Get notes, summary, questions and answers, MCQs, extras, competency-based questions and PDFs of Analytical Chemistry: ICSE Class 10 Chemistry (Concise/Selina). However, the notes should only be treated as references, and changes should be made according to the needs of the students.
Summary
Analytical chemistry helps us understand what substances are made of. One part of this is qualitative analysis, which is about identifying unknown substances. To do this, scientists use chemicals called reagents, which cause specific reactions. Alkalis, such as sodium hydroxide and ammonium hydroxide, are common reagents. When these alkalis are mixed with solutions containing metal salts, solid particles often form. These solids are called precipitates, and their appearance, especially their colour, helps to identify the metal present in the salt.
The original colour of a salt and its solution can also give clues. Salts of elements from certain groups in the periodic table are usually colourless. However, salts of ‘transition elements’ are often coloured. For instance, copper ions in solution are typically blue, while ferrous (iron II) ions are light green, and ferric (iron III) ions are yellowish-brown.
When sodium hydroxide solution is added slowly to different metal salt solutions, distinct changes occur. Calcium salts form a white precipitate that is only slightly soluble if more sodium hydroxide is added. Iron(II) salts produce a dirty green, gelatinous precipitate that does not dissolve in excess sodium hydroxide. Iron(III) salts yield a reddish-brown precipitate, also insoluble in excess. Copper salts give a pale blue precipitate, insoluble in excess. Zinc salts form a white, gelatinous precipitate that dissolves when more sodium hydroxide is added, creating a colourless solution of sodium zincate. Lead salts produce a chalky white precipitate, which also dissolves in excess sodium hydroxide to form sodium plumbite. If ammonium salts are warmed with sodium hydroxide, ammonia gas is released, which has a characteristic sharp smell.
Ammonium hydroxide solution behaves differently in some cases. With calcium salts, no precipitate forms because ammonium hydroxide is a weaker alkali. Iron(II) salts give a dirty green precipitate, and iron(III) salts give a reddish-brown precipitate; both are insoluble in excess ammonium hydroxide. Copper salts initially form a pale blue precipitate. However, if more ammonium hydroxide is added, this precipitate dissolves, and the solution turns a deep blue colour. This deep blue solution is a special identifying test for copper ions. Zinc salts form a white, gelatinous precipitate that dissolves in excess ammonium hydroxide, forming a colourless solution. Lead salts give a chalky white precipitate that is insoluble in excess ammonium hydroxide.
Certain metals, including zinc, aluminium, and lead, can react directly with hot, concentrated alkalis like sodium hydroxide. In these reactions, the metal dissolves, forming a soluble salt (like sodium zincate from zinc) and releasing hydrogen gas.
Some metal oxides and hydroxides show a dual nature; they can react with both acids and strong alkalis. These substances are called amphoteric. Zinc oxide, aluminium oxide, and lead oxide, along with their hydroxides, are examples. They react with acids to form a salt and water, and they also react with strong alkalis to form a different type of salt (like sodium aluminate from aluminium oxide) and water.
Workbook solutions (Concise/Solutions)
Intext Questions and Answers I
1. What do you understand by the following:
(a) Analysis
(b) Qualitative analysis
(c) Reagent
(d) Precipitation
Answer: (a) Analysis is the determination of the chemical components in a given sample.
(b) Qualitative analysis involves the identification of the unknown substances.
(c) A reagent is a substance that reacts with another substance.
(d) Precipitation is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.
2. Write the probable colour of the following salts:
(a) Iron (III) chloride
(b) Potassium nitrate
(c) Ferrous sulphate
(d) Aluminium acetate
Answer: (a) The probable colour of Iron (III) chloride (FeCl₃) solution is yellow.
(b) The probable colour of Potassium nitrate is colourless as it is a salt of a ‘representative element’.
(c) The probable colour of Ferrous sulphate (FeSO₄) is pale green.
(d) The probable colour of Aluminium acetate is colourless as it is a salt of a ‘representative element’.
3. Name the probable cation present based on the following observations:
(a) White precipitate insoluble in NH₄OH but soluble in NaOH
(b) Blue coloured solution.
Answer: (a) The probable cation present when a white precipitate is insoluble in NH₄OH but soluble in NaOH is Pb²⁺ (Lead ion) or Al³⁺ (Aluminium ion). Lead hydroxide (Pb(OH)₂) is a chalky white precipitate insoluble in excess NH₄OH but soluble in excess NaOH forming sodium plumbite. Aluminium hydroxide (Al(OH)₃) forms a gelatinous white precipitate with NaOH which is soluble in excess NaOH forming sodium meta aluminate; with NH₄OH, Al(OH)₃ forms a gelatinous white precipitate which is insoluble in excess NH₄OH.
(b) The probable cation present in a blue coloured solution is Cu²⁺ (Cupric ion).
4. Name the metal hydroxides which are:
(a) Insoluble
(b) Soluble
in (i) Caustic soda solution (ii) Ammonium hydroxide solution.
Answer: (a) Metal hydroxides which are insoluble in (i) Caustic soda solution and (ii) Ammonium hydroxide solution:
Ferrous hydroxide, Fe(OH)₂, is a dirty green precipitate, insoluble in excess NaOH and insoluble in excess NH₄OH.
Ferric hydroxide, Fe(OH)₃, is a reddish brown precipitate, insoluble in excess NaOH and insoluble in excess NH₄OH.
Copper hydroxide, Cu(OH)₂, is a pale blue precipitate, insoluble in excess NaOH.
Lead hydroxide, Pb(OH)₂, is a chalky white precipitate, insoluble in excess NH₄OH.
Calcium hydroxide, Ca(OH)₂, is a white precipitate, sparingly soluble in excess NaOH and no precipitation occurs with NH₄OH.
(b) Metal hydroxides which are soluble in (i) Caustic soda solution and (ii) Ammonium hydroxide solution:
Zinc hydroxide, Zn(OH)₂, is a gelatinous white precipitate, soluble in excess NaOH (forming Sodium zincate) and soluble in excess NH₄OH (forming Tetraammine zinc (II) sulphate or Tetraammine zinc hydroxide).
Lead hydroxide, Pb(OH)₂, is a chalky white precipitate, soluble in excess NaOH (forming Sodium plumbite).
Copper hydroxide, Cu(OH)₂, is a pale blue precipitate, soluble in excess NH₄OH (forming Tetraammine copper (II) sulphate or Tetraammine copper hydroxide, a deep blue solution).
5. What do you observe when ammonium salt is heated with caustic soda solution? Write the word equation.
Answer: When Sodium hydroxide (or any water soluble base) is heated with ammonium salts, ammonia gas is evolved.
The word equation is:
Ammonium chloride + Sodium hydroxide → Sodium chloride + Water + Ammonia
(NH₄)₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O + 2NH₃
6. How will you distinguish NH₄OH solution from NaOH solution?
Answer: To distinguish NH₄OH solution from NaOH solution, you can use a salt solution that reacts differently with each. For example:
Add the unknown alkali drop by drop and then in excess to a solution of Zinc salt (e.g., ZnSO₄).
With NaOH: A gelatinous white precipitate of Zn(OH)₂ forms, which is soluble in excess NaOH.
With NH₄OH: A gelatinous white precipitate of Zn(OH)₂ forms, which is soluble in excess NH₄OH.
Add the unknown alkali drop by drop and then in excess to a solution of Lead salt (e.g., Pb(NO₃)₂).
With NaOH: A chalky white precipitate of Pb(OH)₂ forms, which is soluble in excess NaOH.
With NH₄OH: A chalky white precipitate of Pb(OH)₂ forms, which is insoluble in excess NH₄OH.
Thus, using a lead salt solution can distinguish between NaOH and NH₄OH.
7. Why the alkali is added drop by drop to the salt solution?
Answer: If alkali is added too quickly, it is easy to miss a precipitate that redissolves in excess.
8. Write balanced equations:
(a) Reaction of sodium hydroxide solution with Iron (III) chloride solution.
(b) Copper sulphate solution with ammonium hydroxide solution.
Answer: (a) The balanced equation for the reaction of sodium hydroxide solution with Iron (III) chloride solution is:
FeCl₃ + 3NaOH → Fe(OH)₃↓ + 3NaCl
(yellow) (colourless) (reddish brown ppt.) (colourless)
(b) The balanced equations for Copper sulphate solution with ammonium hydroxide solution are:
CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + (NH₄)₂SO₄
(blue) (Pale blue ppt.) (colourless in solution)
With excess of NH₄OH ppt. dissolves:
Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O
(Tetraammine copper (II) sulphate, DEEP BLUE solution)
OR
Cu(OH)₂ + 4NH₄OH → Cu(NH₃)₄₂ + 4H₂O
(Tetraammine copper hydroxide, DEEP BLUE solution)
Exercise
MCQs
1. The colour of an aqueous solution of copper sulphate is:
(a) Green
(b) Brown
(c) Blue
(d) Yellow
Answer: (c) Blue
2. The colour of the precipitate formed on adding NaOH solution to iron (II) sulphate solution is :
(a) White
(b) Brown
(c) Green
(d) Pale blue
Answer: (c) Green
3. A metal which produces hydrogen on reacting with alkali as well as with acid is :
(a) Iron
(b) Magnesium
(c) Zinc
(d) Copper
Answer: (c) Zinc
4. The salt solution which does not react with ammonium hydroxide is :
(a) Calcium nitrate
(b) Zinc nitrate
(c) Lead nitrate
(d) Copper nitrate
Answer: (a) Calcium nitrate
5. Which of the following is the best reagent to distinguish lead nitrate and zinc nitrate ?
P NaOH solution
Q KOH solution
R NH₄OH solution
(a) Only P
(b) Only Q
(c) Only R
(d) Both P and Q
Answer: (c) Only R
6. Assertion (A): Calcium salt solution does not show any change even after adding an excess of ammonia solution to it.
Reason (R) : The low concentration of hydroxide ion in ammonium hydroxide solution which are unable to precipitate the hydroxide ions of calcium.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
(a) (1) (b) (2) (c) (3) (d) (4)
Answer: (a) (1)
7. Assertion (A): Iron (II) salt solution when reacted with ammonium hydroxide forms a dirty green precipitate.
Reason (R): Iron salts are brown in colour.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
(a) (1) (b) (2) (c) (3) (d) (4)
Answer: (c) (3)
8. Assertion (A): Hydrogen gas is liberated when metals like Zn, Al, Pb react with caustic alkalies.
Reason (R): Alkalies are soluble in water.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
(a) (1) (b) (2) (c) (3) (d) (4)
Answer: (b) (2)
9. Assertion (A): Oxides of most of the metals are basic in nature.
Reason (R): All metal oxides dissolve in water forming alkalies.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
(a) (1) (b) (2) (c) (3) (d) (4)
Answer: (c) (3)
10. Assertion (A): Zinc oxide reacts with acids as well as bases to form salt and water.
Reason (R): Zinc oxide is amphoteric in nature.
(1) Both A and R are true and R is the correct explanation of A.
(2) Both A and R are true but R is not the correct explanation of A.
(3) A is true but R is false.
(4) A is false but R is true.
(a) (1) (b) (2) (c) (3) (d) (4)
Answer: (a) (1)
Very Short Answer Type
1. Name:
(a) two coloured metal ions.
(b) a metal that evolves a gas which burns with a pop sound when boiled with alkali solutions.
(c) two bases which are not alkalis but dissolve in strong alkalis.
(d) a coloured metallic oxide which dissolves in alkalis to yield colourless solutions.
(e) a colourless cation not a representative element.
(f) a yellow monoxide that dissolves in hot and concentrated caustic alkali.
(g) a white, insoluble oxide that dissolves when fused with caustic soda or caustic potash.
(h) a compound containing zinc in the anion.
Answer: (a) Two coloured metal ions are Cu²⁺ (Blue) and Fe²⁺ (Light green) or Fe³⁺ (Yellowish brown).
(b) A metal that evolves a gas which burns with a pop sound when boiled with alkali solutions is Zinc (Zn) or Aluminium (Al) or Lead (Pb). For example, Zn + 2NaOH (Hot and conc.) → Na₂ZnO₂ + H₂.
(c) Two bases which are not alkalis but dissolve in strong alkalis are Zinc hydroxide (Zn(OH)₂) and Aluminium hydroxide (Al(OH)₃) or Lead hydroxide (Pb(OH)₂). These are amphoteric hydroxides.
(d) A coloured metallic oxide which dissolves in alkalis to yield colourless solutions is Lead oxide (PbO), which is yellow and dissolves in NaOH to form colourless Sodium plumbite (Na₂PbO₂).
(e) A colourless cation not a representative element is Zn²⁺ (Zinc ion) or Pb²⁺ (Lead ion).
(f) A yellow monoxide that dissolves in hot and concentrated caustic alkali is Lead oxide (PbO).
(g) A white, insoluble oxide that dissolves when fused with caustic soda or caustic potash is Aluminium oxide (Al₂O₃) or Zinc oxide (ZnO). For example, Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O (when fused, it would be 2Na₃AlO₃ if it’s fused alkali like 2Al + 6NaOH → 2Na₃AlO₃ + 3H₂). More directly, ZnO (white) + 2NaOH → Na₂ZnO₂ + H₂O.
(h) A compound containing zinc in the anion is Sodium zincate (Na₂ZnO₂) or Potassium zincate (K₂ZnO₂).
3. Write the probable colour of the following salts:
(a) Ferrous salts
(b) Ammonium salts
(c) Cupric salts
(d) Calcium salts
(e) Aluminium salts.
Answer: (a) The probable colour of Ferrous salts (Fe²⁺ ion) is pale green (e.g., Ferrous sulphate).
(b) The probable colour of Ammonium salts (NH₄⁺ ion) is colourless.
(c) The probable colour of Cupric salts (Cu²⁺ ion) is blue (e.g., Copper sulphate).
(d) The probable colour of Calcium salts (Ca²⁺ ion) is colourless.
(e) The probable colour of Aluminium salts (Al³⁺ ion) is colourless.
Short Answer Type
1. Name the chloride of a metal which is soluble in excess of ammonium hydroxide. Write the equation for the same.
Answer: Zinc chloride (ZnCl₂) solution would react with NH₄OH to form Zn(OH)₂, which is soluble in excess NH₄OH.
When ammonia solution is added dropwise to zinc chloride solution, a white, gelatinous precipitate of zinc hydroxide forms.
ZnCl₂ + 2NH₄OH → Zn(OH)₂↓ + 2NH₄Cl
When excess ammonia solution is added, the precipitate dissolves, forming a clear, colourless solution.
Zn(OH)₂ + 2NH₄OH + 2NH₄Cl → [Zn(NH₃)₄]Cl₂ + 4H₂O
2. What happens when ammonia solution is added first dropwise and then in excess to the following solutions :
(i) CuSO₄ (ii) ZnSO₄ (iii) FeCl₃.
Write balanced equations for these reactions.
Answer: (i) CuSO₄:
When ammonia solution (NH₄OH) is added dropwise to CuSO₄ solution (blue), a pale blue precipitate of Copper (II) hydroxide, Cu(OH)₂, is formed.
CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + (NH₄)₂SO₄
When excess NH₄OH is added, the pale blue precipitate dissolves to form a deep blue solution of Tetraammine copper (II) sulphate.
Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH → [Cu(NH₃)₄]SO₄ + 4H₂O
OR
Cu(OH)₂ + 4NH₄OH → Cu(NH₃)₄₂ + 4H₂O
(ii) ZnSO₄:
When ammonia solution (NH₄OH) is added dropwise to ZnSO₄ solution (colourless), a white, gelatinous precipitate of Zinc hydroxide, Zn(OH)₂, is formed.
ZnSO₄ + 2NH₄OH → Zn(OH)₂↓ + (NH₄)₂SO₄
When excess NH₄OH is added, the white, gelatinous precipitate dissolves to form a colourless solution of Tetraammine zinc (II) sulphate.
Zn(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH → [Zn(NH₃)₄]SO₄ + 4H₂O
OR
Zn(OH)₂ + 4NH₄OH → Zn(NH₃)₄₂ + 4H₂O
(iii) FeCl₃:
When ammonia solution (NH₄OH) is added dropwise to FeCl₃ solution (yellow solution), a reddish brown precipitate of Ferric hydroxide, Fe(OH)₃, is formed.
FeCl₃ + 3NH₄OH → Fe(OH)₃↓ + 3NH₄Cl
The reddish brown precipitate is insoluble in excess of NH₄OH.
3. What do you observe when caustic soda solution is added, first a little and then in excess, to the following solutions:
(a) FeCl₃
(b) ZnSO₄
(c) Pb(NO₃)₂
(d) CuSO₄
Write balanced equations for the above reactions.
Answer: (a) FeCl₃:
When caustic soda solution (NaOH) is added a little to FeCl₃ solution (yellow), a reddish brown precipitate of Ferric hydroxide, Fe(OH)₃, is formed.
FeCl₃ + 3NaOH → Fe(OH)₃↓ + 3NaCl
When NaOH is added in excess, the reddish brown precipitate remains insoluble.
(b) ZnSO₄:
When caustic soda solution (NaOH) is added a little to ZnSO₄ solution (colourless), a white, gelatinous precipitate of Zinc hydroxide, Zn(OH)₂, is formed.
ZnSO₄ + 2NaOH → Zn(OH)₂↓ + Na₂SO₄
When NaOH is added in excess, the white, gelatinous precipitate dissolves to form a colourless solution of Sodium zincate.
Zn(OH)₂ + 2NaOH (excess) → Na₂ZnO₂ + 2H₂O
(c) Pb(NO₃)₂:
When caustic soda solution (NaOH) is added a little to Pb(NO₃)₂ solution (colourless), a chalky white precipitate of Lead hydroxide, Pb(OH)₂, is formed.
Pb(NO₃)₂ + 2NaOH → Pb(OH)₂↓ + 2NaNO₃
When NaOH is added in excess, the chalky white precipitate dissolves to form a colourless solution of Sodium plumbite.
Pb(OH)₂ + 2NaOH (excess) → Na₂PbO₂ + 2H₂O
(d) CuSO₄:
When caustic soda solution (NaOH) is added a little to CuSO₄ solution (blue), a pale blue precipitate of Copper (II) hydroxide, Cu(OH)₂, is formed.
CuSO₄ + 2NaOH → Cu(OH)₂↓ + Na₂SO₄
When NaOH is added in excess, the pale blue precipitate remains insoluble.
4. What do you observe when freshly precipitated aluminium hydroxide reacts with caustic soda solution ? Give a balanced equation.
Answer: When freshly precipitated aluminium hydroxide reacts with caustic soda solution, the white gelatinous precipitate of aluminium hydroxide dissolves to form a colourless solution of sodium meta aluminate.
The balanced equation is: Al(OH)₃ + NaOH → NaAlO₂ + 2H₂O
5. What is observed when hot concentrated caustic soda solution is added to (a) Zinc and (b) Aluminium ? Write balanced equations.
Answer: (a) Zinc: When hot concentrated caustic soda solution (NaOH) is added to Zinc, the zinc metal reacts to form soluble Sodium zincate and liberates hydrogen gas, which burns with a pop sound.
Zn + 2NaOH (Hot and conc.) → Na₂ZnO₂ + H₂↑
(Sodium zincate (colourless))
(b) Aluminium: When hot concentrated (boiling) caustic soda solution (NaOH) is added to Aluminium, the aluminium metal reacts to form soluble Sodium meta aluminate and liberates hydrogen gas.
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂↑
(Sodium meta aluminate (colourless))
If aluminium reacts with fused alkali (e.g., 6NaOH), it produces sodium aluminate:
2Al + 6NaOH → 2Na₃AlO₃ + 3H₂↑
6. Distinguish by adding : Sodium hydroxide solution or ammonium hydroxide solution to
(a) Calcium salt solution and lead salt solution .
(b) Lead nitrate solution and zinc nitrate solution.
(c) Copper salt solution and ferrous salt solution.
(d) Fe(II) salt solution and Fe(III) salt solution.
(e) Ferrous nitrate and lead nitrate
Answer: (a) Calcium salt solution and lead salt solution:
Using Sodium hydroxide solution:
Calcium salt solution (e.g., Ca(NO₃)₂): Forms a white precipitate of Ca(OH)₂, sparingly soluble in excess NaOH.
Lead salt solution (e.g., Pb(NO₃)₂): Forms a chalky white precipitate of Pb(OH)₂, soluble in excess NaOH.
Using Ammonium hydroxide solution:
Calcium salt solution: No precipitation occurs even with addition of excess NH₄OH.
Lead salt solution: Forms a chalky white precipitate of Pb(OH)₂, insoluble in excess NH₄OH.
Ammonium hydroxide is a better distinguishing reagent here.
(b) Lead nitrate solution and zinc nitrate solution:
Using Sodium hydroxide solution:
Lead nitrate solution: Forms a chalky white precipitate of Pb(OH)₂, soluble in excess NaOH.
Zinc nitrate solution: Forms a gelatinous white precipitate of Zn(OH)₂, soluble in excess NaOH.
(Both are soluble in excess NaOH, so this is not ideal for distinguishing unless observing the nature of precipitate).
Using Ammonium hydroxide solution:
Lead nitrate solution: Forms a chalky white precipitate of Pb(OH)₂, insoluble in excess NH₄OH.
Zinc nitrate solution: Forms a gelatinous white precipitate of Zn(OH)₂, soluble in excess NH₄OH.
(Ammonium hydroxide is a better distinguishing reagent here).
(c) Copper salt solution and ferrous salt solution:
Using Sodium hydroxide solution:
Copper salt solution (e.g., CuSO₄): Forms a pale blue precipitate of Cu(OH)₂, insoluble in excess NaOH.
Ferrous salt solution (e.g., FeSO₄): Forms a dirty green, gelatinous precipitate of Fe(OH)₂, insoluble in excess NaOH.
Using Ammonium hydroxide solution:
Copper salt solution: Forms a pale blue precipitate of Cu(OH)₂, soluble in excess NH₄OH forming a deep blue solution.
Ferrous salt solution: Forms a dirty green precipitate of Fe(OH)₂, insoluble in excess NH₄OH.
(Both reagents can distinguish based on colour of precipitate and solubility in excess for NH₄OH).
(d) Fe(II) salt solution and Fe(III) salt solution:
Using Sodium hydroxide solution:
Fe(II) salt solution (e.g., FeSO₄): Forms a dirty green, gelatinous precipitate of Fe(OH)₂, insoluble in excess NaOH.
Fe(III) salt solution (e.g., FeCl₃): Forms a reddish brown precipitate of Fe(OH)₃, insoluble in excess NaOH.
Using Ammonium hydroxide solution:
Fe(II) salt solution: Forms a dirty green precipitate of Fe(OH)₂, insoluble in excess NH₄OH.
Fe(III) salt solution: Forms a reddish brown precipitate of Fe(OH)₃, insoluble in excess NH₄OH.
(Both reagents distinguish by the colour of the precipitate).
(e) Ferrous nitrate and lead nitrate:
Using Sodium hydroxide solution:
Ferrous nitrate solution: Forms a dirty green, gelatinous precipitate of Fe(OH)₂, insoluble in excess NaOH.
Lead nitrate solution: Forms a chalky white precipitate of Pb(OH)₂, soluble in excess NaOH.
Using Ammonium hydroxide solution:
Ferrous nitrate solution: Forms a dirty green precipitate of Fe(OH)₂, insoluble in excess NH₄OH.
Lead nitrate solution: Forms a chalky white precipitate of Pb(OH)₂, insoluble in excess NH₄OH.
(Sodium hydroxide is a better distinguishing reagent here due to solubility difference in excess).
7. How will you distinguish calcium nitrate and zinc nitrate solution?
Answer: To distinguish between calcium nitrate and zinc nitrate solution, add ammonium hydroxide solution drop by drop and then in excess.
With Calcium nitrate solution: No precipitation occurs even with addition of excess NH₄OH. This is because the concentration of OH⁻ ions from the ionisation of NH₄OH is so low that it cannot precipitate the hydroxide of calcium.
With Zinc nitrate solution: A gelatinous white precipitate of Zn(OH)₂ will form, which is soluble in excess NH₄OH, forming a colourless solution of Tetraammine zinc (II) complex.
Zn(NO₃)₂ + 2NH₄OH → Zn(OH)₂↓ + 2NH₄NO₃
Zn(OH)₂ + 4NH₄OH (excess) → Zn(NH₃)₄₂ + 4H₂O
Long Answer Type
1. You are provided with two reagent bottles marked A and B. One contains NH₄OH solution and the other contains NaOH solution. How will you identify them by a chemical test?
Answer: To identify which bottle contains NH₄OH solution and which contains NaOH solution, a chemical test can be performed using a solution of a lead salt, such as lead nitrate (Pb(NO₃)₂), or a zinc salt, such as zinc sulphate (ZnSO₄).
Test with Lead Nitrate solution:
Take a small amount of lead nitrate solution in two separate test tubes.
To the first test tube, add the solution from bottle A drop by drop, and then in excess.
To the second test tube, add the solution from bottle B drop by drop, and then in excess.
Observations:
If a chalky white precipitate of lead hydroxide (Pb(OH)₂) forms, which is soluble in excess of the added reagent, then that reagent is NaOH solution.
Pb(NO₃)₂ + 2NaOH → Pb(OH)₂↓ + 2NaNO₃
Pb(OH)₂ + 2NaOH (excess) → Na₂PbO₂ + 2H₂O (soluble sodium plumbite)
If a chalky white precipitate of lead hydroxide (Pb(OH)₂) forms, which is insoluble in excess of the added reagent, then that reagent is NH₄OH solution.
Pb(NO₃)₂ + 2NH₄OH → Pb(OH)₂↓ + 2NH₄NO₃ (precipitate insoluble in excess NH₄OH)
2. Write balanced equations for the following conversions
(a) ZnSO₄ →A Zn(OH)₂ →B Na₂ZnO₂.
(b) CuSO₄ →A Cu(OH)₂ →B [Cu(NH₃)₄]SO₄
Answer: (a) ZnSO₄ →A Zn(OH)₂ →B Na₂ZnO₂.
Conversion A: ZnSO₄ to Zn(OH)₂
ZnSO₄ + 2NaOH → Zn(OH)₂↓ + Na₂SO₄
(Reagent A is NaOH solution)
Conversion B: Zn(OH)₂ to Na₂ZnO₂
Zn(OH)₂ + 2NaOH (excess) → Na₂ZnO₂ + 2H₂O
(Reagent B is excess NaOH solution)
(b) CuSO₄ →A Cu(OH)₂ →B [Cu(NH₃)₄]SO₄
Conversion A: CuSO₄ to Cu(OH)₂
CuSO₄ + 2NH₄OH → Cu(OH)₂↓ + (NH₄)₂SO₄
(Reagent A is NH₄OH solution)
Conversion B: Cu(OH)₂ to [Cu(NH₃)₄]SO₄
Cu(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH (excess) → [Cu(NH₃)₄]SO₄ + 4H₂O
(Reagent B is excess NH₄OH solution in the presence of (NH₄)₂SO₄, or simply excess NH₄OH if written as Cu(OH)₂ + 4NH₄OH → Cu(NH₃)₄₂ + 4H₂O, then reaction with H₂SO₄ to get the sulphate, but the direct conversion is shown in the text with (NH₄)₂SO₄ present).
3. (a) What do you understand by amphoteric oxide ?
(b) Give the balanced equations for the reaction with two different amphoteric oxides with a caustic alkali.
(c) Name the products formed.
Answer: (a) Amphoteric oxides and hydroxides are those compounds which react with both acids and alkalis to form salt and water. They exhibit dual character, i.e., they show acidic as well as basic character.
(b) Balanced equations for the reaction of two different amphoteric oxides with a caustic alkali (e.g., NaOH):
1. Zinc oxide (ZnO) with NaOH:
ZnO + 2NaOH → Na₂ZnO₂ + H₂O
2. Aluminium oxide (Al₂O₃) with NaOH:
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
(c) The products formed are:
1. From ZnO and NaOH: Sodium zincate (Na₂ZnO₂) and Water (H₂O).
2. From Al₂O₃ and NaOH: Sodium meta aluminate (NaAlO₂) and Water (H₂O).
4. On adding dilute ammonia solution to a colourless solution of a salt, a white gelatinous precipitate appears. This precipitate however dissolves on addition of excess of ammonia solution.
(a) From the following list, identify which metal salt solution is used above ?
Na, Al, Zn, Pb, Fe
(b) What is the formula of the white gelatinous precipitate obtained ?
(c) Give the balanced equation(s) when sulphate of this metal reacts with ammonia solution in excess.
Answer: (a) From the list Na, Al, Zn, Pb, Fe, the metal salt solution used is a salt of Zinc (Zn).
Sodium (Na) salts do not precipitate with ammonia solution.
Aluminium (Al) salts form a white gelatinous precipitate (Al(OH)₃) with ammonia solution, which is insoluble in excess.
Lead (Pb) salts form a chalky white precipitate (Pb(OH)₂) with ammonia solution, which is insoluble in excess.
Iron (Fe) salts form coloured precipitates (Fe(OH)₂ – dirty green, Fe(OH)₃ – reddish brown) which are insoluble in excess ammonia solution.
Zinc (Zn) salts form a white gelatinous precipitate (Zn(OH)₂) with ammonia solution, which dissolves in excess ammonia solution.
(b) The formula of the white gelatinous precipitate obtained is Zn(OH)₂ (Zinc hydroxide).
(c) The balanced equation(s) when sulphate of this metal (Zinc sulphate, ZnSO₄) reacts with ammonia solution in excess:
Initial precipitation:
ZnSO₄ + 2NH₄OH → Zn(OH)₂↓ + (NH₄)₂SO₄
Dissolution in excess:
Zn(OH)₂ + (NH₄)₂SO₄ + 2NH₄OH (excess) → [Zn(NH₃)₄]SO₄ + 4H₂O
(Tetraammine zinc (II) sulphate, colourless solution)
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