What do you understand by the following:
(a) Analysis
(b) Qualitative analysis
(c) Reagent
(d) Precipitation
Answer
(a) Analysis — Determination of the chemical components in a given sample is called Analysis .
(b) Qualitative analysis — Identification of the unknown substances in a given sample is called Qualitative analysis .
(c) Reagent — A reagent is a substance that reacts with another substances.
(d) Precipitation — The process of formation of an insoluble solid when solutions are mixed is called Precipitation. The solid thus formed is called Precipitate .
Write the probable colour of the following salts:
(a) Iron (III) chloride
(b) Potassium nitrate
(c) Ferrous sulphate
(d) Aluminium acetate
Answer
(a) Iron (III) chloride — Yellow
(b) Potassium nitrate — Colourless
(c) Ferrous sulphate — Pale Green
(d) Aluminium acetate — Colourless
Name the probable cation present based on the following observations:
(a) White precipitate insoluble in NH4 OH but soluble in NaOH.
(b) Blue coloured solution.
Answer
(a) Pb2+
(b) Cu2+
Name the metal hydroxides which are:
(a) Insoluble
(b) Soluble
in
(i) caustic soda solution
(ii) Ammonium hydroxide solution
Answer
Name of the solution Soluble metal hydroxide Insoluble metal hydroxide Caustic soda solution Zn(OH)2 , Pb(OH)2 Fe(OH)2 , Fe(OH)3 , Cu(OH)2 Ammonium hydroxide solution Zn(OH)2 , Cu(OH)2 Fe(OH)2 , Fe(OH)3 , Pb(OH)2
What do you observe when ammonium salt is heated with caustic soda solution? Write the word equation:
Answer
When ammonium salt is heated with caustic soda solution, ammonia gas is evolved.
The word equation is:
Ammonium Salt + Sodium Hydroxide → Δ \xrightarrow{\space\Delta\space} Δ Sodium Salt + Water + Ammonia Gas
How will you distinguish NH4 OH solution from NaOH solution?
Answer
NH4 OH and NaOH can be distinguished by using CuSO4 .
CuSO4 forms a pale blue precipitate which is insoluble in excess of sodium hydroxide and with ammonium hydroxide it forms a pale blue precipitate which dissolves in excess of ammonium hydroxide and forms a deep/inky blue solution.
CuSO 4 blue + 2 NaOH caustic soda - colourless ⟶ Cu(OH) 2 pale blue ppt ↓ + Na 2 SO 4 colourless \underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{caustic soda - colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4} blue CuSO 4 + caustic soda - colourless 2 NaOH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless Na 2 SO 4
CuSO 4 blue + 2 NH 4 OH ⟶ Cu(OH) 2 pale blue ppt ↓ + ( NH 4 ) 2 SO 4 colourless in solution \underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} blue CuSO 4 + 2 NH 4 OH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless in solution ( NH 4 ) 2 SO 4
Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Cu(NH 3 ) 4 ] SO 4 Tetraammine copper (II) sulphate + 4 H 2 O \text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{Tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ copper (II) sulphate Tetraammine [( Cu(NH 3 ) 4 ] SO 4 + 4 H 2 O
Why the alkali is added drop by drop to the salt solution?
Answer
If an alkali is added too quickly, then it is easy to miss a precipitate that redissolves in excess alkali.
Write balanced equation:
(a) Reaction of sodium hydroxide solution with Iron (III) chloride solution
(b) Copper sulphate solution with ammonium hydroxide solution
Answer
(a) When sodium hydroxide solution is added to FeCl3 dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:
FeCl 3 Yellow + 3 NaOH Colourless ⟶ Fe(OH) 3 ↓ Reddish brown ppt + 3 NaCl Colourless \underset{\text{Yellow}}{{\text{FeCl}_3}} + \underset{\text{Colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{Reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{Colourless}}{3\text{NaCl}} Yellow FeCl 3 + Colourless 3 NaOH ⟶ Reddish brown ppt Fe(OH) 3 ↓ + Colourless 3 NaCl
(b) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.
CuSO 4 Blue + 2 NH 4 OH ⟶ Cu(OH) 2 Pale Blue ppt ↓ + ( NH 4 ) 2 SO 4 Colourless in solution \underset{\text{Blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{Pale Blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{Colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} Blue CuSO 4 + 2 NH 4 OH ⟶ Pale Blue ppt Cu(OH) 2 ↓ + Colourless in solution ( NH 4 ) 2 SO 4
On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.
Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Cu(NH 3 ) 4 ] SO 4 Tetraammine copper (II) sulphate + 4 H 2 O \text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{Tetraammine copper (II) sulphate}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ Tetraammine copper (II) sulphate [( Cu(NH 3 ) 4 ] SO 4 + 4 H 2 O
Write the probable colour of the following salts?
(a) Ferrous salts
(b) Ammonium salts
(c) Cupric salts
(d) Calcium salts
(e) Aluminium salts
Answer
(a) Ferrous salts — Pale Green
(b) Ammonium salts — Colourless
(c) Cupric salts — Blue
(d) Calcium salts — Colourless
(e) Aluminium salts — Colourless
Name:
(a) a metallic hydroxide soluble in excess of NH4 OH.
(b) a metallic oxide soluble in excess of caustic soda solution.
(c) a strong alkali.
(d) a weak alkali.
(e) two colourless metal ions.
(f) two coloured metal ions.
(g) a metal that evolves a gas which burns with a pop sound when boiled with alkali solutions.
(h) two bases which are not alkalis but dissolve in strong alkalis.
(i) a coloured metallic oxide which dissolves in alkalis to yield colourless solutions.
(j) a colourless cation not a representative element.
Answer
(a) Copper hydroxide [Cu(OH)2 ]
(b) Zinc oxide [ZnO]
(c) Sodium hydroxide [NaOH]
(d) Ammonium hydroxide [NH4 OH]
(e) Sodium ion [Na+ ] , Potassium ion [K+ ]
(f) Cupric ion [Cu2+ ], Ferrous ion [Fe2+ ]
(g) Aluminium [Al]
(h) Zinc hydroxide [Zn(OH)2 ] and Lead hydroxide [Pb(OH)2 ]
(i) Lead oxide [PbO]
(j) Ammonium ion [NH4 + ]
Write balanced equations for Q.2 (g) and (i).
Answer
2(g):
2 Al + 2 NaOH + 2 H 2 O ⟶ 2 NaAlO 2 sodium meta aluminate [colourless] + 3 H 2 2\text{Al} + 2\text{NaOH} + 2\text{H}_2\text{O}\longrightarrow \underset{\underset{\text{[colourless]}}{\text{sodium meta aluminate}}}{2\text{NaAlO}_2} + 3\text{H}_2 2 Al + 2 NaOH + 2 H 2 O ⟶ [colourless] sodium meta aluminate 2 NaAlO 2 + 3 H 2
2(i):
PbO (yellow) + 2 NaOH ⟶ Na 2 PbO 2 sodium plumbite [colourless, soluble] + H 2 O \underset{\text{(yellow)}}{{\text{PbO}}} + 2\text{NaOH} \longrightarrow \underset{\underset{\text{[colourless, soluble]}}{\text{sodium plumbite}} }{\text{Na}_2\text{PbO}_2} + \text{H}_2\text{O} (yellow) PbO + 2 NaOH ⟶ [colourless, soluble] sodium plumbite Na 2 PbO 2 + H 2 O
What happens when ammonia solution is added first dropwise and then in excess to the following solutions:
(i) CuSO4
(ii) ZnSO4
(iii) FeCl3
Write balanced equations for these reactions.
Answer
(i) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.
CuSO 4 blue + 2 NH 4 OH ⟶ Cu(OH) 2 pale blue ppt ↓ + ( NH 4 ) 2 SO 4 colourless in solution \underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} blue CuSO 4 + 2 NH 4 OH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless in solution ( NH 4 ) 2 SO 4
On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.
Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Cu(NH 3 ) 4 ] SO 4 Tetraammine copper (II) sulphate + 4 H 2 O \text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{Tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ copper (II) sulphate Tetraammine [( Cu(NH 3 ) 4 ] SO 4 + 4 H 2 O
(ii) When ammonia solution is added dropwise to zinc sulphate, a white gelatinous ppt of zinc hydroxide is obtained.
ZnSO 4 colourless solution + 2 NH 4 OH ⟶ Zn(OH) 2 ↓ white gelatinous ppt + ( NH 4 ) 2 SO 4 colourless in solution \underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} colourless solution ZnSO 4 + 2 NH 4 OH ⟶ white gelatinous ppt Zn(OH) 2 ↓ + colourless in solution ( NH 4 ) 2 SO 4
On adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.
Zn(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Zn(NH 3 ) 4 ] SO 4 Tetraammine zinc (II) sulphate + 4 H 2 O \text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Zn(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ zinc (II) sulphate Tetraammine [( Zn(NH 3 ) 4 ] SO 4 + 4 H 2 O
(iii) When ammonia solution is added dropwise to iron (III) chloride, a reddish brown ppt. of Fe(OH)3 is obtained.
FeCl 3 yellow solution + 3 NH 4 OH ⟶ Fe(OH) 3 ↓ reddish brown ppt + 3 NH 4 Cl colourless in solution \underset{\text{yellow solution}}{{\text{FeCl}_3}} + 3\text{NH}_4\text{OH} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless in solution}}{3\text{NH}_4\text{Cl}} yellow solution FeCl 3 + 3 NH 4 OH ⟶ reddish brown ppt Fe(OH) 3 ↓ + colourless in solution 3 NH 4 Cl
Excess of ammonia solution addition doesn't dissolve ppt.
What do you observe when caustic soda solution is added to the following solution, first a little and then in excess :
(a) FeCl3
(b) ZnSO4
(c) Pb(NO3 )2
(d) CuSO4
Write balanced equations for these reactions.
Answer
(a) When caustic soda solution is added to FeCl3 dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:
FeCl 3 yellow + 3 NaOH colourless ⟶ Fe(OH) 3 ↓ reddish brown ppt + 3 NaCl colourless \underset{\text{yellow}}{{\text{FeCl}_3}} + \underset{\text{colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless}}{3\text{NaCl}} yellow FeCl 3 + colourless 3 NaOH ⟶ reddish brown ppt Fe(OH) 3 ↓ + colourless 3 NaCl
(b) When caustic soda solution is added to Zinc sulphate dropwise, a white gelatinous ppt is obtained, which dissolves in excess of NaOH:
ZnSO 4 colourless + 2 NaOH colourless ⟶ Zn(OH) 2 ↓ white gelatinous ppt + Na 2 SO 4 colourless \underset{\text{colourless}}{{\text{ZnSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{ colourless}}{\text{Na}_2\text{SO}_4} colourless ZnSO 4 + colourless 2 NaOH ⟶ white gelatinous ppt Zn(OH) 2 ↓ + colourless Na 2 SO 4
Zn(OH) 2 + 2 NaOH excess ⟶ Na 2 ZnO 2 ↓ colourless + 2 H 2 O {\text{Zn(OH)}_2} + \underset{\text{ excess}}{2\text{NaOH}} \longrightarrow \underset{\text{colourless}}{\text{Na}_2\text{ZnO}_2↓} + 2\text{H}_2\text{O} Zn(OH) 2 + excess 2 NaOH ⟶ colourless Na 2 ZnO 2 ↓ + 2 H 2 O
(c) When caustic soda solution is added to Pb(NO3 )2 dropwise, a chalky white ppt is obtained, which dissolves in excess of NaOH:
Pb(NO 3 ) 2 colourless + 2 NaOH colourless ⟶ Pb(OH) 2 ↓ white ppt + 2 NaNO 3 colourless \underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2↓} + \underset{\text{ colourless}}{2\text{NaNO}_3} colourless Pb(NO 3 ) 2 + colourless 2 NaOH ⟶ white ppt Pb(OH) 2 ↓ + colourless 2 NaNO 3
Pb(OH) 2 + 2 NaOH excess ⟶ Na 2 PbO 2 ↓ sodium plumbite - colourless + 2 H 2 O {\text{Pb(OH)}_2} + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\text{sodium plumbite - colourless}}{\text{Na}_2\text{PbO}_2↓} + 2\text{H}_2\text{O} Pb(OH) 2 + excess 2 NaOH ⟶ sodium plumbite - colourless Na 2 PbO 2 ↓ + 2 H 2 O
(d) When caustic soda solution is added to CuSO4 dropwise, a pale blue ppt is obtained, which is insoluble in excess of NaOH:
CuSO 4 blue + 2 NaOH colourless ⟶ Cu(OH) 2 ↓ pale blue ppt + Na 2 SO 4 colourless \underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2↓} + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4} blue CuSO 4 + colourless 2 NaOH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless Na 2 SO 4
Name the chloride of a metal which is soluble in excess of ammonium hydroxide. Write equation for the same.
Answer
Zinc chloride (ZnCl2 ) is soluble in excess of ammonium hydroxide.
When ammonia solution is added dropwise to zinc chloride solution, a white gelatinous ppt of zinc hydroxide is obtained.
ZnCl 2 colourless solution + 2 NH 4 OH ⟶ Zn(OH) 2 ↓ white gelatinous ppt + 2 NH 4 Cl \underset{\text{colourless solution}}{{\text{ZnCl}_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + 2\text{NH}_4\text{Cl} colourless solution ZnCl 2 + 2 NH 4 OH ⟶ white gelatinous ppt Zn(OH) 2 ↓ + 2 NH 4 Cl
On adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.
Zn(OH) 2 + 2 NH 4 Cl + 2 NH 4 OH ⟶ [ ( Zn(NH 3 ) 4 ] Cl 2 Tetraammine zinc (II) sulphate + 4 H 2 O \text{Zn(OH)}_2 + 2\text{NH}_4\text{Cl} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{Cl}_2} + 4\text{H}_2\text{O} Zn(OH) 2 + 2 NH 4 Cl + 2 NH 4 OH ⟶ zinc (II) sulphate Tetraammine [( Zn(NH 3 ) 4 ] Cl 2 + 4 H 2 O
On adding dilute ammonia solution to a colourless solution of a salt, a white gelatinous precipitate appears. This precipitate however dissolves on addition of excess of ammonia solution. Identify (choose from Na, Al, Zn, Pb, Fe)
(a) Which metal salt solution was used?
(b) What is the formula of the white gelatinous precipitate obtained?
Answer
(a) Zinc (Zn) metal salt solution was used
(b) Zn(OH)2
Name:
(a) A yellow monoxide that dissolves in hot and concentrated caustic alkali.
(b) A white, insoluble oxide that dissolves when fused with caustic soda or caustic potash.
(c) A compound containing zinc in the anion.
Answer
(a) Lead oxide [PbO]
(b) Zinc oxide [ZnO]
(c) Potassium Zincate [K2 ZnO2 ]
Select the correct answer:
Colour of an aqueous solution of copper sulphate is:
Green Brown Blue Yellow Answer
BlueReason — Aqueous solution of copper sulphate is blue.
Select the correct answer:
Colour of the precipitate formed on adding NaOH solution to iron (II) sulphate solution is:
White Brown Green Pale blue Answer
GreenReason — Dirty green precipitate of Ferrous Hydroxide [Fe(OH)2 ] is formed.
Select the correct answer:
A metal which produces hydrogen on reacting with alkali as well as with acid:
Iron Magnesium Zinc Copper Answer
ZincReason — Reactions of Zinc with alkali and acid are shown below: Zn + 2NaOH ⟶ Na2 (ZnO)2 + H2 ↑ Zn + HCl ⟶ ZnCl2 + H2 ↑
Select the correct answer:
The salt solution which does not react with ammonium hydroxide is:
Calcium nitrate Zinc nitrate Lead nitrate Copper nitrate Answer
Calcium NitrateReason — No ppt. occurs even with addition of excess of ammonium hydroxide as the concentration of OH- ions from the ionization of of NH4 OH is so low that it cannot precipitate the hydroxide of calcium.
What do you observe when freshly precipitated aluminum hydroxide reacts with caustic soda solution? Give balanced equation.
Answer
When freshly precipitated aluminum hydroxide reacts with caustic soda solution, a white soluble salt of sodium meta aluminate is obtained.
Al(OH)3 + NaOH ⟶ NaAlO2 [soluble] + 2H2 O
You are provided with two reagent bottles marked A and B. One contains NH4 OH solution and the other contains NaOH solution. How will you identify them by a chemical test?
Answer
Reagent bottles A and B can identified by using Ca(NO3 )2 . When NH4 OH solution is added to Ca(NO3 )2 a white ppt is obtained.
Ca(NO 3 ) 2 colourless + 2 NaOH caustic soda - colourless ⟶ Ca(OH) 2 white ppt ↓ + 2 NaNO 3 colourless \underset{\text{colourless}}{{\text{Ca(NO}_3)_2}} + \underset{\text{caustic soda - colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white ppt}}{\text{Ca(OH)}_2} ↓ + \underset{\text{ colourless}}{2\text{NaNO}_3} colourless Ca(NO 3 ) 2 + caustic soda - colourless 2 NaOH ⟶ white ppt Ca(OH) 2 ↓ + colourless 2 NaNO 3
On the other hand, addition of NH4 OH solution to Ca(NO3 )2 gives no precipitate even when NH4 OH solution is added in excess. Thus, Ca(NO3 )2 can be used to distinguish between NH4 OH and NaOH solution.
Distinguish by adding: Sodium hydroxide solution or Ammonium hydroxide solution to
(a) Calcium salt solution and lead salt solution
(b) Lead nitrate solution and zinc nitrate solution
(c) Copper salt solution and ferrous salt solution
(d) Fe(II) salt solution and Fe(III) salt solution
(e) Ferrous nitrate and lead nitrate
Answer
(a) Ammonium hydroxide on reaction with lead salt solution gives chalky white precipitate of Pb(OH)2 . No precipitation occurs on adding Ammonium hydroxide to Calcium salt solution even when it is added in excess.
Pb(NO 3 ) 2 colourless + 2 NH 4 OH ⟶ Pb(OH) 2 white ppt ↓ + 2 NH 4 NO 3 \underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2} ↓ + 2\text{NH}_4\text{NO}_3 colourless Pb(NO 3 ) 2 + 2 NH 4 OH ⟶ white ppt Pb(OH) 2 ↓ + 2 NH 4 NO 3
(b) When ammonium hydroxide solution is added to each of the compounds, lead nitrate forms a chalky white precipitate of lead hydroxide [Pb(OH)2 ] which is insoluble in excess of ammonium hydroxide.
Pb(NO 3 ) 2 colourless + 2 NH 4 OH ⟶ Pb(OH) 2 white ppt ↓ + 2 NH 4 NO 3 \underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2} ↓ + 2\text{NH}_4\text{NO}_3 colourless Pb(NO 3 ) 2 + 2 NH 4 OH ⟶ white ppt Pb(OH) 2 ↓ + 2 NH 4 NO 3
Whereas a gelatinous white precipitate of zinc hydroxide [Zn(OH)2 ] is formed in case of zinc nitrate, which is soluble in excess of ammonium hydroxide.
Zn(NO3 )2 + 2NH4 OH ⟶ 2NH4 NO3 + Zn(OH)2 ↓
(c) On adding Sodium hydroxide to Copper salt pale blue coloured precipitate is obtained which is insoluble in excess of Sodium hydroxide. Ferrous salt solution gives a dirty green coloured precipitate with Sodium hydroxide which is insoluble in excess of NaOH.
CuSO 4 blue + 2 NaOH colourless ⟶ Cu(OH) 2 ↓ pale blue ppt + Na 2 SO 4 colourless \underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2↓} + \underset{\text{colourless}}{\text{ Na}_2\text{SO}_4} blue CuSO 4 + colourless 2 NaOH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless Na 2 SO 4
FeSO 4 pale green + 2 NaOH colourless ⟶ Fe(OH) 2 ↓ dirty green ppt + Na 2 SO 4 colourless \underset{\text{pale green}}{{\text{FeSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4} pale green FeSO 4 + colourless 2 NaOH ⟶ dirty green ppt Fe(OH) 2 ↓ + colourless Na 2 SO 4
(d) Sodium hydroxide on reaction with Fe(II) salt gives dirty green coloured precipitate, while with Fe(III) salt solution it forms reddish brown precipitate. Both precipitates are insoluble in excess NaOH.
Fe(II) salt:
FeSO 4 pale green solution + 2 NaOH ⟶ Fe(OH) 2 ↓ dirty green ppt + Na 2 SO 4 colourless in solution \underset{\text{pale green solution}}{{\text{FeSO}_4}} + 2\text{NaOH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless in solution}}{\text{Na}_2\text{SO}_4} pale green solution FeSO 4 + 2 NaOH ⟶ dirty green ppt Fe(OH) 2 ↓ + colourless in solution Na 2 SO 4
Fe(III) salt :
FeCl 3 yellow + 3 NaOH colourless ⟶ Fe(OH) 3 ↓ reddish brown ppt + 3 NaCl colourless \underset{\text{yellow}}{{\text{FeCl}_3}} + \underset{\text{colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless}}{3\text{NaCl}} yellow FeCl 3 + colourless 3 NaOH ⟶ reddish brown ppt Fe(OH) 3 ↓ + colourless 3 NaCl
(e) Ammonium hydroxide on reaction with lead nitrate gives a chalky white insoluble precipitate, and with ferrous nitrate forms a dirty green ppt.
Pb(NO 3 ) 2 colourless + 2 NH 4 OH ⟶ Pb(OH) 2 ↓ white ppt + 2 NH 4 NO 3 \underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2↓} + 2\text{NH}_4\text{NO}_3 colourless Pb(NO 3 ) 2 + 2 NH 4 OH ⟶ white ppt Pb(OH) 2 ↓ + 2 NH 4 NO 3
Fe(NO 3 ) 2 colourless + 2 NH 4 OH ⟶ Fe(OH) 2 ↓ dirty green ppt + 2 NH 4 NO 3 colourless \underset{\text{colourless}}{{\text{Fe(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{ colourless}}{2\text{NH}_4\text{NO}_3} colourless Fe(NO 3 ) 2 + 2 NH 4 OH ⟶ dirty green ppt Fe(OH) 2 ↓ + colourless 2 NH 4 NO 3
How will you distinguish calcium nitrate and zinc nitrate solution?
Answer
When ammonium hydroxide (NH4 OH) is added to zinc nitrate solution [Zn(NO3 )2 ], a gelatinous white ppt of zinc hydroxide [Zn(OH)2 ] is obtained which is soluble in excess of NH4 OH.
ZnSO 4 colourless solution + 2 NH 4 OH ⟶ Zn(OH) 2 ↓ white gelatinous ppt + ( NH 4 ) 2 SO 4 colourless in solution \underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} colourless solution ZnSO 4 + 2 NH 4 OH ⟶ white gelatinous ppt Zn(OH) 2 ↓ + colourless in solution ( NH 4 ) 2 SO 4
(With excess NH4 OH ppt. dissolves)
Zn(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Zn(NH 3 ) 4 ] SO 4 Tetraammine zinc (II) sulphate + 4 H 2 O \text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Zn(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ zinc (II) sulphate Tetraammine [( Zn(NH 3 ) 4 ] SO 4 + 4 H 2 O
On the other hand, calcium nitrate solution [Ca(NO3 )2 ] does not give any ppt. even when excess of ammonium hydroxide is added.
What is observed when hot concentrated caustic soda solution is added to
(a) Zinc
(b) Aluminium?
Write balanced equations.
Answer
(a) When hot concentrated caustic soda solution is added to zinc, soluble salt of sodium zincate [Na2 ZnO2 ] is formed and hydrogen gas is liberated.
The balanced equation is:
Zn + 2 NaOH hot and conc. ⟶ Na 2 ZnO 2 sodium zincate colourless + H 2 ↑ \text{Zn} + \underset{\text{hot and conc.}}{2\text{NaOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium zincate}}}{\text{Na}_2\text{ZnO}_2} + \text{H}_2↑ Zn + hot and conc. 2 NaOH ⟶ colourless sodium zincate Na 2 ZnO 2 + H 2 ↑
(b) When hot concentrated caustic soda solution is added to aluminium, soluble salt of sodium meta aluminate [NaAlO2 ] is formed and hydrogen gas is liberated.
2 Al + 2 NaOH + 2 H 2 O ⟶ 2 NaAlO 2 sodium meta aluminate colourless + 3 H 2 ↑ 2\text{Al} + 2\text{NaOH} + 2\text{H}_2\text{O} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium meta aluminate}}}{2\text{NaAlO}_2} + 3\text{H}_2↑ 2 Al + 2 NaOH + 2 H 2 O ⟶ colourless sodium meta aluminate 2 NaAlO 2 + 3 H 2 ↑
(a) What do you understand by amphoteric oxide?
(b) Give the balanced equations for the reaction with two different amphoteric oxides with a caustic alkali.
(c) Name the products formed.
Answer
(a) Amphoteric oxides and hydroxides are those compounds which react with both acids and alkalis to form salt and water.
(b) Balanced equations for the reaction of Zinc Oxide and Lead Oxide with Caustic Soda are given below:
ZnO + 2NaOH ⟶ Na2 ZnO2 + H2 O
PbO + 2NaOH ⟶ Na2 PbO2 + H2 O
(c) Sodium zincate [Na2 ZnO2 ] and sodium plumbite [Na2 PbO2 ] are the products formed.
Write balanced equations for the following conversions:
(a) Zn(SO) 4 → A Zn(OH) 2 → B Na 2 ZnO 2 \text{Zn(SO)}_4 \xrightarrow{\text{A}} \text{Zn(OH)}_2 \xrightarrow{\text{B}} \text{Na}_2\text{ZnO}_2 Zn(SO) 4 A Zn(OH) 2 B Na 2 ZnO 2
(b) Cu(SO) 4 → A Cu(OH) 2 → B [Cu(NH 3 ) 4 ] SO 4 \text{Cu(SO)}_4 \xrightarrow{\text{A}} \text{Cu(OH)}_2 \xrightarrow{\text{B}} \text{[Cu(NH}_3)_4]\text{SO}_4 Cu(SO) 4 A Cu(OH) 2 B [Cu(NH 3 ) 4 ] SO 4
Answer
(a) When sodium hydroxide solution is added dropwise to zinc sulphate, a white gelatinous ppt of zinc hydroxide is obtained.
ZnSO 4 colourless + 2 NaOH colourless ⟶ Zn(OH) 2 white gelatinous ppt ↓ + Na 2 SO 4 colourless \underset{\text{colourless}}{{\text{ZnSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2} ↓ + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4} colourless ZnSO 4 + colourless 2 NaOH ⟶ white gelatinous ppt Zn(OH) 2 ↓ + colourless Na 2 SO 4
On adding excess of NaOH solution, the ppt dissolves and a colourless solution is obtained.
Zn(OH) 2 + 2 NaOH excess ⟶ Na 2 ZnO 2 sodium zincate colourless + 2 H 2 O \text{Zn(OH)}_2 + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium zincate}}}{\text{Na}_2\text{ZnO}_2} + 2\text{H}_2\text{O} Zn(OH) 2 + excess 2 NaOH ⟶ colourless sodium zincate Na 2 ZnO 2 + 2 H 2 O
(b) When ammonia solution is added dropwise to copper sulphate, a pale blue ppt of copper hydroxide is obtained.
CuSO 4 blue + 2 NH 4 OH ⟶ Cu(OH) 2 pale blue ppt ↓ + ( NH 4 ) 2 SO 4 colourless in solution \underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4} blue CuSO 4 + 2 NH 4 OH ⟶ pale blue ppt Cu(OH) 2 ↓ + colourless in solution ( NH 4 ) 2 SO 4
On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.
Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ [ ( Cu(NH 3 ) 4 ] SO 4 tetraammine copper (II) sulphate + 4 H 2 O \text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O} Cu(OH) 2 + ( NH 4 ) 2 SO 4 + 2 NH 4 OH ⟶ copper (II) sulphate tetraammine [( Cu(NH 3 ) 4 ] SO 4 + 4 H 2 O