CBSE Class 10 Chemistry Qualitative Analysis of Salts MCQs Set 15

Answer: (a) dilute hydrochloric acid

Explanation:
1. Dilute HCl precipitates the chlorides AgCl, Hg\( _2 \)Cl\( _2 \) and PbCl\( _2 \), all of which are white.
2. The chlorides of other common metals are soluble, so these three metals are separated first.

Teacher's Note:
1. A group reagent precipitates a whole group of metal ions together.
2. Chlorides of silver, mercury(I) and lead are the insoluble chlorides.

Answer: (a) lead chloride

Explanation:
1. Lead chloride is sparingly soluble in cold water but quite soluble in hot water.
2. Silver chloride and mercurous chloride do not dissolve even in hot water.

Teacher's Note:
1. This helps to separate lead from the other two.
2. The precipitate re-forms on cooling.

Answer: (c) dissolves in ammonium hydroxide

Explanation:
1. AgCl is decomposed by light to give silver, which colours it violet.
2. It dissolves in ammonia, forming the soluble complex Ag(NH\( _3 \))\( _2 \)Cl, and is insoluble in dilute nitric acid.

Teacher's Note:
1. Photographic films use silver halides for this reason.
2. Mercurous chloride does not dissolve in ammonia.

Answer: (a) finely divided mercury is formed

Explanation:
1. Hg\( _2 \)Cl\( _2 \) reacts with ammonia to give a mixture of mercury(II) amino-chloride and finely divided black mercury.
2. The black colour distinguishes it from silver and lead chlorides.

Teacher's Note:
1. The name calomel is used for mercurous chloride.
2. Lead chloride neither dissolves nor turns black with ammonia.

Answer: (d) golden spangles

Explanation:
1. Pb(NO\( _3 \))\( _2 \) + 2KI \( \rightarrow \) PbI\( _2 \) + 2KNO\( _3 \).
2. Lead iodide is much more soluble in hot water than in cold, so it crystallises on cooling as shining golden spangles.

Teacher's Note:
1. This is a delicate test for lead.
2. Silver iodide and mercurous iodide behave differently.

Answer: (d) hydrogen sulphide in the presence of dilute hydrochloric acid

Explanation:
1. The sulphides of this group are insoluble even in dilute acid.
2. The acid keeps the concentration of sulphide ions low, so that sulphides of zinc, manganese and later groups do not precipitate.

Teacher's Note:
1. The precipitates are HgS, PbS and CuS.
2. Lead is mostly removed already as chloride in the first group.

Answer: (d) a pale blue precipitate that dissolves to a deep blue solution

Explanation:
1. A pale blue precipitate of basic copper sulphate forms first.
2. In excess ammonia it dissolves, forming the deep blue tetraammine copper(II) complex, which is a delicate test for copper.

Teacher's Note:
1. A deep blue solution means copper.
2. Zinc hydroxide also dissolves in excess ammonia but gives a colourless solution.

Answer: (b) a yellow then scarlet precipitate that dissolves in excess potassium iodide

Explanation:
1. HgCl\( _2 \) + 2KI \( \rightarrow \) HgI\( _2 \) + 2KCl; mercuric iodide is first yellow and then scarlet red.
2. With excess KI it dissolves as the complex K\( _2 \)HgI\( _4 \).

Teacher's Note:
1. Adding KOH and ammonia to this complex gives Nessler's reagent.
2. The red iodide identifies mercury.

Answer: (a) reddish-brown ferric hydroxide and white aluminium hydroxide

Explanation:
1. Fe\( ^{3+} \) gives reddish-brown Fe(OH)\( _3 \), and Al\( ^{3+} \) gives gelatinous white Al(OH)\( _3 \).
2. Ammonium chloride is added to limit the concentration of OH\( ^- \) so that hydroxides of later groups do not precipitate.

Teacher's Note:
1. The reagent for this group is NH\( _4 \)Cl and NH\( _4 \)OH.
2. Ferrous iron needs special tests.

Answer: (c) ammonium thiocyanate (sulphocyanide)

Explanation:
1. Fe\( ^{3+} \) + CNS\( ^- \) forms the blood-red complex ion.
2. Ferrous salts give no such colour.

Teacher's Note:
1. The colour disappears on adding mercuric chloride.
2. This is a very delicate test for ferric ions.

Answer: (d) potassium ferricyanide

Explanation:
1. Fe\( ^{2+} \) reacts with ferricyanide to give Turnbull's blue.
2. With ferric chloride, ferricyanide gives only a brown solution.

Teacher's Note:
1. Ferrous with ferricyanide and ferric with ferrocyanide both give deep blue products.
2. The ferric-ferrocyanide product is Prussian blue.

Answer: (d) potassium ferrocyanide

Explanation:
1. 3K\( _4 \)Fe(CN)\( _6 \) + 4FeCl\( _3 \) \( \rightarrow \) Fe\( _4 \)[Fe(CN)\( _6 \)]\( _3 \) + 12KCl, which is Prussian blue.
2. Ferrous salts give only a white or pale blue precipitate with ferrocyanide.

Teacher's Note:
1. Prussian blue is used as a pigment.
2. The reagent names look alike; ferro- refers to Fe(II) in the reagent.

Answer: (a) it is amphoteric and forms sodium aluminate

Explanation:
1. Al(OH)\( _3 \) behaves as an acid towards a strong base: Al(OH)\( _3 \) + NaOH \( \rightarrow \) NaAlO\( _2 \) + 2H\( _2 \)O.
2. Its solubility in both acids and alkalis shows its amphoteric nature.

Teacher's Note:
1. Zinc hydroxide behaves in the same way.
2. Ferric hydroxide does not dissolve in excess alkali.

Answer: (a) white and flesh (pink) coloured

Explanation:
1. H\( _2 \)S passed into an ammoniacal solution of zinc salt gives white ZnS; manganese gives flesh-coloured MnS.
2. The medium is made alkaline because these sulphides dissolve in acids.

Teacher's Note:
1. NH\( _4 \)Cl is added first so that the hydroxides are not precipitated.
2. Nickel and cobalt sulphides are black.

Answer: (d) sodium zincate

Explanation:
1. Zn(OH)\( _2 \) + 2NaOH \( \rightarrow \) Na\( _2 \)ZnO\( _2 \) + 2H\( _2 \)O.
2. Passing H\( _2 \)S through the zincate solution again gives white zinc sulphide.

Teacher's Note:
1. Zinc hydroxide is amphoteric.
2. The white ZnS confirms zinc.

Answer: (c) ammonium carbonate in the presence of ammonium chloride and ammonium hydroxide

Explanation:
1. The white carbonates CaCO\( _3 \) and BaCO\( _3 \) are precipitated.
2. Ammonium chloride stops magnesium from being precipitated and NH\( _4 \)OH keeps the solution slightly alkaline.

Teacher's Note:
1. The carbonates dissolve in acids with effervescence.
2. Strontium is also in this group.

Answer: (c) a brick-red flame

Explanation:
1. Calcium gives brick red, barium apple green, sodium golden yellow and potassium lilac (violet).
2. Chlorides are used because they are volatile.

Teacher's Note:
1. Moisten the salt with concentrated HCl before the test.
2. Observe the potassium flame through blue glass to cut out sodium yellow.

Answer: (d) a light (apple) green flame

Explanation:
1. Barium vapour emits green light in the flame.
2. Seen through an indigo prism or blue glass, the green colour still appears green.

Teacher's Note:
1. Copper salts give a bluish-green flame.
2. Strontium gives crimson red.

Answer: (c) barium chloride but not with calcium chloride

Explanation:
1. BaCl\( _2 \) + K\( _2 \)CrO\( _4 \) \( \rightarrow \) BaCrO\( _4 \) + 2KCl; barium chromate is insoluble.
2. Calcium chromate is soluble, so no precipitate forms.

Teacher's Note:
1. The precipitate dissolves in dilute HCl.
2. This distinguishes barium from calcium.

Answer: (c) barium chloride but not with calcium chloride

Explanation:
1. Barium sulphate is extremely insoluble and precipitates even from very dilute solutions.
2. Calcium sulphate is appreciably soluble, so it does not precipitate from dilute solution.

Teacher's Note:
1. In concentrated solutions both precipitate.
2. The solubility of sulphates decreases down the group.

Answer: (c) ammonium oxalate

Explanation:
1. CaCl\( _2 \) + (NH\( _4 \))\( _2 \)C\( _2 \)O\( _4 \) \( \rightarrow \) CaC\( _2 \)O\( _4 \) + 2NH\( _4 \)Cl.
2. Calcium oxalate is insoluble even in very dilute solutions.

Teacher's Note:
1. Calcium oxalate is formed in kidney stones.
2. Add ammonia to neutralise acid first.

Answer: (d) a white precipitate of magnesium ammonium phosphate

Explanation:
1. NH\( _4 \)Cl prevents the precipitation of magnesium hydroxide.
2. The Mg\( ^{2+} \) ions then form white crystalline MgNH\( _4 \)PO\( _4 \) with the phosphate.

Teacher's Note:
1. The precipitate forms slowly in dilute solution.
2. This is the usual test for magnesium.

Answer: (c) turns red litmus blue and gives dense white fumes with HCl

Explanation:
1. NH\( _4 \)Cl + NaOH \( \rightarrow \) NaCl + H\( _2 \)O + NH\( _3 \).
2. Ammonia is alkaline and forms white fumes of NH\( _4 \)Cl with hydrogen chloride.

Teacher's Note:
1. Nessler's reagent gives a brown precipitate with ammonia.
2. Turmeric paper also turns brown in ammonia.

Answer: (c) soluble calcium bicarbonate is formed

Explanation:
1. CO\( _2 \) + Ca(OH)\( _2 \) \( \rightarrow \) CaCO\( _3 \) + H\( _2 \)O (milky).
2. With excess CO\( _2 \): CaCO\( _3 \) + CO\( _2 \) + H\( _2 \)O \( \rightarrow \) Ca(HCO\( _3 \))\( _2 \), which is soluble.

Teacher's Note:
1. Boiling the clear solution makes it milky again.
2. This is the test for carbon dioxide.