Frank Brothers Solutions for ICSE Class 10 Chemistry Chapter 4 Analytical Chemistry

Chapter 4. Analytical Chemistry

Solution 1:
1. Cuprous salts = Colourless
2. Cupric salts = Blue
3. Aluminium salts = Colourless
4. Ferrous salts = Light green
5. Ferric salts = Yellow
6. Calcium salts = Colourless
In simple words: Different metal salts have their own characteristic colors when dissolved in water. This helps chemists identify which metal is present in a solution.

📝 Teacher's Note: Use test tubes with actual salt solutions to demonstrate these colors. Students remember better when they see the vibrant blue of copper sulfate or the pale green of iron sulfate firsthand.

🎯 Exam Tip: Make a color chart and memorize it. Questions often ask you to identify metals based on their salt colors, so knowing this table perfectly is essential.

Solution 2:
(a) Ferrous ion and Ferric ion.
(i) \( \text{FeSO}_4 + 2\text{NH}_4\text{OH} \rightarrow \text{Fe(OH)}_2 \downarrow + (\text{NH}_4)_2\text{SO}_4 \)
Fe(OH)\( _2 \) forms dirty green precipitates.
(i) \( \text{Fe}_2(\text{SO}_4)_3 + 6\text{NH}_4\text{OH} \rightarrow 2\text{Fe(OH)}_3 \downarrow + 3(\text{NH}_4)_2\text{SO}_4 \)
Fe(OH)\( _3 \) forms reddish brown precipitates.
(b) Zinc ion and Lead ion.
(i) \( \text{ZnSO}_4 + 2\text{NH}_4\text{OH} \rightarrow \text{Zn(OH)}_2 \downarrow + (\text{NH}_4)_2\text{SO}_4 \)
Zn(OH)\( _2 \) forms white gelatinous precipitates. In the presence of excess of ammonium hydroxide these precipitates get dissolved.
(ii) \( \text{Pb(NO}_3)_2 + 2\text{NH}_4\text{OH} \rightarrow \text{Pb(OH)}_2 \downarrow + 2\text{NH}_4\text{NO}_3 \)
Pb(OH)\( _2 \) forms white precipitates. This precipitate is insoluble in the presence of excess of ammonium hydroxide
Concept Insight: Ammonium hydroxide forms insoluble hydroxides when treated with certain metallic salt solutions. The insoluble hydroxides thus formed get precipitated in the form of a precipitate and may be identified by their distinct colours.
In simple words: When ammonium hydroxide is added to metal salts, it forms colored precipitates that help us identify different metals. The color and behavior with excess reagent tells us which metal is present.

📝 Teacher's Note: Demonstrate the difference between Fe²⁺ (dirty green) and Fe³⁺ (reddish brown) precipitates side by side. Emphasize that zinc hydroxide dissolves in excess NH₄OH while lead hydroxide does not.

🎯 Exam Tip: Remember the key difference - zinc hydroxide dissolves in excess NH₄OH but lead hydroxide remains insoluble. This distinguishes Zn²⁺ from Pb²⁺ ions.

Solution 3:
By use of Ammonium hydroxide we can identify the ions PbCO\( _3 \), ZnCO\( _3 \) and CaCO\( _3 \) as:
(i) \( \text{PbCO}_3 + 2\text{NH}_4\text{OH} \rightarrow \text{Pb(OH)}_2 \downarrow + 2\text{NH}_4\text{NO}_3 \)
Pb(OH)\( _2 \) forms white precipitate which are insoluble in excess ammonium hydroxide.
(ii) \( \text{ZnCO}_3 + \text{NH}_4\text{OH} \rightarrow \text{Zn(OH)}_2 \downarrow + 2\text{Na}_2\text{CO}_3 \)
Zn(OH)\( _2 \) forms white gelatinous precipitate which are soluble in excess ammonium hydroxide.
(iii) \( \text{CaCO}_3 + \text{NH}_4\text{OH} \rightarrow \text{Ca(OH)}_2 + 2\text{Na}_2\text{CO}_3 \)
No precipitation of Ca(OH)\( _2 \) occurs even with addition of excess of NH\( _4 \)OH. Because the concentration of hydroxide ion from ammonium hydroxide is so low that it cannot precipitate the hydroxide of calcium.
Concept Insight: Some precipitated metallic hydroxides by ammonium hydroxide become soluble hydroxides when treated with excess of ammonium hydroxide due to the formation of a soluble complex salt in the presence of excess of ammonium hydroxide.
In simple words: Different carbonates react differently with ammonium hydroxide - lead gives insoluble white precipitate, zinc gives soluble white precipitate, and calcium shows no precipitation at all.

📝 Teacher's Note: Show students that calcium carbonate doesn't precipitate because NH₄OH is a weak base. Use this to explain the concept of relative strength of bases and their ability to precipitate different hydroxides.

🎯 Exam Tip: Key distinguishing feature: PbCO₃ → insoluble white ppt, ZnCO₃ → soluble white ppt in excess NH₄OH, CaCO₃ → no precipitation. Learn this pattern.

Solution 4:
K\( _2 \)SO\( _4 \).
In simple words: The compound is potassium sulfate, which is a common salt used in fertilizers and chemical reactions.

📝 Teacher's Note: This appears to be a simple identification answer. Connect it to the broader context of salt analysis and help students understand the systematic approach to identifying unknown salts.

🎯 Exam Tip: When writing chemical formulas, always check that the charges balance. K has +1 charge, SO₄ has -2 charge, so you need 2 K atoms.

Solution 5:
(a) Addition of caustic soda to FeCl\( _3 \) solution:
FeCl\( _3 \) + 3NaOH \( \rightarrow \) Fe(OH)\( _3 \downarrow \) + 3NaCl
Reddish Brown ppt.
(b) Addition of caustic soda to ZnSO\( _4 \) solution:
ZnSO\( _4 \) + 2NaOH \( \rightarrow \) Zn(OH)\( _2 \downarrow \) + Na\( _2 \)SO\( _4 \)
White gelatinous ppt.
(c) Addition of caustic soda to Pb(NO\( _3 \))\( _2 \) solution:
Pb(NO\( _3 \))\( _2 \) + 2NaOH \( \rightarrow \) Pb(OH)\( _2 \downarrow \) + 2NaNO\( _3 \)
White ppt.
(d) Addition of caustic soda to CuSO\( _4 \) solution:
CuSO\( _4 \) + 2NaOH \( \rightarrow \) Cu(OH)\( _2 \downarrow \) + Na\( _2 \)SO\( _4 \)
Pale blue ppt.
In simple words: When sodium hydroxide (caustic soda) is added to different metal salts, it forms different colored precipitates that help identify the metal present.

📝 Teacher's Note: Demonstrate these reactions in test tubes to show the distinct colors. Emphasize that NaOH is a stronger base than NH₄OH, so it precipitates hydroxides more readily.

🎯 Exam Tip: Remember the colors: Fe³⁺ gives reddish brown, Cu²⁺ gives pale blue, Zn²⁺ and Pb²⁺ both give white (but zinc's is gelatinous). Color coding helps in quick identification.

Solution 6:
The reaction of freshly precipitated aluminium hydroxide with caustic soda solution is as:
Al(OH)\( _3 \) + NaOH \( \rightarrow \) NaAlO\( _2 \) + 2H\( _2 \)O
Sodium aluminate
(White)
In simple words: Aluminum hydroxide dissolves in sodium hydroxide to form a clear solution of sodium aluminate, showing aluminum's amphoteric nature.

📝 Teacher's Note: This is a perfect example to explain amphoteric behavior. Aluminum hydroxide acts as an acid when it reacts with the strong base NaOH. Compare with previous reactions where metals just formed precipitates.

🎯 Exam Tip: Key point: Al(OH)₃ dissolves in excess NaOH, unlike other metal hydroxides that remain as precipitates. This distinguishes aluminum from other metals.

Solution 7:
Amphoteric oxides: Amphoteric oxides are those compounds, which react with both acids and alkalis to form salt and water.
For example: Oxides of Aluminium, zinc and lead are amphoteric in nature.
Balanced equations for the reaction of different amphoteric oxides with a caustic alkali:
Amphoteric oxide + Alkali \( \rightarrow \) Salt + Water
(a) ZnO + 2NaOH \( \rightarrow \) Na\( _2 \)ZnO\( _2 \) + H\( _2 \)O
White          Sodium zincate
                     (Colourless)
(b) Al\( _2 \)O\( _3 \) + 2NaOH \( \rightarrow \) NaAlO\( _2 \) + 2H\( _2 \)O
White          Sodium aluminate
                     (White)
(c) PbO + 2NaOH \( \rightarrow \) Na\( _2 \)PbO\( _2 \) + H\( _2 \)O
Yellow          Sodium plumbite
                     (colourless)
In simple words: Amphoteric oxides are special compounds that can react with both acids and bases, like aluminum oxide, zinc oxide, and lead oxide.

📝 Teacher's Note: Demonstrate with zinc oxide - show it dissolving in both HCl (forming ZnCl₂) and NaOH (forming Na₂ZnO₂). This dual behavior is what makes these oxides "amphoteric."

🎯 Exam Tip: Remember the three main amphoteric oxides: Al₂O₃, ZnO, and PbO. They react with both acids and bases to form salts, unlike normal oxides which react with only one type.

Solution 8:
(a) Zn(OH)\( _2 \) from Pb(OH)\( _2 \): Ammonium hydroxide (NH\( _4 \)OH) solution can separate Zn(OH)\( _2 \) from Pb(OH)\( _2 \) as Zn(OH)\( _2 \) precipitates are dissolved in excess of NH\( _4 \)OH solution while Pb(OH)\( _2 \) precipitates are insoluble in excess of NH\( _4 \)OH solution.
(b) CaO from PbO: Sodium hydroxide solution can separate CaO from PbO as CaO precipitates are sparingly soluble in excess of NaOH solution while PbO precipitates are soluble in excess of NaOH solution.
(c) CuO from ZnO: Sodium hydroxide solution can separate CuO from ZnO as CuO precipitates remains insoluble in excess of NaOH solution while ZnO precipitates are soluble in excess of NaOH solution.
In simple words: Different metal compounds can be separated by using their different solubility behaviors with excess amounts of reagents like ammonium hydroxide or sodium hydroxide.

📝 Teacher's Note: Use test tubes to show selective dissolution. Mix precipitates and then add excess reagent to show how one dissolves while the other remains. This teaches the principle of selective separation.

🎯 Exam Tip: Learn the solubility patterns: Zn compounds dissolve in excess NH₄OH and NaOH, Pb compounds only dissolve in excess NaOH, Cu compounds don't dissolve in either excess reagent.

Solution 9:
Examples of amphoteric hydroxides are: Zn(OH)\( _2 \), Al(OH)\( _3 \).
In simple words: Zinc hydroxide and aluminum hydroxide are special because they can act like both acids and bases depending on what they react with.

📝 Teacher's Note: Connect this to the previous solution about amphoteric oxides. Show students that both the oxides and hydroxides of these metals show amphoteric behavior. It's a characteristic property of these elements.

🎯 Exam Tip: Two main amphoteric hydroxides to remember: Zn(OH)₂ and Al(OH)₃. They dissolve in both strong acids and strong bases, forming different types of salts.

Solution 10:
(a) The powdered metal added to sodium hydroxide solution is Aluminium.
(b) The gas evolved is hydrogen.
(c) The salt present in the colorless solution is sodium aluminate (NaAlO\( _2 \)).
Concept Insight: The alloy of aluminium metal i.e. duraluminium finds use in the construction of aircrafts. Reaction of aluminium metal with sodium hydroxide is as:
2Al + 2NaOH + 2H\( _2 \)O \( \rightarrow \) 2NaAlO\( _2 \) + 3H\( _2 \)
In simple words: When aluminum metal is added to sodium hydroxide solution, it dissolves and produces hydrogen gas, forming a colorless solution of sodium aluminate.

📝 Teacher's Note: This is an unusual reaction - most metals don't dissolve in bases. Use this to highlight aluminum's special amphoteric nature. The hydrogen gas evolution can be tested with a burning splint.

🎯 Exam Tip: Remember this unique reaction of aluminum with NaOH. Unlike other metals that react with acids to give H₂, aluminum reacts with bases too, proving its amphoteric character.

Solution 11:
(a) Zn(OH)\( _2 \)
(b) Na\( _2 \)O
(c) NaOH
(d) NH\( _4 \)OH
(e) Cu\( ^{2+} \), Mn\( ^{2+} \)
(f) Zn(OH)\( _2 \) and Pb(OH)\( _2 \)
(g) PbO
In simple words: This appears to be a list of chemical formulas for various compounds including hydroxides, oxides, and ions commonly encountered in analytical chemistry.

📝 Teacher's Note: This seems to be answers to multiple questions about different chemical species. Review each formula with students and discuss their properties, especially focusing on solubility patterns and colors.

🎯 Exam Tip: When writing chemical formulas, always double-check the charges and make sure they balance. Practice writing formulas for common hydroxides, oxides, and salts frequently.

Solution 12:
Ammonia solution is NH\( _3 \) + H\( _2 \)O \( \rightarrow \) NH\( _4 \)OH
The general reaction of NH\( _4 \)OH with metal salt solutions is:
Salt solution + Ammonium hydroxide \( \rightarrow \) Metal hydroxide + salt formed in solution
(a) Dropwise addition of NH\( _4 \)OH:
CuSO\( _4 \) + 2NH\( _4 \)OH \( \rightarrow \) Cu(OH)\( _2 \) + (NH\( _4 \))\( _2 \)SO\( _4 \)
Blue                  Pale blue ppt.
With excess of NH\( _4 \)OH the precipitate of copper(II) hydroxide dissolves as:
Cu(OH)\( _2 \) + (NH\( _4 \))\( _2 \)SO\( _4 \) + 2NH\( _4 \)OH \( \rightarrow \) [Cu(NH\( _3 \))\( _4 \)]SO\( _4 \) + 4H\( _2 \)O
               In excess     Tetramine copper(II)
                                      Sulphate
                                  (Deep blue solution)
(b) Dropwise addition of NH\( _4 \)OH:
ZnSO\( _4 \) + 2NH\( _4 \)OH \( \rightarrow \) Zn(OH)\( _2 \) + (NH\( _4 \))\( _2 \)SO\( _4 \)
Colourless                White ppt.
With excess of NH\( _4 \)OH the precipitate of zinc(II) hydroxide dissolves as:
Zn(OH)\( _2 \) + (NH\( _4 \))\( _2 \)SO\( _4 \) + 2NH\( _4 \)OH \( \rightarrow \) [Zn(NH\( _3 \))\( _4 \)]SO\( _4 \) + 4H\( _2 \)O
               In excess     Tetramine zinc(II)
                                      Sulphate
                                  (colourless solution)
(c) Dropwise addition of NH\( _4 \)OH:
FeCl\( _3 \) + 3NH\( _4 \)OH \( \rightarrow \) Fe(OH)\( _3 \) + 3NH\( _4 \)Cl
Yellow                  (Dirty green ppt.)
With excess of NH\( _4 \)OH, the precipitate does not dissolve.
In simple words: When ammonium hydroxide is added dropwise to metal salts, precipitates form. Some dissolve in excess NH₄OH forming complex salts, while others remain insoluble.

📝 Teacher's Note: Demonstrate the dramatic color change when copper hydroxide dissolves in excess ammonia to form the deep blue tetramine complex. This is one of the most striking reactions in qualitative analysis.

🎯 Exam Tip: Key distinction: Cu²⁺ and Zn²⁺ hydroxides dissolve in excess NH₄OH forming colored/colorless complex ions respectively, but Fe³⁺ hydroxide remains insoluble. This is used to separate these ions.

Solution 13:
The chloride of a metal which is soluble in excess of ammonium hydroxide is zinc chloride i.e. ZnCl\( _2 \).
In simple words: Among metal chlorides, zinc chloride is special because when treated with excess ammonium hydroxide, the initially formed precipitate dissolves to form a clear solution.

📝 Teacher's Note: Connect this to the previous solution about zinc hydroxide formation and dissolution. Emphasize that this property is used to identify and separate zinc from other metals in qualitative analysis.

🎯 Exam Tip: Remember the sequence: ZnCl₂ + NH₄OH → Zn(OH)₂ (white precipitate) → excess NH₄OH → [Zn(NH₃)₄]Cl₂ (colorless solution). This behavior is unique to zinc among common metals.

Solution 14:

(a) Zinc (Zn) metal salt solution was used.

(b) The formula of white gelatinous precipitate is \( \text{Zn(OH)}_2 \).

Concept Insight: \( \text{ZnSO}_4 + 2\text{NH}_4\text{OH} \rightarrow \text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 \)

With excess of \( \text{NH}_4\text{OH} \) the precipitate of zinc(II) hydroxide dissolves as:

\( \text{Zn(OH)}_2 + (\text{NH}_4)_2\text{SO}_4 + 2\text{NH}_4\text{OH} \rightarrow [\text{Zn(NH}_3)_4]\text{SO}_4 + 4\text{H}_2\text{O} \)

📝 Teacher's Note: Show students how zinc hydroxide first forms as a white precipitate, then dissolves to form a complex ion when excess ammonia is added. This demonstrates both precipitation and complex formation in one reaction.

🎯 Exam Tip: Always mention both the initial precipitation and the dissolution in excess - this shows complete understanding of the two-step process and gets full marks.

Solution 15:

1. PbO

2. \( \text{Al}_2\text{O}_3 \)

3. \( \text{Na}_2\text{ZnO}_2 \)

📝 Teacher's Note: These are the products when metals are heated with sodium hydroxide - lead oxide, aluminum oxide, and sodium zincate respectively. Connect this to amphoteric behavior of metals.

🎯 Exam Tip: Remember the formulas exactly as written - \( \text{Na}_2\text{ZnO}_2 \) for sodium zincate is a common formula to get wrong in exams.

Solution 16:

1. transition, \( \text{Cr}^{3+} \), \( \text{Fe}^{2+} \), \( \text{MnO}_4^{4-} \)

2. \( \text{Zn(OH)}_2 \)

3. \( \text{NH}_4\text{Cl} \)

4. \( \text{Al}_2\text{O}_3 \), Al

5. \( \text{NH}_4\text{OH} \)

📝 Teacher's Note: These answers cover transition metal ions, hydroxides, salts, and oxides. Point out the oxidation states in the transition metal ions to students.

🎯 Exam Tip: Write the charges correctly for ions like \( \text{Cr}^{3+} \) and \( \text{Fe}^{2+} \) - the charge notation is essential for full marks.

Solution 1992-1:

1. Addition of KCN

2. Addition of excess of NaOH

3. Addition of excess of \( \text{NH}_4\text{OH} \)

📝 Teacher's Note: These are three different reagents that can be used to distinguish between metal ions based on their different reactions. Emphasize the word "excess" - it's crucial for complete reactions.

🎯 Exam Tip: Always specify "excess" when writing NaOH or \( \text{NH}_4\text{OH} \) as reagents - this shows you understand the need for complete reaction.

Solution 1993-1:

(a) Zinc nitrate solution from calcium nitrate solution:

(i) \( \text{Zn(NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Zn(OH)}_2 + 2\text{NaNO}_3 \)

On further addition of NaOH, \( \text{Zn(OH)}_2 \) dissolves.

(ii) \( \text{Ca(NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Ca(OH)}_2 + 2\text{NaNO}_3 \)

\( \text{Ca(OH)}_2 \) precipitates are sparingly soluble in excess of sodium hydroxide.

(b) Iron (II) chloride from iron (III) chloride

(i) \( \text{FeCl}_2 + 2\text{NaOH} \rightarrow \text{Fe(OH)}_2 + 2\text{NaCl} \)

\( \text{Fe(OH)}_2 \) precipitates are dirty green gelatinous in nature.

(i) \( \text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl} \)

\( \text{Fe(OH)}_3 \) precipitates are reddish brown in colour.

(c) Lead hydroxide from magnesium hydroxide.

When sodium hydroxide is added, lead hydroxide is dissolved in it but when sodium hydroxide is added to magnesium hydroxide, there is no visible reaction i.e. it remains insoluble.

📝 Teacher's Note: Focus on the color differences - dirty green for Fe(II) vs reddish brown for Fe(III). Also emphasize solubility differences - zinc and lead hydroxides dissolve in excess NaOH while calcium and magnesium do not.

🎯 Exam Tip: Always mention the colors of precipitates and their behavior in excess reagent - this shows complete observation skills and earns maximum marks.

Solution 1995-1:

1. The metal ion present in solution A is \( \text{Pb}^{2+} \).

2. The cation present in solution B is \( \text{Cu}^{2+} \). The probable colour of solution B is blue.

📝 Teacher's Note: Connect the blue color to the presence of copper(II) ions - this is a classic identification method in qualitative analysis.

🎯 Exam Tip: Always state both the ion and its characteristic color when identifying metal ions from their solutions.

Solution 1996-1:

When sodium hydroxide solution is added to zinc sulphate solution, till it is in excess white gelatinous precipitates of \( \text{Zn(OH)}_2 \) are formed and due to the excess of sodium hydroxide these ppt. get dissolved immidiately:

\( \text{ZnSO}_4 + 2\text{NaOH} \rightarrow \text{Zn(OH)}_2 + \text{Na}_2\text{SO}_4 \)

white gelatinous ppt.

\( \text{Zn(OH)}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + 2\text{H}_2\text{O} \)

colourless

📝 Teacher's Note: Demonstrate this reaction in class showing the temporary white precipitate that quickly dissolves. This illustrates the amphoteric nature of zinc hydroxide.

🎯 Exam Tip: Write both equations to show the complete process - formation of precipitate followed by dissolution. This demonstrates thorough understanding.

Solution 1996-2:

The solutions for the tests will be prepared by dissolving the given powders separately in water.

1. Solution of Calcium carbonate:

Calcium carbonate is \( \text{CaCO}_3 \) and contains \( \text{Ca}^{2+} \) ions. Sodium hydroxide solution NaOH can be used to identify \( \text{Ca}^{2+} \) since its addition to calcium carbonate solution will give white precipitates of \( \text{Ca(OH)}_2 \) which are sparingly soluble in excess of NaOH.

1. Solution of Lead carbonate:

Lead carbonate is \( \text{PbCO}_3 \) and contains \( \text{Pb}^{2+} \) ions. Ammonium hydroxide solution \( \text{NH}_4\text{OH} \) can be used to identify \( \text{Pb}^{2+} \) since its addition to lead carbonate solution will give white precipitates of \( \text{Pb(OH)}_2 \) which are insoluble in excess of \( \text{NH}_4\text{OH} \).

2. Solution of Zinc carbonate:

Zinc carbonate is \( \text{ZnCO}_3 \) and contains \( \text{Zn}^{2+} \) ions. Sodium hydroxide solution NaOH can be used to identify \( \text{Zn}^{2+} \) since its addition to zinc carbonate solution will give white gelatinous precipitates of \( \text{Zn(OH)}_3 \) which are soluble in excess of NaOH.

📝 Teacher's Note: Emphasize the different solubilities - calcium hydroxide is sparingly soluble, lead hydroxide is insoluble in \( \text{NH}_4\text{OH} \), while zinc hydroxide dissolves completely in excess NaOH.

🎯 Exam Tip: Always mention the specific reagent needed for each identification and whether the precipitate dissolves in excess - this shows complete analytical thinking.

Solution 1996-3:

For the reaction that will take place when copper sulphate solution is added to sodium hydroxide solution the equation is as:

\( \text{CuSO}_4 + 2\text{NaOH} \rightarrow \text{Cu(OH)}_2 + \text{Na}_2\text{SO}_4 \)

📝 Teacher's Note: Point out that copper hydroxide is blue in color and does not dissolve in excess NaOH, unlike zinc hydroxide. This helps students distinguish between different metal hydroxides.

🎯 Exam Tip: Always balance the chemical equation correctly - count atoms on both sides to ensure the equation is properly balanced.

Solution 1997-1:

a. (i) Sodium Hydroxide

(ii) Ammonium Hydroxide

📝 Teacher's Note: Use this table to teach systematic qualitative analysis - show how different reagents give different patterns of results that help identify unknown substances.

🎯 Exam Tip: Create clear tables like this in exams - it shows organized thinking and makes it easy for examiners to award marks for correct observations.

Solution 1998-1:

(a) Sodium hydroxide solution gives dirty green coloured precipitates with iron(II) sulphate solution.

With iron(III) sulphate solution sodium hydroxide solution gives reddish brown precipitates.

(b) When barium chloride solution is added to iron(II) sulphate solution it gives white precipitate of \( \text{BaSO}_4 \).

(c) Sodium carbonate + hydrochloric acid = sodium chloride + water + carbon dioxide

\( \text{NaCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \)

Sodium sulphite + hydrochloric acid \( \rightarrow \) sodium chloride + water + hydrogen sulphide

\( 2\text{Na}_2\text{SO}_3 + 4\text{HCl} \rightarrow 4\text{NaCl} + 6\text{H}_2\text{O} + 2\text{H}_2\text{S} \)

Production of Foul smelling hydrogen sulphide gas will easily help to distinguish between sodium carbonate and sodium sulphite.

📝 Teacher's Note: Emphasize the smell test for distinguishing gases - \( \text{CO}_2 \) is odorless while \( \text{H}_2\text{S} \) has a characteristic rotten egg smell. This is a practical identification method.

🎯 Exam Tip: Always mention the distinctive properties like smell, color, or solubility that help distinguish between similar compounds - this shows practical knowledge.

Solution 1999-1:

(i) Sodium chloride solution and sodium nitrate solution.

Add freshly prepared ferrous sulphate solution to the two solutions. Then by the side of the test tube, pour concentrated sulphuric acid to each slowly. The one in which brown ring appears is sodium nitrate solution while the other is sodium chloride solution.

(ii) Sodium sulphate solution and sodium chloride solution.

(iii) Zinc nitrate solution from calcium nitrate solution:

a. \( \text{ZnNO}_3 + 2\text{NaOH} \rightarrow \text{Zn(OH)}_2 + \text{NaNO}_3 \)

White gelatinous ppt.

On further addition of NaOH, \( \text{Zn(OH)}_2 \) dissolves.

b. \( \text{Ca(NO}_3)_2 + 2\text{NaOH} \rightarrow \text{Ca(OH)}_2 + 2\text{NaNO}_3 \)

White ppt.

\( \text{Ca(OH)}_2 \) precipitates are sparingly soluble in excess of sodium hydroxide.

📝 Teacher's Note: Teach the brown ring test for nitrates systematically - it's a classic qualitative analysis technique. Show students how to layer the sulfuric acid carefully to get the ring at the interface.

🎯 Exam Tip: For the brown ring test, always mention "freshly prepared" ferrous sulfate and describe the layering technique - these details show proper laboratory knowledge.

Solution 2000-1:

(i) \( \text{FeCl}_2 + 2\text{NaOH} \rightarrow \text{Fe(OH)}_2 + 2\text{NaCl} \)

(ii) \( 2\text{NaOH} + \text{Cl}_2 \rightarrow \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O} \)

(iii) \( \text{Zn} + 2\text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{H}_2 \)

(iv) \( \text{SO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_3 + \text{H}_2\text{O} \)

📝 Teacher's Note: These equations show different types of reactions with sodium hydroxide - precipitation, disproportionation, amphoteric reaction, and acid-base neutralization respectively.

🎯 Exam Tip: Balance each equation carefully and check that charges balance in ionic equations - this prevents common mistakes in chemical equations.

Solution 2001-1:

(i) Neutral litmus solution turns blue in colour when added to alkaline solution.

(ii) \( \text{Fe}_2(\text{SO}_4)_3 + 6\text{NH}_4\text{OH} \rightarrow 2\text{Fe(OH)}_3 + 3(\text{NH}_4)_2\text{SO}_4 \)

Yellow                        Reddish brown

(iii) \( \text{Pb(NO}_3)_2 + 2\text{NaCl} \rightarrow \text{PbCl}_2 + 2\text{NaNO}_3 \)

(iv) Nothing is observed since ethane is a saturated hydrocarbon.

(v) Sulfur burns with a blue flame concomitant with formation of sulfur dioxide, notable for its peculiar suffocating odor.

📝 Teacher's Note: Connect the color changes to the formation of different compounds - yellow \( \text{Fe}_2(\text{SO}_4)_3 \) forms reddish brown \( \text{Fe(OH)}_3 \). Also explain why saturated hydrocarbons don't react with bromine water.

🎯 Exam Tip: Always mention "no reaction" for saturated hydrocarbons with bromine water - this shows you understand the difference between saturated and unsaturated compounds.

Solution 2003-1:

(i) \( \text{Zn}^{2+} \) ions on addition of \( \text{NH}_4\text{OH} \) forms white precipitates of \( \text{Zn(OH)}_2 \) which further dissolves in excess of \( \text{NH}_4\text{OH} \).

On the other hand, \( \text{Pb}^{2+} \) ions do form \( \text{Pb(OH)}_2 \) with ammonium hydroxides but these precipitates do not dissolve in excess of \( \text{NH}_4\text{OH} \).

(ii)

📝 Teacher's Note: Teach students to observe color changes during heating and cooling of carbonates - these are distinctive identification features for different metal carbonates.

🎯 Exam Tip: Always specify "on cooling" when describing the final color of heated carbonates - the color may change during heating but the question asks for the final state.

Solution 2003-2

Answer:

(i) Sodium hydroxide when added to zinc sulphate gives gelatinous white precipitate which dissolves in excess of sodium hydroxide.

\( ZnSO_4 + 2NaOH \rightarrow Zn(OH)_2 + Na_2SO_4 \)

\( Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 \)

(ii) When ammonium hydroxide is added in small quantity to copper sulphate solution, it gives blue precipitate of \( Cu(OH)_2 \).

\( CuSO_4 + 2NH_4OH \rightarrow Cu(OH)_2 + (NH_4)_2SO_4 \)

Blue                     Pale blue ppt.

When ammonium hydroxide is added in excess, the blue precipitate dissolves giving deep blue solution of tetra amine copper sulphate.

\( Cu(OH)_2 + (NH_4)_2SO_4 + 2NH_4OH \rightarrow [Cu(NH_3)_4]SO_4 + 4H_2O \)

In excess Tetrammine copper(II) Sulphate (Deep blue solution)

(iii) Curdy white precipitate of \( AgCl \) formed by reaction between hydrochloric acid and silver nitrate solution, dissolves in excess of \( NH_4OH \).

\( AgNO_3 + HCl \rightarrow AgCl + HNO_3 \)

\( AgCl + 2NH_4OH \rightarrow Ag(NH_3)_2Cl + 2H_2O \)

(iv) Starch paper turns blue black:

\( 2KI + Cl_2 \rightarrow 2KCl + I_2 \)

\( I_2 \) reacts with starch to give blue black colour.

The chlorine liberates iodine from \( KI \) and then it is decolourised.

(v) Pink colour of \( KMnO_4 \) is discharged.

\( 2KMnO_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 5O + 3H_2O \)          - (1)

\( SO_2 + H_2O + O \rightarrow H_2SO_4 \) x 5                                             - (2)

\( 2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4 \)

📝 Teacher's Note: Use simple precipitation and redox reactions to demonstrate these chemical tests. Show students how excess reagent can reverse or modify the initial reaction products. Visual demonstrations work best for color changes.

🎯 Exam Tip: Always write balanced chemical equations for each step. Mention the color of precipitates and solutions clearly as these are key identifying features examiners look for.

Solution 2004-1

Answer:

📝 Teacher's Note: Demonstrate this table practically in the lab. Students should observe that amphoteric hydroxides like Zn(OH)₂ and Pb(OH)₂ dissolve in excess NaOH, while basic hydroxides remain insoluble.

🎯 Exam Tip: Remember that zinc and lead hydroxides are amphoteric - they dissolve in excess NaOH. This is a common distinction question in exams.

Solution 2005-1

Answer:

1. B and E (Iron (II) sulphate and Magnesium sulphate)

2. C and F (Iron (III) chloride and Zinc chloride)

3. D (Lead nitrate)

4. A (Copper nitrate)

5. F (Zinc chloride)

📝 Teacher's Note: This appears to be a matching question. Help students understand the characteristic properties of each salt - color, solubility, and reaction patterns with common reagents.

🎯 Exam Tip: Learn the distinctive colors and properties of transition metal salts. Iron(II) gives pale green, Iron(III) gives yellow-brown, Copper(II) gives blue solutions.

Solution 2006-1

Answer:

📝 Teacher's Note: This is an excellent matching exercise for chemical properties. Ensure students understand the specific test for each substance and can identify compounds based on their characteristic reactions.

🎯 Exam Tip: Learn the key identifying tests: Chlorine turns starch-iodide paper blue, Nitrates produce brown NO₂ gas with concentrated H₂SO₄, Iron(II) gives dirty green precipitate with NaOH.

Solution 2009-1

Answer: C (Aluminium oxide)

📝 Teacher's Note: This appears to be a single-answer question. Emphasize to students the importance of understanding the context and properties that make aluminium oxide the correct choice.

🎯 Exam Tip: When given multiple choice answers, eliminate obviously incorrect options first, then focus on the distinguishing properties of the remaining choices.

Solution 2009-2

Answer:

1. P is Ferric chloride

2. Q is an ammonium salt

3. R is ferrous sulphate

📝 Teacher's Note: This appears to be a compound identification question. Help students learn to identify compounds based on their chemical behavior and characteristic properties in different reactions.

🎯 Exam Tip: Learn the characteristic colors and reactions: Ferric chloride is yellow-brown, ferrous sulphate is pale green, ammonium salts release ammonia gas when heated with alkali.

Solution 2009-3

Answer:

1. When \( BaCl_2 \) solution is added to the given solution \( ZnSO_4 \) gives a white precipitate while no precipitate is obtained with \( ZnCl_2 \) solution.

2. When NaOH solution is added to the given solution, iron (II) chloride gives dirty green precipitate while reddish brown precipitate is obtained with iron(III) chloride.

📝 Teacher's Note: These are classic distinguishing tests. The first distinguishes between sulphate and chloride ions using barium chloride. The second distinguishes between iron(II) and iron(III) using the different colored hydroxide precipitates.

🎯 Exam Tip: Remember the test for sulphates: BaCl₂ gives white precipitate with sulphates but no precipitate with chlorides. For iron salts: Fe²⁺ gives dirty green, Fe³⁺ gives reddish brown precipitate with NaOH.