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Chemistry - Analytical Chemistry

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Analytical Chemistry involves the determination of the chemical components of a given sample. In Grade 10 ICSE, the focus is on qualitative analysis using NaOHNaOH and NH4OHNH_4OH.

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Color of Salts: Most salts of representative elements (Group 1, 2, 13-17) are colorless, while salts of transition elements like Cu2+Cu^{2+} (blue), Fe2+Fe^{2+} (dirty green), and Fe3+Fe^{3+} (reddish-brown) are characteristically colored.

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Action of Sodium Hydroxide (NaOHNaOH): Used to identify cations. For example, Ca2+Ca^{2+} forms a white precipitate of Ca(OH)2Ca(OH)_2 which is insoluble in excess NaOHNaOH.

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Action of Ammonium Hydroxide (NH4OHNH_4OH): Used to distinguish between cations. For instance, Cu2+Cu^{2+} forms a pale blue precipitate that dissolves in excess NH4OHNH_4OH to form a deep blue (inky blue) solution of [Cu(NH3)4]SO4[Cu(NH_3)_4]SO_4.

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Amphoteric Oxides/Hydroxides: Oxides and hydroxides of metals like Zn,Pb,Zn, Pb, and AlAl react with both acids and strong alkalis. For example, ZnOZnO reacts with NaOHNaOH to form sodium zincate (Na2ZnO2Na_2ZnO_2).

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Action of Alkalis on Metals: Certain metals like Al,Zn,Al, Zn, and PbPb react with boiling concentrated NaOHNaOH or KOHKOH to liberate H2H_2 gas and form soluble salt complexes.

📐Formulae

Fe2++2OH−→Fe(OH)2↓(Dirty Green)Fe^{2+} + 2OH^- \rightarrow Fe(OH)_2 \downarrow (Dirty\ Green)

Fe3++3OH−→Fe(OH)3↓(Reddish Brown)Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow (Reddish\ Brown)

Cu2++2OH−→Cu(OH)2↓(Pale Blue)Cu^{2+} + 2OH^- \rightarrow Cu(OH)_2 \downarrow (Pale\ Blue)

Zn(OH)2+2NaOH→Na2ZnO2+2H2O(Soluble Sodium Zincate)Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O (Soluble\ Sodium\ Zincate)

Pb(OH)2+2NaOH→Na2PbO2+2H2O(Soluble Sodium Plumbite)Pb(OH)_2 + 2NaOH \rightarrow Na_2PbO_2 + 2H_2O (Soluble\ Sodium\ Plumbite)

Al2O3+2NaOH→2NaAlO2+H2O(Sodium Meta−aluminate)Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O (Sodium\ Meta-aluminate)

Zn+2NaOH→Na2ZnO2+H2↑Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow

💡Examples

Problem 1:

A substance 'X' is a salt of a transition metal. When NaOHNaOH solution is added to a solution of 'X', a dirty green precipitate is formed. Name the cation present in 'X' and write the chemical equation.

Solution:

The cation present is Fe2+Fe^{2+} (Ferrous ion). The equation is: FeSO4+2NaOH→Fe(OH)2↓+Na2SO4FeSO_4 + 2NaOH \rightarrow Fe(OH)_2 \downarrow + Na_2SO_4

Explanation:

The dirty green color of the precipitate is a specific characteristic of Iron(II) hydroxide, which is insoluble in excess Sodium Hydroxide.

Problem 2:

How can you distinguish between ZnSO4ZnSO_4 and Ca(NO3)2Ca(NO_3)_2 using NaOHNaOH solution?

Solution:

Add NaOHNaOH dropwise to both solutions. Ca(NO3)2Ca(NO_3)_2 forms a white precipitate of Ca(OH)2Ca(OH)_2 which remains insoluble in excess NaOHNaOH. ZnSO4ZnSO_4 forms a gelatinous white precipitate of Zn(OH)2Zn(OH)_2 which dissolves in excess NaOHNaOH to form a clear solution of sodium zincate.

Explanation:

Zinc is amphoteric and reacts with excess alkali to form a soluble complex, whereas Calcium is a group 2 metal and its hydroxide is not amphoteric.

Problem 3:

What is observed when excess NH4OHNH_4OH is added to a solution containing Cu2+Cu^{2+} ions?

Solution:

A pale blue precipitate of Cu(OH)2Cu(OH)_2 is first formed. On adding excess NH4OHNH_4OH, the precipitate dissolves to form a clear, deep blue (inky blue) solution of tetrammine copper(II) sulphate: Cu(OH)2+(NH4)2SO4+2NH4OH→[Cu(NH3)4]SO4+4H2OCu(OH)_2 + (NH_4)_2SO_4 + 2NH_4OH \rightarrow [Cu(NH_3)_4]SO_4 + 4H_2O

Explanation:

The formation of the soluble complex [Cu(NH3)4]2+[Cu(NH_3)_4]^{2+} is responsible for the deep blue color, a definitive test for Copper(II) ions.