Analytical Chemistry

Reagent bottles A and B can identified by using Ca(NO₃)₂.
When NH₄OH solution is added to Ca(NO₃)₂ a white ppt is obtained.

$\underset{\text{colourless}}{{\text{Ca(NO}_3)_2}} + \underset{\text{caustic soda - colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white ppt}}{\text{Ca(OH)}_2} ↓ + \underset{\text{ colourless}}{2\text{NaNO}_3}$

On the other hand, addition of NH₄OH solution to Ca(NO₃)₂ gives no precipitate even when NH₄OH solution is added in excess. Thus, Ca(NO₃)₂ can be used to distinguish between NH₄OH and NaOH solution.

(a) When sodium hydroxide solution is added dropwise to zinc sulphate, a white gelatinous ppt of zinc hydroxide is obtained.

$\underset{\text{colourless}}{{\text{ZnSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2} ↓ + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4}$

On adding excess of NaOH solution, the ppt dissolves and a colourless solution is obtained.

$\text{Zn(OH)}_2 + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium zincate}}}{\text{Na}_2\text{ZnO}_2} + 2\text{H}_2\text{O}$

(b) When ammonia solution is added dropwise to copper sulphate, a pale blue ppt of copper hydroxide is obtained.

$\underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.

$\text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

(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 ⟶ Na₂ZnO₂ + H₂O

PbO + 2NaOH ⟶ Na₂PbO₂ + H₂O

(c) Sodium zincate [Na₂ZnO₂] and sodium plumbite [Na₂PbO₂] are the products formed.

(a) Zinc (Zn) metal salt solution was used

(b) Zn(OH)₂

(c) The balanced equations are given below:

$\underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

With excess of NH₄OH ppt. dissolves

$\text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

Blue

Reason — Aqueous solution of copper sulphate is blue.

Green

Reason — Dirty green precipitate of Ferrous Hydroxide [Fe(OH)₂] is formed.

Zinc

Reason — Reactions of Zinc with alkali and acid are shown below:
Zn + 2NaOH ⟶ Na₂(ZnO)₂ + H₂↑
Zn + HCl ⟶ ZnCl₂ + H₂↑

Calcium Nitrate

Reason — No ppt. occurs even with addition of excess of ammonium hydroxide as the concentration of OH^- ions from the ionization of of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.

Both P and Q

Reason — On adding NaOH or KOH to lead nitrate, a white precipitate of lead hydroxide [Pb(OH)₂] is formed, which is insoluble in excess NaOH or KOH.

On adding NaOH or KOH to zinc nitrate, a white precipitate of zinc hydroxide [Zn(OH)₂] is formed, which is soluble in excess NaOH or KOH, forming a clear solution.

Even though NH₄OH can separate the two, it is not the most effective reagent because, although it produces a white precipitate with both, lead hydroxide does not dissolve in excess NH₄OH, while zinc hydroxide does.

Both A and R are true and R is the correct explanation of A.

Explanation — Calcium salt will not show precipitation even with addition of excess of NH₄OH. This is because the concentration of OH^- ions from the ionization of NH₄OH is so low that it cannot precipitate the hydroxide of calcium. Hence both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

A is true but R is false.

Explanation — When Iron (II) salt (Green colour) solution are reacted with ammonium hydroxide, a dirty green insoluble precipitate is formed. Hence the assertion (A) is true.

$\underset{\text{green}}{{\text{FeSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

Iron (II) salts are Green and Iron (III) salts are yellow in colour so reason (R) is false.

Both A and R are true but R is not the correct explanation of A.

Explanation — Certain metals like zinc, aluminium and lead react with hot concentrated caustic alkalis (NaOH, KOH) to give the corresponding soluble salt and liberate hydrogen.

$\text{Zn} + \underset{\text{hot and conc.}}{2\text{NaOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium zincate}}}{\text{Na}_2\text{ZnO}_2} + \text{H}_2↑$

$\text{Zn} + \underset{\text{hot and conc.}}{2\text{KOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{potassium zincate}}}{\text{K}_2\text{ZnO}_2} + \text{H}_2↑$

Hence the assertion (A) is true.

Alkalis are bases that are soluble in water. When dissolved in water, alkalis release hydroxide ions (OH⁻), making the solution basic (pH > 7). Hence reason (R) is true but it does not support assertion (A).

A is true but R is false.

Explanation — Oxides of most of the metals are basic in nature. They dissolve in water forming hydroxides (or alkalis).

For example:

$\underset{\text{sodium oxide}}{{\text{Na}_2\text{O}}} + \text{H}_2\text{O} \longrightarrow \underset{\text{sodium hydroxide}}{2\text{NaOH}}$

Hence the assertion (A) is true.

Reason (R) is false because, not all metal oxides dissolve in water to form alkalis. A few metallic oxides and hydroxides exhibit dual character, i.e., they show acidic as well as basic character. They are said to be amphoteric in nature.
For example : Copper(II) oxide (CuO)

CuO + H₂O ⟶ No reaction

Both A and R are true and R is the correct explanation of A.

Explanation — Zinc oxide reacts with both acids and concentrated alkalis (NaOH and KOH) forming salt and water.

ZnO + 2HCl ⟶ ZnCl₂ + H₂O

ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

Hence the assertion (a) is true.

Reason (R) is true because zinc oxide is amphoteric in nature, reacts with both acids and concentrated alkalis forming salt and water. Hence Both A and R are true and reason (R) is the correct explanation of assertion(A).

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.

$\underset{\text{colourless solution}}{{\text{ZnCl}_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + 2\text{NH}_4\text{Cl}$

On adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.

$\text{Zn(OH)}_2 + 2\text{NH}_4\text{Cl} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{Cl}_2} + 4\text{H}_2\text{O}$

(i) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.

$\underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.

$\text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{Tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

(ii) When ammonia solution is added dropwise to zinc sulphate, a white gelatinous ppt of zinc hydroxide is obtained.

$\underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

On adding excess of ammonia solution, the ppt dissolves and a colourless solution is obtained.

$\text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

(iii) When ammonia solution is added dropwise to iron (III) chloride, a reddish brown ppt. of Fe(OH)₃ is obtained.

$\underset{\text{yellow solution}}{{\text{FeCl}_3}} + 3\text{NH}_4\text{OH} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless in solution}}{3\text{NH}_4\text{Cl}}$

Excess of ammonia solution addition doesn't dissolve ppt.

(a) When caustic soda solution is added to FeCl₃ dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:

$\underset{\text{yellow}}{{\text{FeCl}_3}} + \underset{\text{colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless}}{3\text{NaCl}}$

(b) When caustic soda solution is added to Zinc sulphate dropwise, a white gelatinous ppt is obtained, which dissolves in excess of NaOH:

$\underset{\text{colourless}}{{\text{ZnSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{ colourless}}{\text{Na}_2\text{SO}_4}$

${\text{Zn(OH)}_2} + \underset{\text{ excess}}{2\text{NaOH}} \longrightarrow \underset{\text{colourless}}{\text{Na}_2\text{ZnO}_2↓} + 2\text{H}_2\text{O}$

(c) When caustic soda solution is added to Pb(NO₃)₂ dropwise, a chalky white ppt is obtained, which dissolves in excess of NaOH:

$\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2↓} + \underset{\text{ colourless}}{2\text{NaNO}_3}$

${\text{Pb(OH)}_2} + \underset{\text{excess}}{2\text{NaOH}} \longrightarrow \underset{\text{sodium plumbite - colourless}}{\text{Na}_2\text{PbO}_2↓} + 2\text{H}_2\text{O}$

(d) When caustic soda solution is added to CuSO₄ dropwise, a pale blue ppt is obtained, which is insoluble in excess of NaOH:

$\underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2↓} + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4}$

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] + 2H₂O

(a) When hot concentrated caustic soda solution is added to zinc, soluble salt of sodium zincate [Na₂ZnO₂] is formed and hydrogen gas is liberated.

The balanced equation is:

$\text{Zn} + \underset{\text{hot and conc.}}{2\text{NaOH}} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium zincate}}}{\text{Na}_2\text{ZnO}_2} + \text{H}_2↑$

(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.

$2\text{Al} + 2\text{NaOH} + 2\text{H}_2\text{O} \longrightarrow \underset{\underset{\text{colourless}}{\text{sodium meta aluminate}}}{2\text{NaAlO}_2} + 3\text{H}_2↑$

(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.

$\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2} ↓ + 2\text{NH}_4\text{NO}_3$

(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.

$\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2} ↓ + 2\text{NH}_4\text{NO}_3$

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 + 2NH₄OH ⟶ 2NH₄NO₃ + 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.

$\underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2↓} + \underset{\text{colourless}}{\text{ Na}_2\text{SO}_4}$

$\underset{\text{pale green}}{{\text{FeSO}_4}} + \underset{\text{colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4}$

(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:

$\underset{\text{pale green solution}}{{\text{FeSO}_4}} + 2\text{NaOH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{colourless in solution}}{\text{Na}_2\text{SO}_4}$

Fe(III) salt :

$\underset{\text{yellow}}{{\text{FeCl}_3}} + \underset{\text{colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{colourless}}{3\text{NaCl}}$

(e) Ammonium hydroxide on reaction with lead nitrate gives a chalky white insoluble precipitate, and with ferrous nitrate forms a dirty green ppt.

$\underset{\text{colourless}}{{\text{Pb(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white ppt}}{\text{Pb(OH)}_2↓} + 2\text{NH}_4\text{NO}_3$

$\underset{\text{colourless}}{{\text{Fe(NO}_3)_2}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{dirty green ppt}}{\text{Fe(OH)}_2↓} + \underset{\text{ colourless}}{2\text{NH}_4\text{NO}_3}$

When ammonium hydroxide (NH₄OH) 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 NH₄OH.

$\underset{\text{colourless solution}}{{\text{ZnSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{white gelatinous ppt}}{\text{Zn(OH)}_2↓} + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

(With excess NH₄OH ppt. dissolves)

$\text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{zinc (II) sulphate}}{\text{Tetraammine}}}{[(\text{Zn(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

On the other hand, calcium nitrate solution [Ca(NO₃)₂] does not give any ppt. even when excess of ammonium hydroxide is added.

(a) Cupric ion [Cu²⁺], Ferrous ion [Fe²⁺]

(b) Aluminium [Al]

(c) Zinc hydroxide [Zn(OH)₂] and Lead hydroxide [Pb(OH)₂]

(d) Lead oxide [PbO]

(e) Ammonium ion [NH₄⁺]

(f) Lead oxide [PbO]

(g) Zinc oxide [ZnO]

(h) Potassium Zincate [K₂ZnO₂]

(a) Ferrous salts — Pale Green

(b) Ammonium salts — Colourless

(c) Cupric salts — Blue

(d) Calcium salts — Colourless

(e) Aluminium salts — Colourless

(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.

(a) Iron (III) chloride — Yellow

(b) Potassium nitrate — Colourless

(c) Ferrous sulphate — Pale Green

(d) Aluminium acetate — Colourless

(a) Pb²⁺

(b) Cu²⁺

When ammonium salt is heated with caustic soda solution, ammonia gas is evolved.

The word equation is:

Ammonium Salt + Sodium Hydroxide $\xrightarrow{\space\Delta\space}$ Sodium Salt + Water + Ammonia Gas

NH₄OH 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.

$\underset{\text{blue}}{{\text{CuSO}_4}} + \underset{\text{caustic soda - colourless}}{2\text{NaOH}} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless}}{\text{Na}_2\text{SO}_4}$

$\underset{\text{blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{pale blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

$\text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\underset{\text{copper (II) sulphate}}{\text{Tetraammine}}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$

If an alkali is added too quickly, then it is easy to miss a precipitate that redissolves in excess alkali.

(a) When sodium hydroxide solution is added to FeCl₃ dropwise, a reddish brown ppt is obtained, which is insoluble in excess of NaOH:

$\underset{\text{Yellow}}{{\text{FeCl}_3}} + \underset{\text{Colourless}}{3\text{NaOH}} \longrightarrow \underset{\text{Reddish brown ppt}}{\text{Fe(OH)}_3↓} + \underset{\text{Colourless}}{3\text{NaCl}}$

(b) When ammonia solution is added dropwise to cupper sulphate, a pale blue ppt of copper hydroxide is obtained.

$\underset{\text{Blue}}{{\text{CuSO}_4}} + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{Pale Blue ppt}}{\text{Cu(OH)}_2} ↓ + \underset{\text{Colourless in solution}}{(\text{NH}_4)_2\text{SO}_4}$

On adding excess of ammonia solution, the ppt dissolves and a deep blue solution is obtained.

$\text{Cu(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \longrightarrow \underset{\text{Tetraammine copper (II) sulphate}}{[(\text{Cu(NH}_3)_4]\text{SO}_4} + 4\text{H}_2\text{O}$