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Question

Write the observations, and balanced equations for the following reactions:
(i) Sodium hydroxide is added dropwise to a solution of zinc sulphate, till it is in excess.
(ii) Ammonium hydroxide is added first in a small quantity, and then in excess, to a solution of copper sulphate.
(iii) Excess of ammonium hydroxide is added to a substance, obtained by adding hydrochloric acid to silver nitrate solution.
(iv) Moist starch iodide paper is placed at the mouth of a test-tube containing chlorine gas.
(v) A paper dipped in potassium permanganate solution is placed at the mouth of a test-tube, containing sulphur dioxide gas.

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Solution

(i) When sodium hydroxide is added, in drops, to a solution of zinc sulphate, white gelatinous precipitate of zinc hydroxide is formed, which forms a soluble complex sodium zincate in excess of sodium hydroxide.

ZnSO4 + 2NaOH Zn(OH)2 + Na2SO4Zinc sulphate Sodium Zinc hydroxide Sodium sulphate hydroxide (White ppt.) (colourless)Zn(OH)2 + 2NaOH (excess) Na2ZnO2 + 2H2OZinc hydroxide Sodium zincate (colourless)

(ii) When ammonium hydroxide is added in a small quantity to copper sulphate solution, pale blue precipitate of cupric hydroxide is formed. By adding ammonium hydroxide in excess, the formed precipitate dissolves to form deep blue solution of tetraamine copper (II) sulphate.
CuSO4 + 2 NH4OH Cu(OH)2 + (NH4)2SO4Copper sulphate Ammonium Cupric hydroxide (Blue) hydroxide (Pale blue ppt)
Cu(OH)2 + (NH4)2SO4 + 2 NH4OH [Cu(NH3)4]SO4 + 4 H2O Cupric hydroxide (In excess) Tetrammine copper(II) (Pale blue ppt) sulphate (deep blue)

(iii) Silver nitrate on reaction with hydrochloric acid forms a white precipitate of silver chloride. When excess of ammonium hydroxide is added to silver chloride, it forms soluble diammine silver complex.
AgNO3 + HCl AgCl + HNO3Silver nitrate Hydrochloric Silver chloride acid (White solid) AgCl + 2NH4OH [Ag(NH3)2]Cl + 2H2OSilver chloride excess Diammine silver complex(white solid) (soluble)

(iv) The iodide paper turns blue because chlorine gas oxidises some of the iodide ions on the starch paper to iodine molecules. Further, these iodine molecules react with starch to form a complex, which gives blue colour.
Cl2 + 2 KI I2 + 2 KCl Chlorine Potassium Iodide Iodine Potassium chloridegas (present on starch paper)

(v) The purple colour of the paper dipped in KMnO4 gets de-colourised when sulphur dioxide gas comes in contact with it, because SO2 is a reducing agent and reduces potassium permanganate.
2KMnO4 + 5SO2 +2H2O 2MnSO4 + K2SO4 + 2H2SO4Potassium Sulphur Manganesepermangamate dioxide sulphate(purple colour) (colourless)

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