solution:
(a) Conc. H₂SO₄ is a covalent compound. Hence does not contain H⁺ ions. Dilute H₂SO₄ contains H₃O⁺ which reacts with Zn and liberates H₂.

         H₂SO₄ + H₂O ————→ 2H₃O⁺ + SO₄^2–

         Zn + 2H₃O⁺ ————→ Zn²⁺ + 2H₂O + H₂O ↑

(b) Fe is more electropositive than Cu, hence it displaces copper form CuSO₄ solution.

         Fe_(s) + CuSO_(aq) ————→ FeSO_4(aq) + Cu_(s)

Problem 15: An aqueous solution containing one mole of HgI₂ and two moles of NaI is orange in colour. On addition of excess NaI the solution becomes colourless. The orange colour reappears on subsequent addition of NaOCl. Explain with equations.

Solution:

         Hgl₂ + 2Nal ————→           Na₂[Hgl₄]

         ^{(orange)                           coloured due to residual Hgl2)}

 

          Hgl₂ + Nal(excess) ————→  Na₂[Hgl₄]

          ^{(orange)                                        (colourless because there is no residual Hgl2)}

 

         2Na₂[Hgl₄] + 2NaOCl + H₂O ————→ 2Hgl₂ + NaCl + 4NaOH + 2Nal₃

                                                                          ^(orange)