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+CO₂→ CaCO₃↓+H₂O; the white $CaCO_3$CaCO₃ 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 $NH_4OH$NH₄OH forming the deep blue $[Cu(NH_3)_4]^{2+}$[Cu(NH₃)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 $NH_4OH$NH₄OH. $Zn^{2+}$Zn^2+ dissolves in both, distinguishing the two.

Answer: (a) Copper carbonate. $CuCO_3\xrightarrow{\Delta}CuO+CO_2\uparrow$CuCO₃→ (Δ)CuO+CO₂↑: green copper carbonate leaves black copper(II) oxide, and $CO_2$CO₂ turns lime water milky. Zinc carbonate (white) leaves a residue that is yellow when hot and white when cold; nitrates give brown fumes.

Answer: (a) Adding dilute $HCl$HCl first removes carbonate/sulphite, so only acid-insoluble white $BaSO_4$BaSO₄ 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 ($CO_2$CO₂ forms white $CaCO_3$CaCO₃). $CO_2$CO₂ also extinguishes a burning splint, whereas hydrogen does not turn lime water milky.

1. Litmus: the acid turns blue litmus red; the alkali turns red litmus blue.

2. Phenolphthalein: the alkali turns it pink; it stays colourless in the acid.

(Also: the acid gives brisk effervescence of $CO_2$CO₂ with sodium carbonate; methyl orange turns pink/red in the acid and yellow in the alkali.)

The brisk effervescence with dilute acid giving a gas that turns lime water milky shows the salt is a carbonate ($CO_3^{2-}$CO₃^2-); the gas is carbon dioxide ($CO_2$CO₂).

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow$CaCO₃+2HCl→ CaCl₂+H₂O+CO₂↑ (in the salt)
$Ca(OH)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O$Ca(OH)₂+CO₂→ CaCO₃↓+H₂O (milkiness)

Disappearance of milkiness with excess $CO_2$CO₂ (soluble bicarbonate forms):
$CaCO_3+H_2O+CO_2\rightarrow Ca(HCO_3)_2$CaCO₃+H₂O+CO₂→ Ca(HCO₃)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+AgNO₃→ AgCl↓+NaNO₃

Effect of $NH_4OH$NH₄OH: the white precipitate is readily soluble in ammonium hydroxide, forming a soluble complex. This solubility distinguishes chloride from a sulphate (whose $BaSO_4$BaSO₄ is insoluble) and confirms $Cl^-$Cl^-.

Anion: nitrate ($NO_3^-$NO₃^-). Reddish-brown fumes ($NO_2$NO₂) with oxygen (rekindles a glowing splint) on heating show a nitrate.

Cation: lead ($Pb^{2+}$Pb^2+). A residue that is reddish-brown when hot and yellow when cold is lead(II) oxide, $PbO$PbO. The white precipitate with dilute $HCl$HCl is lead chloride, $PbCl_2$PbCl₂.

Salt: lead nitrate, $Pb(NO_3)_2$Pb(NO₃)₂. The crackling (decrepitation) on heating is also typical of lead nitrate crystals.

(Zinc nitrate would instead leave a residue that is yellow when hot and white when cold, and would give no precipitate with dilute $HCl$HCl.)

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.

$NH_4^+$NH₄^+: no precipitate; on warming, ammonia gas is evolved (turns red litmus blue): $NH_4Cl+NaOH\rightarrow NaCl+H_2O+NH_3\uparrow$NH₄Cl+NaOH→ NaCl+H₂O+NH₃↑.

(i) Ammonium hydroxide ($NH_4OH$NH₄OH):

• $Zn^{2+}$Zn^2+: white ppt of $Zn(OH)_2$Zn(OH)₂ that dissolves in excess $NH_4OH$NH₄OH (forms $[Zn(NH_3)_4]^{2+}$[Zn(NH₃)₄]^2+).
• $Pb^{2+}$Pb^2+: white ppt of $Pb(OH)_2$Pb(OH)₂ that is insoluble in excess $NH_4OH$NH₄OH.

(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→ PbI₂↓+2K^+.

(iii) Dilute hydrochloric acid:

• $Zn^{2+}$Zn^2+: no precipitate (zinc chloride is soluble).
• $Pb^{2+}$Pb^2+: a white precipitate of $PbCl_2$PbCl₂ forms (soluble in hot water, reappears on cooling): $Pb^{2+}+2Cl^-\rightarrow PbCl_2\downarrow$Pb^2++2Cl^-→ PbCl₂↓.

Any of these clearly distinguishes the two cations; the yellow $PbI_2$PbI₂ and white $PbCl_2$PbCl₂ are especially characteristic of lead.

(i) The gas is chlorine, $Cl_2$Cl₂; X is manganese dioxide, $MnO_2$MnO₂.

(ii) $MnO_2+4HCl\xrightarrow{\Delta}MnCl_2+2H_2O+Cl_2\uparrow$MnO₂+4HCl→ (Δ)MnCl₂+2H₂O+Cl₂↑

(iii) Y is copper(II) oxide, $CuO$CuO: $CuO+2HCl\rightarrow CuCl_2+H_2O$CuO+2HCl→ CuCl₂+H₂O. Copper(II) chloride gives the blue-green solution.

(iv) Chlorine is a bleaching agent: with moisture it forms hypochlorous acid, which releases oxygen and bleaches the dye.