Practical Work — ICSE Class 10 Chemistry Important Questions

Answer: (c) Carbon dioxide.

$Ca(OH)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O$Ca(OH)_2+CO_2→ CaCO_3+H_2O; the white $CaCO_3$CaCO_3 turns lime water milky.

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

$Cu^{2+}$Cu^2+ gives a pale blue $Cu(OH)_2$Cu(OH)_2 ppt that dissolves in excess $NH_4OH$NH_4OH forming the deep blue $[Cu(NH_3)_4]^{2+}$[Cu(NH_3)_4]^2+ complex.

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

$Pb^{2+}$Pb^2+ gives a white ppt soluble in excess $NaOH$NaOH (amphoteric) but insoluble in excess $NH_4OH$NH_4OH. $Zn^{2+}$Zn^2+ dissolves in both, distinguishing the two.

Answer: (b) Copper(II) sulphate pentahydrate.

$CuSO_4\cdot5H_2O\xrightarrow{\Delta}CuSO_4+5H_2O$CuSO_4·5H_2O→ (Δ)CuSO_4+5H_2O; blue crystals turn white and the water given off (turning anhydrous $CuSO_4$CuSO_4 blue) confirms water of crystallisation.

Answer: (a) Adding dilute $HCl$HCl first removes carbonate/sulphite, so only acid-insoluble white $BaSO_4$BaSO_4 confirms sulphate; R correctly explains A.

Hydrogen: bring a burning splint to the mouth of the tube - hydrogen burns with a 'pop' sound (and a pale blue flame).

Carbon dioxide: pass the gas through lime water - it turns milky ($CO_2$CO_2 forms white $CaCO_3$CaCO_3). $CO_2$CO_2 also extinguishes a burning splint, whereas hydrogen does not turn lime water milky.

Iron(III) chloride: a reddish-brown precipitate of $Fe(OH)_3$Fe(OH)_3 forms, which is insoluble in excess $NaOH$NaOH.

Zinc sulphate: a white precipitate of $Zn(OH)_2$Zn(OH)_2 forms, which dissolves in excess $NaOH$NaOH (amphoteric) to give a colourless sodium zincate solution.

The brisk effervescence with dilute acid giving a gas that turns lime water milky shows the salt is a carbonate ($CO_3^{2-}$CO_3^2-); the gas is carbon dioxide ($CO_2$CO_2).

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow$CaCO_3+2HCl→ CaCl_2+H_2O+CO_2 (in the salt)
$Ca(OH)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O$Ca(OH)_2+CO_2→ CaCO_3+H_2O (milkiness)

Disappearance of milkiness with excess $CO_2$CO_2 (soluble bicarbonate forms):
$CaCO_3+H_2O+CO_2\rightarrow Ca(HCO_3)_2$CaCO_3+H_2O+CO_2→ Ca(HCO_3)_2

Test for chloride ($Cl^-$Cl^-): To the salt solution add a little dilute nitric acid, then silver nitrate solution.

Observation: a white precipitate of silver chloride forms, which turns grey-violet on standing in light.
$NaCl+AgNO_3\rightarrow AgCl\downarrow+NaNO_3$NaCl+AgNO_3→ AgCl+NaNO_3

Effect of $NH_4OH$NH_4OH: the white precipitate is readily soluble in ammonium hydroxide, forming a soluble complex. This solubility distinguishes chloride from a sulphate (whose $BaSO_4$BaSO_4 is insoluble) and confirms $Cl^-$Cl^-.

Reddish-brown fumes on heating indicate a nitrate ($NO_2$NO_2 evolved). A residue that is yellow when hot and white on cooling is characteristic of zinc oxide ($ZnO$ZnO), so the cation is $Zn^{2+}$Zn^2+. A white precipitate with dilute $HCl$HCl is not given by zinc chloride; so this points to a mixed observation - the white precipitate with dilute HCl actually confirms $Pb^{2+}$Pb^2+ (forming $PbCl_2$PbCl_2).

Given the yellow-hot/white-cold residue is the stronger diagnostic for zinc, and lead nitrate also gives brown fumes, the salt is best identified as lead(II) nitrate, $Pb(NO_3)_2$Pb(NO_3)_2: cation $Pb^{2+}$Pb^2+ (white $PbCl_2$PbCl_2 ppt with dilute $HCl$HCl, soluble in hot water), anion nitrate ($NO_3^-$NO_3^-, brown $NO_2$NO_2 fumes).
$2Pb(NO_3)_2\xrightarrow{\Delta}2PbO+4NO_2\uparrow+O_2\uparrow$2Pb(NO_3)_2→ (Δ)2PbO+4NO_2+O_2; $Pb^{2+}+2Cl^-\rightarrow PbCl_2\downarrow$Pb^2++2Cl^-→ PbCl_2.

Action of $NaOH$NaOH solution on cations:

$Cu^{2+}$Cu^2+: pale blue precipitate of $Cu(OH)_2$Cu(OH)_2; insoluble in excess.

$Fe^{2+}$Fe^2+: dirty green precipitate of $Fe(OH)_2$Fe(OH)_2; insoluble in excess (turns brown on standing as it oxidises).

$Fe^{3+}$Fe^3+: reddish-brown precipitate of $Fe(OH)_3$Fe(OH)_3; insoluble in excess.

$Zn^{2+}$Zn^2+: white precipitate of $Zn(OH)_2$Zn(OH)_2; soluble in excess giving colourless sodium zincate.

$Pb^{2+}$Pb^2+: white precipitate of $Pb(OH)_2$Pb(OH)_2; soluble in excess giving sodium plumbite.

$Ca^{2+}$Ca^2+: white precipitate (sparingly), $Ca(OH)_2$Ca(OH)_2; insoluble in excess.

$NH_4^+$NH_4^+: no precipitate; on warming, ammonia gas is evolved (turns red litmus blue): $NH_4Cl+NaOH\rightarrow NaCl+H_2O+NH_3\uparrow$NH_4Cl+NaOH→ NaCl+H_2O+NH_3.

(i) Ammonium hydroxide ($NH_4OH$NH_4OH):

• $Zn^{2+}$Zn^2+: white ppt of $Zn(OH)_2$Zn(OH)_2 that dissolves in excess $NH_4OH$NH_4OH (forms $[Zn(NH_3)_4]^{2+}$[Zn(NH_3)_4]^2+).
• $Pb^{2+}$Pb^2+: white ppt of $Pb(OH)_2$Pb(OH)_2 that is insoluble in excess $NH_4OH$NH_4OH.

(ii) Potassium iodide (KI) solution:

• $Zn^{2+}$Zn^2+: no yellow precipitate (zinc iodide is soluble/colourless).
• $Pb^{2+}$Pb^2+: a bright yellow precipitate of lead iodide forms: $Pb^{2+}+2KI\rightarrow PbI_2\downarrow+2K^+$Pb^2++2KI→ PbI_2+2K^+.

(iii) Dilute hydrochloric acid:

• $Zn^{2+}$Zn^2+: no precipitate (zinc chloride is soluble).
• $Pb^{2+}$Pb^2+: a white precipitate of $PbCl_2$PbCl_2 forms (soluble in hot water, reappears on cooling): $Pb^{2+}+2Cl^-\rightarrow PbCl_2\downarrow$Pb^2++2Cl^-→ PbCl_2.

Any of these clearly distinguishes the two cations; the yellow $PbI_2$PbI_2 and white $PbCl_2$PbCl_2 are especially characteristic of lead.

(i) Gas Z is ammonia ($NH_3$NH_3) - pungent, turns red litmus blue, gives white fumes with conc. $HCl$HCl.

(ii) The ammonium ion ($NH_4^+$NH_4^+); ammonium salts release ammonia on warming with $NaOH$NaOH.

(iii) $NH_4Cl+NaOH\xrightarrow{\Delta}NaCl+H_2O+NH_3\uparrow$NH_4Cl+NaOH→ (Δ)NaCl+H_2O+NH_3

(iv) White fumes of ammonium chloride:
$NH_3+HCl\rightarrow NH_4Cl$NH_3+HCl→ NH_4Cl