Analytical Chemistry: Uses of Ammonium Hydroxide and Sodium Hydroxide — ICSE Class 10 Chemistry Important Questions

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 $NH_4OH$NH_4OH 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 $NH_4OH$NH_4OH (forming a soluble complex). Lead hydroxide dissolves only in excess $NaOH$NaOH, not in excess $NH_4OH$NH_4OH.

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

Calcium hydroxide is white and insoluble in excess of both reagents, unlike zinc (soluble in excess of both) and lead (soluble only in excess $NaOH$NaOH). Copper is ruled out as it gives a blue, not white, precipitate.

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.

Add ammonium hydroxide solution to each in turn. Both first give a white precipitate of the metal hydroxide.

On adding excess ammonium hydroxide: the white precipitate with zinc chloride dissolves to give a colourless solution (soluble complex), whereas the white precipitate with lead nitrate remains insoluble. This difference distinguishes the two solutions.

(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:
$FeSO_4 + 2NaOH \rightarrow Fe(OH)_2\downarrow + Na_2SO_4$FeSO_4 + 2NaOH → Fe(OH)_2 + Na_2SO_4
(dirty-green precipitate)

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

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

P — copper sulphate. With $NH_4OH$NH_4OH 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 $NH_4OH$NH_4OH).

Thus P = $CuSO_4$CuSO_4, Q = $ZnSO_4$ZnSO_4, R = $Pb(NO_3)_2$Pb(NO_3)_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 $NH_4OH$NH_4OH 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 $NH_4OH$NH_4OH 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 $NH_4OH$NH_4OH 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 $NH_4OH$NH_4OH).
• C = $Cu^{2+}$Cu^2+ (pale-blue precipitate; deep-blue solution with excess $NH_4OH$NH_4OH).
• D = $Pb^{2+}$Pb^2+ (white precipitate, soluble in excess $NaOH$NaOH but insoluble in excess $NH_4OH$NH_4OH).

(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.