Chemical Analysis

Properties of Sulphur Dioxide (SO₂)

The reaction between dilute HCl and sodium sulphite (Na₂SO₃) releases sulphur dioxide gas (SO₂).

• Chemical Nature: SO₂ acts as a powerful reducing agent.
• Visual Test: When exposed to a paper wetted with acidified potassium dichromate (K₂Cr₂O₇), it reduces the orange dichromate ion (Cr₂O₇²⁻) into the green chromium (III) ion (Cr³⁺).

Formation of Insoluble Sulphides

Let's analyze the products of each reaction:

• (a) H₂S + Lead (II) acetate: Produces Lead (II) sulphide (PbS), which is highly insoluble and forms a distinct black precipitate.
• (b), (c), and (d): These are all purely redox reactions that yield soluble aqueous products. They are characterized by color changes (e.g., removing iodine's color, turning dichromate green, or decolorizing permanganate), but no solid precipitates are formed.

Halide Reactions with Conc. H₂SO₄

Hot, concentrated sulfuric acid acts as both an acid and an oxidizing agent. Its reaction depends on the halide:

• Salt X (NaCl): Releases HCl gas. HCl is highly stable and difficult for sulfuric acid to oxidize further.
• Salt Y (NaI or NaBr): Releases HI or HBr gas. These are stronger reducing agents and are partially oxidized by the hot H₂SO₄ (e.g., HI rapidly oxidizes to release striking violet iodine vapors).

Identifying Anions by Precipitation Profiles

Let's evaluate the given chemical tests to find the anions:

• Test for X (Barium Chloride): BaCl₂ precipitates both phosphates and sulphates as white solids. However, Barium phosphate dissolves in dilute HCl, whereas Barium sulphate does not. Since X dissolves, X is Phosphate.
• Test for Y (Lead (II) Acetate): Lead acetate produces a white precipitate of Lead sulphate (PbSO₄) when added to sulphates. (If it were a sulphide, it would be black PbS). Thus, Y is Sulphate.

Identifying the Calcium Cation

We are looking for a metal cation that forms a white precipitate with BOTH sulfates and carbonates.

• With Sulfuric Acid: Calcium (Ca²⁺) reacts to form Calcium sulfate (CaSO₄), a white precipitate. (Note: Mg²⁺, Fe²⁺, and Cu²⁺ form soluble sulfates, meaning no precipitate).
• With Sodium Carbonate: Calcium reacts to form Calcium carbonate (CaCO₃), another classic white precipitate.

Limiting Reactant & Excess Concentration

Calculate the molar amounts to determine what is left over after the reaction completes.

1. Initial Moles:
   Moles H₂SO₄ = 0.2 L × 0.2 M = 0.04 mol.
   Moles Ca(OH)₂ = 0.3 L × 0.2 M = 0.06 mol.
2. Reaction & Excess: The balanced equation shows a 1:1 reacting ratio. Thus, 0.04 mol of acid reacts with exactly 0.04 mol of base. The unreacted base left over is: 0.06 - 0.04 = 0.02 mol Ca(OH)₂.
3. New Concentration: The final mixture volume is 200 mL + 300 mL = 500 mL (0.5 L).
   New Molarity = Moles / Volume = 0.02 mol / 0.5 L = 0.04 M.

Equilibrium Constant Expressions

A fundamental rule of writing Equilibrium Constants (Kc) is that pure solids (s) and pure liquids (l) are entirely excluded from the expression because their concentrations remain strictly constant.

• Reaction (b) Analysis: Both CaCO₃ (the reactant) and CaO (one of the products) are solids.
• Writing the Formula: Eliminating the solids leaves only the gaseous product. Therefore, the expression simplifies drastically to: Kc = [CO₂].

Reactivity with Dilute HCl

• With Na₂SO₃: HCl reacts to release sulfur dioxide gas (SO₂), which turns acidified potassium dichromate paper green:
  Na2SO3 + 2HCl → 2NaCl + H2O + SO2↑
• With Na₃PO₄: HCl cannot displace phosphate because phosphoric acid is more stable than hydrochloric acid (no observable gas is evolved).

No reaction occurs with pairs (b) or (c). Both salts in pair (d) produce CO₂ gas, making them indistinguishable with HCl alone.

Analysis of Distinguishing Factors

Gas Removal Chemistry

To selectively trap and eliminate H₂S and SO₂ gases:

1. H₂S removal: Reacts with Lead (II) acetate to precipitate solid, black Lead sulphide (PbS):
   (CH3COO)2Pb + H2S → PbS(s) + 2CH3COOH
2. SO₂ removal: Reacts with acidified potassium dichromate, which oxidizes SO₂ and reduces Cr(VI) to Cr(III), turning green:
   K2Cr2O7 + 3SO2 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O

Precipitation Reactions

Let's check the reactions for Calcium chloride (CaCl₂):

• Reaction with AgNO₃:
  CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s)↓ (white ppt) + Ca(NO3)2(aq)
• Reaction with (NH₄)₂CO₃:
  CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s)↓ (white ppt) + 2NH4Cl(aq)

Both conditions are satisfied.

Step-by-Step Reaction Mechanism

When excess sodium hydroxide (NaOH) is added to the mixture containing Fe³⁺ and Al³⁺:

1. Iron(III) reaction: Fe³⁺ reacts with OH^- to form a reddish-brown precipitate of Iron(III) hydroxide, which is insoluble in excess NaOH:
   Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)↓
2. Aluminium reaction: Al³⁺ initially precipitates as Al(OH)₃, but being amphoteric, it dissolves in excess NaOH to form soluble sodium meta-aluminate:
   Al(OH)3(s) + OH-(aq) → AlO2-(aq) + 2H2O(l)
3. Spectators: Na⁺ and Cl^- ions remain in solution unchanged.

Step 1: Calculate KOH Molarity

Step 2: Titration Math

Reaction: H2SO4 + 2KOH → K2SO4 + 2H2O

Gravimetric Analysis Steps

Detailed Step-by-Step Analysis

1. Identifying Salt X (Carbonate or Bicarbonate)

• Observation: A gas evolves that does not change the color of acidified potassium permanganate (KMnO₄).
• Chemical Deduction: Acidified KMnO₄ is a purple oxidizing agent. Gases like SO₂ or H₂S are reducing agents and would decolorize it. However, Carbon dioxide (CO₂) cannot be oxidized further and does not affect KMnO₄.
• Anion Identity: Salt X contains either Carbonate (CO₃^2-) or Bicarbonate (HCO₃^-).
• The Reaction Box:


CO₃^2-(s) + 2H⁺(aq) → H₂O(l) + CO₂(g) ↑



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2. Identifying Salt Y (Thiosulphate)

• Observation: A suspension forms in the solution and the evolved gas affects (decolorizes) the acidified potassium permanganate solution.
• Chemical Deduction: The formation of a yellow suspension of elemental sulfur (S) alongside the evolution of sulfur dioxide gas (SO₂) is the unique signature test for the thiosulphate anion. The SO₂ gas reduces the purple KMnO₄ to a colorless solution.
• Anion Identity: Salt Y contains Thiosulphate (S₂O₃^2-).
• The Reaction Box:


S₂O₃^2-(s) + 2H⁺(aq) → S(s) ↓ (yellow) + SO₂(g) ↑ + H₂O(l)



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3. Identifying Salt Z (Stable Acid Anion)

• Observation: No reaction occurs (no color change, no gas evolved).
• Chemical Deduction: Dilute hydrochloric acid (HCl) can only displace anions derived from less stable acids. A lack of reaction means the anion belongs to a group derived from more stable acids that HCl cannot displace.
• Anion Identity: Salt Z contains an anion from the conc. H₂SO₄ group (Cl^-, Br^-, I^-, NO₃^-) or the BaCl₂ group (SO₄^2-, PO₄^3-).

Detailed Step-by-Step Analysis

1. Identifying Solution W (Sodium sulphide)

• Observation: A black precipitate forms when lead(II) nitrate is added.
• Chemical Deduction: Lead ions (Pb²⁺) react immediately with sulphide ions (S^2-) to form an insoluble black precipitate of lead(II) sulphide (PbS).
• Anion Identity: Solution W must contain Sulphide (S^2-), narrowing our choices down to options B and C.
• The Reaction Box:


Pb²⁺(aq) + S^2-(aq) → PbS(s) ↓ (Black)

2. Identifying Solution Z (Sodium thiosulphate)

• Observation: It decolorizes the distinctive brown color of the iodine solution.
• Chemical Deduction: Iodine solution (I₂) is widely used as a standard volumetric reagent to detect thiosulphate (S₂O₃^2-). Thiosulphate reduces brown iodine molecules into colorless iodide ions (I^-).
• Anion Identity: Solution Z must be Sodium thiosulphate (Na₂S₂O₃), confirming option B as the only correct match.
• The Reaction Box:


2Na₂S₂O₃(aq) + I₂(aq) (Brown) → Na₂S₄O₆(aq) + 2NaI(aq) (Colorless)

3. Verifying Solution X (Ammonium carbonate)

• Observation: A precipitate forms when calcium hydroxide solution is introduced.
• Chemical Deduction: Calcium ions (Ca²⁺) from the hydroxide link up with carbonate ions (CO₃^2-) from the ammonium carbonate to yield a cloudy white precipitate of calcium carbonate (CaCO₃).
• The Reaction Box:


Ca²⁺(aq) + CO₃^2-(aq) → CaCO₃(s) ↓ (White)

4. Verifying Solution Y (Potassium bicarbonate)

• Observation: No precipitate forms upon the initial addition of calcium hydroxide solution.
• Chemical Deduction: All bicarbonate salts (HCO₃^-) are completely soluble in water, meaning no immediate precipitation occurs compared to the direct mixing of a carbonate salt with calcium.

1. Understanding the Two Gases

• Nitric oxide gas (NO): A colorless gas that is easily oxidized because nitrogen is in a +2 oxidation state.
• Hydrogen chloride gas (HCl): A colorless, acidic gas that dissolves readily in water to form hydrochloric acid.

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2. Why Option (a) Cannot Distinguish Between Them

• When passing either NO or HCl gas into a Sodium chloride solution (NaCl), no visible chemical reaction occurs.
• HCl gas will simply dissolve into the aqueous solution without producing any precipitate, color change, or distinct gas evolution.
• Therefore, it is completely ineffective for distinguishing between them.

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3. How the Other Options Successfully Distinguish the Gases

• (b) Exposing each separately to air:
  • NO reacts immediately with oxygen at the mouth of the tube to form reddish-brown fumes of nitrogen dioxide (NO₂).
  • HCl does not form reddish-brown fumes.


2NO(g) + O₂(g) → 2NO₂(g) (Reddish-brown)


• (c) Glass rod wet with ammonia:
  • HCl gas reacts instantly with ammonia gas (NH₃) to form dense white fumes of ammonium chloride (NH₄Cl).
  • NO does not react to form white fumes.


HCl(g) + NH₃(g) → NH₄Cl(s) (Dense White Fumes)


• (d) Acidified potassium permanganate solution:
  • Since NO acts as a reducing agent, it reduces the purple permanganate ions (MnO₄^-) and decolorizes the solution.
  • HCl gas does not show this decolorization effect under standard testing conditions.

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Detailed Step-by-Step Analysis

1. Identifying Acid (1) - Sulphuric Acid

• Observation: Orange vapors evolve when hydrogen bromide gas (HBr) is passed into hot concentrated acid.
• Chemical Deduction: Hot concentrated Sulphuric acid (H₂SO₄) acts as a strong oxidizing agent. It oxidizes colorless hydrogen bromide gas into orange-red bromine vapors (Br₂).
• The Reaction Box:


2HBr(g) + H₂SO₄(conc.) →^Δ SO₂(g) + 2H₂O(l) + Br₂(g) (Orange Vapors)



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2. Identifying Acid (2) - Hydrochloric Acid

• Observation: Dense white clouds form when a glass rod wet with ammonia solution is exposed to the acid vapors.
• Chemical Deduction: Volatile Hydrochloric acid (HCl) releases hydrogen chloride gas, which instantly combines with ammonia gas (NH₃) to form solid ammonium chloride particles suspended as white clouds.
• The Reaction Box:


HCl(g) + NH₃(g) → NH₄Cl(s) (White Clouds)



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3. Identifying Acid (3) - Phosphoric Acid

• Observation: A white precipitate forms upon adding barium nitrate solution.
• Chemical Deduction: Barium nitrate (Ba(NO₃)₂) reacts with phosphate ions from Phosphoric acid (H₃PO₄) to yield an insoluble white precipitate of barium phosphate (Ba₃(PO₄)₂). Nitric acid (Option d) would not react with barium nitrate.
• The Reaction Box:


2H₃PO₄(aq) + 3Ba(NO₃)₂(aq) → Ba₃(PO₄)₂(s) ↓ (White) + 6HNO₃(aq)

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Detailed Step-by-Step Analysis

1. Identifying the Secret Reagent

• The Clue: The question asks for the reagent used in the confirmatory test for one of the cations in the fifth analytical group.
• Fifth Group Cation: The main cation studied in this group is Calcium (Ca²⁺).
• The Confirmatory Test: The confirmatory test for Ca²⁺ involves adding dilute sulphuric acid (H₂SO₄), which forms a distinctive white precipitate of calcium sulphate (CaSO₄).
• Conclusion: The secret reagent we must use to test the salts is dilute H₂SO₄.

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2. Why Option (c) is the Correct Choice

• Anion Detection: Dilute H₂SO₄ is a strong, stable acid that can easily displace less stable acid anions, acting exactly like the dilute HCl group reagent. Therefore, it can detect the Nitrite (NO₂^-) anion in Pb(NO₂)₂ by evolving nitrous fumes.
• Cation Detection: Simultaneously, the sulphate ions (SO₄^2-) react with Lead(II) ions (Pb²⁺) to form a highly insoluble white precipitate. This dual action makes it a perfect detection test for the salt.
• The Combined Reaction Box:


Pb(NO₂)₂(aq) + H₂SO₄(dil.) → PbSO₄(s) ↓ (White) + 2HNO₂(aq)


• The produced nitrous acid (HNO₂) spontaneously decomposes to release nitric oxide gas (NO), which turns into reddish-brown NO₂ gas fumes at the mouth of the tube:


2NO(g) + O₂(g) → 2NO₂(g) ↑ (Reddish-Brown)



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3. Why the Other Options Are Incorrect

• (a) MgCl₂: Dilute acids cannot displace the chloride (Cl^-) ion; concentrated H₂SO₄ with heating is mandatory.
• (b) Fe₂(SO3)₃: This salt is chemically impossible/unstable in reality because Iron(III) (Fe³⁺) is an oxidizing agent that instantly reacts with the reducing sulphite (SO₃^2-) anion via an internal redox pathway.
• (d) Ba₃(PO4)₂: Barium phosphate is a highly insoluble salt that does not dissolve or react with dilute acids.

Step 1: Molar mass of Na₂CO₃·10H₂O = 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286 g/mol

Step 2: Moles = 14.3/286 = 0.05 mol in 500 mL → Molarity = 0.05/0.5 = 0.1 M

Step 3: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Moles of Na₂CO₃ in 25 mL = 0.1 × 0.025 = 0.0025 mol

Moles of HCl needed = 2 × 0.0025 = 0.005 mol

Concentration of HCl = 0.005/0.025 = 0.2 M

• Salt X (NaCl): Reacts with conc. H₂SO₄ to evolve colorless HCl gas.
• Salt Y (NaI): Reacts with conc. H₂SO₄ to evolve violet iodine (I₂) vapors, which turn paper wet with starch blue.

1. First, add Hydrochloric acid (HCl): This precipitates Pb²⁺ selectively as insoluble white PbCl₂, which is separated by filtration.
2. Next, add Sulphuric acid (H₂SO₄) to the filtrate: This precipitates Ca²⁺ as insoluble white CaSO₄.
3. The Cu²⁺ remains dissolved as soluble CuSO₄ in the solution, achieving full separation.

• Moles of AgCl precipitated = (0.963 g)/(143.5 g/mol) = 0.00671 moles.
• Since 1 mole of MCl yields 1 mole of AgCl, moles of MCl = 0.00671 moles.
• Molar mass of MCl = (0.5 g)/(0.00671 mol) = 74.5 g/mol.
• Atomic mass of M = 74.5 - 35.5 (Cl) = 39 g/mol (which is Potassium, K).

Step 1: Analyze the Clues

• The anion is X²⁻ (divalent). This eliminates Nitrite (NO₂⁻).
• The gas (Y) "oxidizes with normal oxidizing agents", which means Y is a reducing agent. This eliminates Carbonate (CO₂), as carbon is already in its highest oxidation state (+4).

Step 2: Evaluate Remaining Options

Sulphite (SO₃²⁻):

• Reaction: SO₃²⁻ + 2H⁺ → H₂O + SO₂↑
• Gas (Y) is SO₂.
• SO₂ is a known reducing agent and has a pungent odor.
• It famously reduces acidified potassium dichromate from orange (Cr₂O₇²⁻) to green (Cr³⁺). This matches the description perfectly.

Sulphide (S²⁻):

• Reaction: S²⁻ + 2H⁺ → H₂S↑
• Gas (Y) is H₂S.
• H₂S has a strong, unpleasant "rotten egg" smell, not "odorless" as described in option C.

Therefore, the only option that fits all criteria is B.

A good distinguishing test must give a clearly different result for each substance. Let's analyze the reagents:

Since hot concentrated sulfuric acid produces brown fumes with both nitrate and nitrite, it is not a preferred method for telling them apart.

The described reactions are characteristic tests for halide and nitrate ions using concentrated sulfuric acid (H₂SO₄).

Therefore, Acid (X) is concentrated H₂SO₄, and the salts contain iodide, bromide, and nitrate ions, matching option (D).

Step 1: Analyze the Diagram

Barium chloride (BaCl₂) reacts with aqueous solutions X and Y to form two different precipitates (A and B). The crucial clue is that both precipitates are soluble in dilute acids (like HCl).

Step 2: Evaluate Barium Precipitates

Step 3: Eliminate Options

• Options (a), (b), and (c) contain either Na₂SO₄ as a reactant (which would form insoluble BaSO₄) or list BaSO₄ as a product. Since the diagram specifically states both precipitates are soluble in dilute acid, any option involving sulfate is incorrect.
• Option (d) pairs Na₂CO₃ (forms BaCO₃, soluble) and Ba₃(PO₄)₂ (soluble). Both meet the condition perfectly.

Step 1: Analyze the Clues

• Clue 1: The acid derived from Salt A detects the cation of Salt B.
• Clue 2: The acid derived from Salt B detects the anion of Salt A.

Step 2: Test Option (c)

Let's assign A = KCl and B = PbSO₄.

Salt A is KCl.
The acid of Salt A is Hydrochloric Acid (HCl).
Does HCl detect the cation of Salt B (Pb²⁺)?
Yes! Dilute HCl is the group reagent for Analytical Group I cations (Ag⁺, Pb²⁺, Hg₂²⁺). It reacts to form a white precipitate: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Salt B is PbSO₄.
The acid of Salt B is Sulfuric Acid (H₂SO₄).
Does concentrated H₂SO₄ detect the anion of Salt A (Cl⁻)?
Yes! Concentrated H₂SO₄ is the group reagent for halides. It reacts with solid chlorides to evolve acidic HCl gas: 2KCl + H₂SO₄ → K₂SO₄ + 2HCl↑

Since both conditions hold perfectly, Option (c) is the correct answer.

Step 1: Write the balanced chemical equation.

From the equation, 1 mole of acid reacts with 2 moles of base (nₐ = 1, n_b = 2).

Step 2: Use the titration formula.

(Mₐ × 32) / 1 = (M_b × 16) / 2

32 Mₐ = 8 M_b

Mₐ = (8 / 32) M_b = ¼ M_b

Therefore, the concentration of the sulphuric acid is a quarter of the concentration of the potassium hydroxide.

Step 1: Convert solubility to Molarity (mol/L).
Solubility given = 4.4 × 10⁻⁵ mol / 100 mL.
To find mol / 1000 mL (1 Liter), multiply by 10:
Molar Solubility (s) = 4.4 × 10⁻⁴ mol/L.

Step 2: Write the dissociation equation and Ksp expression.
MgCO₃(s) ⇌ Mg²⁺(aq) + CO₃²⁻(aq)
Ksp = [Mg²⁺][CO₃²⁻] = (s)(s) = s².

Step 3: Calculate Ksp.
Ksp = (4.4 × 10⁻⁴)² = 19.36 × 10⁻⁸ = 1.936 × 10⁻⁷.

Step 1: Identify Salts X and Y

Same cation, but X is soluble in water and Y is not. Both react with HCl to release the same gas Z.

This describes bicarbonate (X) and carbonate (Y) salts where Z = CO₂.

Example: X = Ca(HCO₃)₂ (soluble), Y = CaCO₃ (insoluble in water but dissolves in acids)

Step 2: Test Each Option

Option (b): Heating bicarbonate produces CO₂ and carbonate:

This perfectly matches option (b).

Analysis of Each Salt

Salt A: CaCl₂

CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl↑

HCl is a colorless gas, difficult to oxidize (Cl⁻ is the weakest reducing halide ion) ✓

Salt B: MgI₂

Conc. H₂SO₄ oxidizes HI to I₂ vapors (purple-violet colored vapors):

MgI₂ + H₂SO₄ → MgSO₄ + 2HI

8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O

The resulting iodine solution (I₂ water) can be used to detect first-group anions (like sulfide, sulfite ions) ✓

Salt C: Ag₃PO₄

Ag₃PO₄ + H₂SO₄ → Ag₂SO₄ + H₃PO₄

H₃PO₄ is a non-volatile acid, so no gas is released ✓

Why Not Other Options?

• Option (a) PbSO₄ is incorrect for C? Actually PbSO₄ is insoluble but the question's "no gas" works. However, option (b) is the model answer favored due to the specific behavior of MgI₂ producing detectable iodine vapors.

Concept

For equal masses and equal volumes: M = (mass/M.W.)/V

Higher molar mass → fewer moles → lower concentration

Molar Masses

Ca(OH)₂ has the highest molar mass → lowest concentration.

Solubility Analysis with H₂SO₄

Why Other Options Fail

• Na₂CO₃: precipitates ALL three as carbonates

• NaNO₃: all nitrates are soluble, no separation

• HCl: only Pb²⁺ removed as PbCl₂

Step 1: Find Molarity of BaCl₂ Solution

n(BaCl₂·2H₂O) = 45 / 244 = 0.1844 mol

M = 0.1844 / 0.500 L = 0.369 M

Step 2: Moles BaCl₂ in 84.2 mL

n(BaCl₂) = 0.369 × 0.0842 = 0.0311 mol

Step 3: Moles of Na₂SO₄ (1:1 ratio with BaCl₂)

n(Na₂SO₄) = 0.0311 mol

Step 4: Find Molar Mass of Hydrated Salt

M(hydrate) = 10 g / 0.0311 mol = 322 g/mol

Step 5: Solve for X

322 = 142 + 18X

18X = 180 → X = 10 ✓

Formula: Na₂SO₄·10H₂O

Testing X₂Y₃

X₂Y₃ → 2X³⁺ + 3Y²⁻

If s = 1×10⁻⁴: [X³⁺] = 2s; [Y²⁻] = 3s

Ksp = (2s)² × (3s)³ = 4s² × 27s³ = 108 s⁵

= 108 × (10⁻⁴)⁵ = 108 × 10⁻²⁰ = 1.08 × 10⁻¹⁸ ✓

Verification

This matches the given Ksp exactly, confirming the formula is X₂Y₃.

Both Na₂CO₃ and NaNO₂ dissolve in water with no visible difference, so water cannot distinguish them. KMnO₄/H⁺ is decolorised by NaNO₂ (reducing agent) but not by Na₂CO₃. Dilute acids release CO₂ from carbonate, differentiating them.

NaCl + Pb(NO₃)₂ → PbCl₂↓ (white precipitate, sparingly soluble).

BaCl₂ is fully soluble → no precipitate with Ba²⁺.

Na₂SO₄ and Na₂CO₃ form precipitates with both Pb²⁺ and Ba²⁺, so they cannot be used to distinguish.

Na₂SO₃ + HCl → SO₂↑ (gas) ✔ | Ag₂SO₃ is a white precipitate ✔.

AgBr is pale yellow (not black); AgI is yellow; AgCl is white — so options (a) and (d) are wrong.

Ag₂S is black, so (c) is also wrong — sulfide with HCl gives H₂S, not a typical acid–gas scenario that matches.

NaI + conc. H₂SO₄ → I₂ vapour (violet).

I₂ + Na₂S₂O₃ → 2NaI + Na₂S₄O₆ (colour disappears — standard iodometric reaction).

n(NaOH) = 0.010 L × 1 M = 0.010 mol

V(HCl) needed = n / M = 0.010 / 2 = 0.005 L = 5 mL

Final burette reading = 6.5 + 5.0 = 11.5 mL

Mass of FeCl_x = 5.34 × 0.6008 = 3.208 g

Mass of 6H₂O lost = 5.34 − 3.208 = 2.132 g → n(H₂O) = 2.132/18 = 0.11844 mol

n(FeCl_x) = 0.11844/6 = 0.01974 mol

M(FeCl_x) = 3.208/0.01974 = 162.5 g/mol

56 + 35.5x = 162.5 → 35.5x = 106.5 → x = 3

[NO] = 0.15 M, [O₂] = 0.15 M, [NO₂] = 0.25 M

Q = [NO₂] / ([NO][O₂]^½) = 0.25 / (0.15 × √0.15) = 0.25 / 0.0581 ≈ 4.30

Q (4.30) < K_c (6.33) → reaction proceeds forward to increase [NO₂] and reach equilibrium.

The correct option is b).

CO₂ is already fully oxidised (C = +4), so it does NOT oxidise further in air. Options (a), (c), and (d) all describe MgCO₃, which IS a product.

The correct option is d).

The goal is to keep Cu²⁺ in solution while removing Ag⁺, Pb²⁺ and Ca²⁺.

• Step 1 (NaCl): Cl⁻ precipitates AgCl↓ (white) + PbCl₂↓ (white). Cu²⁺ + Ca²⁺ stay in solution.
• Step 2 (Filtration): Remove AgCl & PbCl₂.
• Step 3 (MgSO₄): SO₄²⁻ precipitates Ca²⁺ as CaSO₄↓ (sparingly soluble). CuSO₄ stays soluble.
• Step 4 (Filtration): Final solution contains only the copper cation. ✓

Why not (a)? Adding H₂S precipitates CuS↓ — that would REMOVE copper from the solution, leaving Ca²⁺ behind, which is the opposite of what is required.

The correct option is b).

• Salt X (an iodide): Hot conc. H₂SO₄ oxidises I⁻ to violet I₂ vapour, which turns starch paper blue-black. With AgNO₃: I⁻ → AgI↓ (yellow), which is insoluble in ammonia. ✓
• Salt Y (a carbonate): BaCl₂ gives BaCO₃ (white, dissolves in dilute HCl with effervescence). With AgNO₃: CO₃²⁻ → Ag₂CO₃↓ (pale yellow), which dissolves in ammonia. ✓

Key clue: starch turning blue = iodine ⇒ X is an iodide (AgI yellow, NOT a chloride).

The correct option is a) n = 2.

The correct option is b).

The diprotic acid at 0.5 M supplies 1 eq/L of H⁺; the monoacidic base KOH at 1 M supplies 1 eq/L of OH⁻ — so equal volumes neutralise exactly. ✓

The correct option is c).

Only K₂CO₃ reacts: K₂CO₃ + MgSO₄ → MgCO₃↓ + K₂SO₄ (KCl gives no precipitate).

Correction note: 0.0595 × 138 = 8.21 g (not 3.81 g) — this gives % KCl = 31.55%.

The correct option is c).

🎯 The Core Chemical Rule: To distinguish between two salts, adding dilute hydrochloric acid (HCl) must yield contrasting visual observations. This usually means a reaction occurs with only one salt, or one forms a unique insoluble precipitate while the other stays completely dissolved.

✅ Why Option (c) is Correct:
• With Sodium Carbonate (Na_2CO3): The salt dissolves smoothly with effervescence from escaping CO₂ gas, producing a completely clear, fully soluble sodium chloride (NaCl) solution.
• With Lead(II) Bicarbonate (Pb(HCO₃)₂): It similarly generates CO₂ gas bubbles, but the liberated lead ions (Pb²⁺) instantly bond with the chloride ions (Cl^-) from the test acid. This triggers the sudden precipitation of a dense white precipitate of lead(II) chloride (PbCl₂ ↓) because Pb²⁺ belongs to Cation Analytical Group I.

❌ Why Other Options Fail:
• Option (a): Both salts are soluble sodium compounds that react identically to produce a clear solution alongside CO₂ gas.
• Option (b): HCl cannot displace the stable sulfate anion from sodium sulfate, nor does it react with sodium chloride. Absolutely no reaction takes place.
• Option (d): Both barium carbonate and magnesium bicarbonate dissolve cleanly in HCl, yielding identical CO₂ gas bubbles and uniform, transparent liquid phases.

🎯 The Core Chemical Rule: An unknown salt is identified by treating its solid form with a strong acid to detect volatile acid radicals (anions), and treating its aqueous solution to detect basic radicals (cations) that produce characteristic insoluble precipitates.

✅ Analyzing Anion and Cation Tests for Salt (X):
• 1. Anion Determination (Acid Radical): When hot concentrated sulfuric acid (H_2SO4) is added to solid calcium bromide (CaBr₂), hydrogen bromide gas (HBr) is initially evolved. The hot concentrated acid then partially oxidizes HBr, releasing a mixture of gases consisting of reddish-brown bromine vapors (Br₂) and sulfur dioxide gas (SO₂).
• 2. Cation Determination (Basic Radical): When dilute sulfuric acid is added to the aqueous solution of the salt, calcium cations (Ca²⁺) instantly combine with sulfate anions (SO₄^2-). This triggers the formation of a distinct white precipitate of calcium sulfate (CaSO₄ ↓), which is insoluble in dilute acids.

❌ Why Other Options Fail:
• NaI: While the iodide anion produces a gaseous mixture (I₂ and SO₂), the sodium cation (Na⁺) forms highly soluble salts. It will never create a precipitate with dilute sulfuric acid.
• CaCO₃: Calcium carbonate is an insoluble rock-solid salt in water. It cannot be used to prepare a clear initial salt solution, and it dissolves instantly with vigorous effervescence when exposed to dilute acids.
• Pb(SO₃)₂: The sulfite group (SO₃^2-) reacts readily with weak dilute acids and does not require hot concentrated conditions. It evolves only a single gas (SO₂) instead of a gaseous mixture.

🎯 The Core Chemical Rule: Barium chloride (BaCl₂) is used as a main reagent to precipitate specific anions. The key to distinguishing between the resulting barium precipitates lies in their solubility; some dissolve cleanly in dilute hydrochloric acid (HCl), while others remain completely insoluble.

✅ Analyzing Precipitates and Solubility:
• With Phosphate (PO₄^3-): Barium chloride reacts with phosphate solutions to form a white precipitate of barium phosphate [Ba₃(PO₄)₂ ↓]. This precipitate dissolves smoothly in dilute hydrochloric acid.
• With Carbonate (CO₃^2-): Barium chloride reacts with soluble carbonate solutions to form a white precipitate of barium carbonate [BaCO₃ ↓]. This precipitate also dissolves readily in dilute acids with accompanying carbon dioxide (CO₂) gas evolution.

❌ Why Other Options Fail:
• Sulfate (SO₄^2-): Barium chloride reacts with sulfate groups to yield a white precipitate of barium sulfate [BaSO₄ ↓]. However, this precipitate is strictly insoluble in dilute hydrochloric acid. This immediately eliminates options (a), (b), and (d).
• Nitrate (NO₃^-): All nitrate salts are highly soluble in water, so no precipitate forms when mixed with barium chloride solution.

🎯 The Core Chemical Rule: An unknown cation is identified by observing its precipitation behavior with specific reagents. Cations are classified into analytical groups based on the solubility rules of their chlorides, sulfates, acetates, or other salts.

✅ Testing the Cation with the Three Reagents:
• With Hydrochloric Acid (HCl): It forms a **white precipitate**. This indicates that the cation belongs to Analytical Group I, which precipitates as insoluble chlorides. This directly matches the lead(II) cation, forming lead(II) chloride (PbCl₂ ↓).
• With Sodium Sulfate (Na₂SO₄): It yields a **white precipitate** due to the combination of the cation with sulfate ions, forming lead(II) sulfate (PbSO₄ ↓), a well-known insoluble white solid.
• With Acetic Acid (CH₃COOH): **No precipitate** forms because lead(II) acetate [Pb(CH₃COO)₂] is highly soluble in water, leaving the solution perfectly clear, which fully aligns with the diagram.

❌ Why Other Options Fail:
• Cu²⁺ & Fe²⁺: Both copper(II) chloride (CuCl₂) and iron(II) chloride (FeCl₂) are highly soluble in water and will not form a precipitate with hydrochloric acid.
• Ca²⁺: Calcium chloride (CaCl₂) is extremely soluble in water, meaning no precipitate occurs upon adding hydrochloric acid.

🎯 The Core Chemical Rule: To separate a mixture of anions smoothly via fractional precipitation, each added reagent must selectively precipitate exactly **one single remaining anion** at a time without interfering with the others.

✅ Analyzing the Separation Step-by-Step:
• Step 1 — Adding Copper Nitrate [Cu(NO₃)₂]: Copper ions (Cu²⁺) selectively react with the sulphide ions (S^2-) to precipitate **black copper(II) sulphide** (CuS ↓). Meanwhile, copper chloride and copper sulfate remain fully soluble. After filtering out the CuS precipitate, the solution contains only sulphate and chloride.
• Step 2 — Adding Calcium Hydroxlide [Ca(OH)₂]: Calcium ions (Ca²⁺) selectively form a **white precipitate of calcium sulphate** (CaSO₄ ↓). On the other hand, calcium chloride (CaCl₂) is extremely soluble. Filtering out this precipitate leaves only chloride ions behind in the mixture.
• Step 3 — Adding Silver Nitrate [AgNO₃]: Finally, silver ions (Ag⁺) precipitate the last remaining anion, chloride (Cl^-), as a **white precipitate of silver chloride** (AgCl ↓).

❌ Why Order (c) Fails: If you add silver nitrate first, the Ag⁺ ions would unselectively precipitate **all three anions at the same time** because silver sulphide (Ag_2S), silver chloride (AgCl), and silver sulphate (Ag_2SO4) are all highly insoluble solids, ruining the separation process completely.

n(HCl)=7.3/36.5=0.2 mol → n(CaCO₃)=0.1 mol → mass=10 g. Purity=80%, impurities=20%.

Exp1: acid:base = 2 mmol:1 mmol = 2:1. Exp2: base=5 mmol → acid=10 mmol → 10/40 = 0.250 M.