Analytical Chemistry Class 10 ICSE Important Questions with Answers

Answer: (b) $Fe^{3+}$Fe^3+.

Iron(III) salts give a reddish-brown precipitate of iron(III) hydroxide, $Fe(OH)_3$Fe(OH)₃, with sodium hydroxide, which is insoluble in excess. Iron(II) gives a dirty-green precipitate instead.

Answer: (c) $Cu^{2+}$Cu^2+.

Copper(II) ions give a pale-blue precipitate of $Cu(OH)_2$Cu(OH)₂; it is insoluble in excess $NaOH$NaOH but dissolves in excess $NH_4OH$NH₄OH to form a deep-blue soluble complex, which is characteristic of copper.

Answer: (b) $Zn^{2+}$Zn^2+.

Zinc hydroxide, $Zn(OH)_2$Zn(OH)₂, is white and dissolves in excess $NaOH$NaOH (forming sodium zincate) and also in excess $NH_4OH$NH₄OH (forming a soluble complex). Lead hydroxide dissolves only in excess $NaOH$NaOH, not in excess $NH_4OH$NH₄OH.

Answer: (b) Ammonium chloride.

This is the special action of sodium hydroxide on ammonium salts: $NH_4Cl+NaOH\xrightarrow{\Delta}NaCl+H_2O+NH_3\uparrow$NH₄Cl+NaOH→ (Δ)NaCl+H₂O+NH₃↑. Ammonia is the only common alkaline gas, so it turns moist red litmus blue.

Answer: (a) Both statements are true and R correctly explains A. $Pb(OH)_2$Pb(OH)₂ is white and amphoteric; excess $NaOH$NaOH dissolves it to form soluble sodium plumbite, so the precipitate redissolves.

(i) Iron(II) salt: a dirty-green precipitate of iron(II) hydroxide, $Fe(OH)_2$Fe(OH)₂, is formed.

(ii) Copper(II) salt: a pale-blue precipitate of copper(II) hydroxide, $Cu(OH)_2$Cu(OH)₂, is formed.

(i) $2Al+2NaOH+2H_2O\rightarrow2NaAlO_2+3H_2\uparrow$2Al+2NaOH+2H₂O→2NaAlO₂+3H₂↑ (sodium meta-aluminate)

(ii) $ZnO+2NaOH\rightarrow Na_2ZnO_2+H_2O$ZnO+2NaOH→ Na₂ZnO₂+H₂O (sodium zincate)

They show the amphoteric nature of aluminium, zinc (and lead) and their oxides and hydroxides: they react with both acids and alkalis.

(i) Zinc sulphate: a white precipitate of $Zn(OH)_2$Zn(OH)₂ forms; on adding excess $NaOH$NaOH it dissolves to give a colourless solution of sodium zincate.

(ii) Iron(III) chloride: a reddish-brown precipitate of $Fe(OH)_3$Fe(OH)₃ forms; it is insoluble in excess $NaOH$NaOH.

(iii) Calcium chloride: a white precipitate of $Ca(OH)_2$Ca(OH)₂ forms (often only with a fairly concentrated solution); it is insoluble in excess $NaOH$NaOH.

(i) Ferrous sulphate:
$FeSO_4 + 2NaOH \rightarrow Fe(OH)_2\downarrow + Na_2SO_4$FeSO₄ + 2NaOH → Fe(OH)₂↓ + Na₂SO₄
(dirty-green precipitate)

(ii) Copper sulphate:
$CuSO_4 + 2NaOH \rightarrow Cu(OH)_2\downarrow + Na_2SO_4$CuSO₄ + 2NaOH → Cu(OH)₂↓ + Na₂SO₄
(pale-blue precipitate)

(iii) Zinc hydroxide in excess $NaOH$NaOH:
$Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$Zn(OH)₂ + 2NaOH → Na₂ZnO₂ + 2H₂O
(white precipitate dissolves to form soluble sodium zincate)

P — copper sulphate. With $NH_4OH$NH₄OH it first gives a pale-blue precipitate of $Cu(OH)_2$Cu(OH)₂ which dissolves in excess to form a characteristic deep-blue soluble complex.

Q — zinc sulphate. It gives a white precipitate of $Zn(OH)_2$Zn(OH)₂ which dissolves in excess ammonium hydroxide to form a colourless soluble complex.

R — lead nitrate. It gives a white precipitate of $Pb(OH)_2$Pb(OH)₂ which is insoluble in excess ammonium hydroxide (lead hydroxide dissolves only in excess $NaOH$NaOH, not in $NH_4OH$NH₄OH).

Thus P = $CuSO_4$CuSO₄, Q = $ZnSO_4$ZnSO₄, R = $Pb(NO_3)_2$Pb(NO₃)₂.

The white hydroxides of zinc and lead are amphoteric and therefore dissolve in excess sodium hydroxide, whereas the coloured hydroxides of copper and iron do not.

(a) Both reagents supply $OH^{-}$OH^- ions that precipitate metal cations as their hydroxides. Since each metal hydroxide has a characteristic colour and a characteristic solubility in excess reagent, the colour of the precipitate together with its behaviour in excess $NaOH$NaOH or $NH_4OH$NH₄OH allows the cation to be identified.

(b) (i) $Cu^{2+}$Cu^2+ vs $Fe^{3+}$Fe^3+: With $NaOH$NaOH, copper gives a pale-blue precipitate and iron(III) a reddish-brown precipitate; the different colours distinguish them. Also, with excess $NH_4OH$NH₄OH the copper precipitate dissolves to a deep-blue solution while the iron(III) precipitate does not.

(ii) $Zn^{2+}$Zn^2+ vs $Pb^{2+}$Pb^2+: Both give white precipitates soluble in excess $NaOH$NaOH. Use ammonium hydroxide: in excess $NH_4OH$NH₄OH the zinc precipitate dissolves, but the lead precipitate remains insoluble.

(iii) $Fe^{2+}$Fe^2+ vs $Fe^{3+}$Fe^3+: With $NaOH$NaOH, iron(II) gives a dirty-green precipitate of $Fe(OH)_2$Fe(OH)₂ while iron(III) gives a reddish-brown precipitate of $Fe(OH)_3$Fe(OH)₃; the colours distinguish them.

(i) Cations present:

• A = $Fe^{3+}$Fe^3+ (reddish-brown, insoluble in both reagents).
• B = $Zn^{2+}$Zn^2+ (white precipitate, soluble in excess of both $NaOH$NaOH and $NH_4OH$NH₄OH).
• C = $Cu^{2+}$Cu^2+ (pale-blue precipitate; deep-blue solution with excess $NH_4OH$NH₄OH).
• D = $Pb^{2+}$Pb^2+ (white precipitate, soluble in excess $NaOH$NaOH but insoluble in excess $NH_4OH$NH₄OH).

(ii) In A the precipitate is iron(III) hydroxide, $Fe(OH)_3$Fe(OH)₃, which is reddish-brown.

(iii) The copper(II) ion, $Cu^{2+}$Cu^2+ (salt C), forms the deep-blue soluble complex with excess ammonium hydroxide.