KnowledgeBoat Logo
|

Chemistry

Identify the cation [positive ion] and anion [negative ion] in - A, B and C. Also identify P, Q, R, S, T, U, V, W.

(a) Substance 'A' is water soluble and gives a curdy white precipitate 'P' with silver nitrate solution. 'P' is soluble in ammonium hydroxide but insoluble in dil. HNO3. Substance 'A' reacts with ammonium hydroxide solution to give a white precipitate 'Q' soluble in excess of NH4OH.

(b) A solution of substance 'B' is added to barium chloride solution. A white ppt. 'R' is formed, insoluble in dil. HCl or HNO3. A dirty green ppt. 'S' is formed on addition of ammonium hydroxide to a solution of 'B' and the precipitate is insoluble in excess of ammonium hydroxide.

(c) Substance 'C' is a coloured, crystalline salt which on heating decomposes leaving a black residue 'T'. On addition of copper turnings and conc. H2SO4 to 'C' a coloured acidic gas 'U' is evolved on heating. A solution of 'C' is added to NaOH soln. until in excess. A pale blue ppt. 'P' is obtained insoluble in excess of NaOH. A solution of 'C' is then added to NH4OH soln. in excess to give an inky blue solution 'V'. A solution of 'C' is warmed and hydrogen sulphide gas is passed through it. A black ppt. 'W' appears.

Practical Chemistry

9 Likes

Answer

(a) A is ZnCl2

Cation and anion in ZnCl2 : Zn2+ and Cl-

ZnCl2 + 2AgNO3 ⟶ 2AgCl ↓ [curdy white ppt.] + Zn(NO3)2

P is AgCl

The precipitate AgCl is soluble in ammonium hydroxide but insoluble in dil. HNO3.

ZnCl2 + 2NH4OH ⟶ 2NH4Cl + Zn(OH)2 ↓ [white ppt.]

2NH4Cl + Zn(OH)2 + 2NH4OH [excess] ⟶ 4H2O + [Zn(NH3)2]Cl2

Q is Zn(OH)2

(b) B is ferrous sulphate (FeSO4)

Cation and anion in FeSO4 : Fe2+ and SO42-

FeSO4 + BaCl2 ⟶ BaSO4 ↓ [white ppt.] + 2NaCl

R is BaSO4

FeSO4 + 2NH4OH ⟶ (NH4)2SO4 + Fe(OH)2 ↓

S is Fe(OH)2

(c) C is copper nitrate Cu(NO3)2

Cation and anion in Cu(NO3)2 : Cu2+ and NO31-

Black residue (T) : CuO - copper oxide

2Cu(NO3)2[Copper nitrate (blue)]⟶Δ2CuO[copper oxide - black]+4NO2[Nitrogen dioxide]+O2[Oxygen]\underset{\text{[Copper nitrate (blue)]}}{2\text{Cu(NO}_3)_2} \overset{\Delta}{\longrightarrow} \underset{\text{[copper oxide - black]}}{2\text{CuO}} + \underset{\text{[Nitrogen dioxide]}}{4\text{NO}_2} + \underset{\text{[Oxygen]}}{\text{O}_2}

U is Nitrogen dioxide (NO2)

P is Copper [II] hydroxide Cu(OH)2

Cu(NO3)2 + 2NaOH ⟶ 2NaNO3 + Cu(OH)2 ↓

V is [Cu(NH3)4]SO4

Cu(NO3)2 + 2NH4OH ⟶ Cu(OH)2 ↓ + 2NH4NO3

2Cu(OH)2 + 2NH4NO3 + 2NH4OH ⟶ 2Cu(NH3)4NO3 + 4H2O

W is CuS

Cu(NO3)2 + H2S ⟶ CuS + 2HNO3

Answered By

5 Likes

Related Questions