Chapter 6: Equilibrium Application Questions | chemistry Class 11 CBSE

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Chapter 6: Equilibrium Application Questions | chemistry Class 11 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A soda water bottle is kept in a refrigerator and then opened at room temperature. With reference to Henry's law and gas-liquid equilibrium, explain why it fizzes much more vigorously when opened at room temperature than when opened directly from the refrigerator.

Soda water contains CO2 dissolved under high pressure, maintaining the equilibrium CO2(gas) ⇌ CO2(in solution); according to Henry's law, the mass of dissolved gas is proportional to the pressure of gas above the solvent.
When the bottle is opened, CO2 escapes to establish a new equilibrium corresponding to the lower partial pressure of CO2 in the atmosphere, causing fizzing as dissolved CO2 bubbles out.
At higher room temperature, the solubility of CO2 in water decreases (Henry's law constant changes with temperature; dissolved gas concentration decreases as temperature increases), so there is more excess dissolved CO2 that needs to escape.
When opened from the refrigerator (low temperature), CO2 is more soluble in the cold water, so less CO2 escapes per unit time, resulting in gentler fizzing; at room temperature, solubility is lower and the equilibrium shift requires more CO2 to escape, resulting in vigorous fizzing.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In the manufacture of sulphuric acid by the contact process, the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is carried out using V2O5 catalyst at about 450°C. Explain why a catalyst is necessary even though Kc = 1.7 × 1026 at room temperature, and why the yield decreases at very high temperatures.

Although Kc = 1.7 × 1026 indicates that the equilibrium greatly favours product formation at room temperature, the oxidation of SO2 to SO3 is extremely slow at low temperatures due to high activation energy; a catalyst (V2O5 or Pt) is needed to make the reaction proceed at a practical rate.
The catalyst provides an alternative lower-energy pathway that increases both forward and reverse rates equally; it does not alter the equilibrium composition or Kc, but allows the equilibrium to be reached much faster.
The reaction is exothermic (ΔH < 0); increasing temperature shifts the equilibrium to the left by Le Chatelier's principle, decreasing Kc and reducing the yield of SO3. Very high temperatures thus favour the reactants.
A compromise temperature of ~450°C is used to balance adequate reaction rate (which would otherwise be too slow at lower temperatures) with acceptable equilibrium yield; this is a classic trade-off between kinetics and thermodynamics in industrial chemistry.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A student prepares a blood sample for analysis and notes that human blood has a pH of 7.4. Calculate the concentration of H+ ions in blood. What type of buffer system maintains this pH?

Using the pH formula: pH = –log[H+]; therefore [H+] = 10–pH = 10–7.4 = antilog(–7.4) = 3.98 × 10–8 M ≈ 4.0 × 10–8 M. Since pH > 7, blood is basic (alkaline), reflecting the normal physiological condition.
Blood pH is maintained primarily by the carbonic acid-bicarbonate buffer system: H2CO3 ⇌ H+ + HCO3–; when excess acid enters the blood, HCO3– (conjugate base) neutralizes it; when excess base enters, H2CO3 donates a proton to counteract the rise in pH.
The Henderson-Hasselbalch equation applies: pH = pKa(H2CO3) + log([HCO3–]/[H2CO3]) = 7.4; with pKa ≈ 6.1, the ratio [HCO3–]/[H2CO3] ≈ 20:1 in normal blood, maintained by the lungs (regulating CO2 levels) and kidneys (regulating HCO3–).
Any deviation from pH 7.4 (acidosis if pH < 7.35, alkalosis if pH > 7.45) indicates metabolic or respiratory malfunction; this underscores why buffer equilibria are central to clinical medicine and biochemistry.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A chemist studying the formation of iron thiocyanate complex [Fe(SCN)]2+ in a solution observes that the red colour intensifies upon addition of KSCN but fades when oxalic acid is added. Explain these observations using Le Chatelier's principle and the concept of ionic equilibrium.

The equilibrium is: Fe3+(aq) + SCN–(aq) ⇌ [Fe(SCN)]2+(aq) (yellow) (colourless) (deep red). Le Chatelier's principle predicts that any change in concentration of a reactant or product shifts the equilibrium to counteract the change.
Addition of KSCN increases [SCN–]; this disturbs the equilibrium, making Qc < Kc, so the forward reaction is favoured to consume the excess SCN– and form more [Fe(SCN)]2+, intensifying the red colour.
Addition of oxalic acid (H2C2O4) introduces oxalate ions that react with Fe3+ to form the stable complex [Fe(C2O4)3]3–, effectively removing free Fe3+ from the equilibrium. By Le Chatelier's principle, the equilibrium shifts left to replenish Fe3+ by dissociating [Fe(SCN)]2+, decreasing its concentration and causing the red colour to fade.
This experiment beautifully demonstrates Le Chatelier's principle in ionic equilibria: adding a reactant shifts equilibrium forward (more product), while removing a reactant (by complexation) shifts equilibrium backward (less product).

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 5

Applied Concept

A geologist finds that limestone (CaCO3) deposits dissolve more rapidly in acidic rainwater. With reference to the equilibrium CaCO3(s) ⇌ Ca2+(aq) + CO32–(aq), explain the effect of acid rain (H+ ions) on solubility, using Le Chatelier's principle.

CaCO3 is a sparingly soluble salt with Ksp = 2.8 × 10–9. In pure water, solubility is limited by this equilibrium. The carbonate ion (CO32–) is the anion of carbonic acid, a weak acid.
When acid rain (containing H+ ions) falls on limestone, H+ ions react with CO32– ions: CO32– + H+ → HCO3–, and further: HCO3– + H+ → H2CO3 → H2O + CO2(g). This effectively removes CO32– from solution.
By Le Chatelier's principle, removal of CO32– from the dissolution equilibrium shifts it to the right to replenish CO32–, dissolving more CaCO3. This increases the solubility of CaCO3 far beyond its Ksp value in neutral water.
The process continues as long as acid is available, which is why acid rain gradually erodes limestone monuments, statues, and geological formations. This is a real-world example of how pH affects the solubility of sparingly soluble salts, as described by the equation S = {Ksp([H+] + Ka)/Ka}1/2.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A pharmaceutical company needs to prepare a buffer solution of pH 9.25 for a drug formulation. Using your knowledge of the ammonia-ammonium chloride buffer system, explain how this buffer is prepared and how it resists pH change.

A basic buffer of pH ≈ 9.25 can be prepared by dissolving equimolar amounts of ammonia (NH3) and ammonium chloride (NH4Cl) in water. This works because the pKa of NH4+ = 14 – pKb = 14 – 4.75 = 9.25; when [NH3] = [NH4+], pH = pKa = 9.25 from the Henderson-Hasselbalch equation.
Resistance to acid addition: when HCl (H+ ions) is added, H+ reacts with NH3 (the base component): NH3 + H+ → NH4+. This consumes H+ without significant change in [NH3]/[NH4+] ratio, so pH changes very little.
Resistance to base addition: when NaOH (OH– ions) is added, OH– reacts with NH4+ (the acid component): NH4+ + OH– → NH3 + H2O. This consumes OH– without significant change in the ratio, so pH again changes very little.
The buffer capacity is greatest when [NH3] = [NH4+] (i.e., pH = pKa = 9.25); the pH is unaffected by dilution because the ratio [NH4+]/[NH3] remains unchanged, consistent with the Henderson-Hasselbalch equation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

In a gravimetric analysis, a chemist wants to precipitate all Ag+ ions from solution as AgCl by adding HCl. After adding excess HCl, will the precipitate re-dissolve? Explain using common ion effect and solubility product (Ksp of AgCl = 1.8 × 10–10).

AgCl is a sparingly soluble salt: AgCl(s) ⇌ Ag+(aq) + Cl–(aq), Ksp = [Ag+][Cl–] = 1.8 × 10–10. In pure water, S = √Ksp = √(1.8 × 10–10) = 1.34 × 10–5 mol L–1.
When excess HCl is added, [Cl–] increases significantly (common ion effect). By Le Chatelier's principle, the equilibrium shifts to the left, causing more AgCl to precipitate. [Ag+] = Ksp/[Cl–]; with [Cl–] = 0.1 M, [Ag+] = 1.8 × 10–9 M — nearly all Ag+ has precipitated.
Excess HCl does not dissolve the precipitate; rather, it suppresses solubility further. The common ion effect reduces [Ag+] to negligible levels, ensuring complete precipitation of silver for accurate gravimetric estimation.
This is the basis of gravimetric analysis: the common ion effect (Le Chatelier's principle applied to solubility equilibria) is exploited to achieve nearly complete precipitation. In practice, a slight excess of the precipitating reagent is added but very large excess is avoided to prevent formation of soluble complexes like AgCl2–.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A student dissolves 0.1 mol of acetic acid in water to make 1 L solution. The pH is found to be 2.87. Calculate the degree of ionization and Ka. Then predict what happens to pH and degree of ionization on adding 0.1 mol of sodium acetate to this solution.

From pH = 2.87: [H+] = 10–2.87 = 1.35 × 10–3 M = [CH3COO–]. Initial concentration c = 0.1 M; degree of ionization α = [H+]/c = 1.35 × 10–3/0.1 = 0.0135 = 1.35%. Ka = [H+][CH3COO–]/[CH3COOH] = (1.35 × 10–3)2/(0.1 – 1.35 × 10–3) ≈ (1.35 × 10–3)2/0.0987 ≈ 1.85 × 10–5.
On adding 0.1 mol sodium acetate: this is a common ion (CH3COO–) which shifts equilibrium CH3COOH ⇌ H+ + CH3COO– to the left, suppressing ionization. [CH3COO–] ≈ 0.1 M from the salt, so by Henderson-Hasselbalch: pH = pKa + log([salt]/[acid]) = 4.74 + log(0.1/0.1) = 4.74.
The degree of ionization decreases dramatically; [H+] = Ka × [HA]/[A–] = 1.74 × 10–5 × 0.1/0.1 = 1.74 × 10–5 M, so α = 1.74 × 10–5/0.1 = 0.0174% — much less than the original 1.35%.
This illustrates the common ion effect: sodium acetate provides CH3COO– ions that suppress acid ionization, raising pH from 2.87 to 4.74 while the solution simultaneously becomes a buffer, demonstrating the dual role of common ion addition.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

With reference to the equilibrium N2O4(g) ⇌ 2NO2(g), explain what will be observed if a sealed tube containing the equilibrium mixture at room temperature is (a) immersed in hot water and (b) placed in a freezing mixture.

The reaction N2O4(g) ⇌ 2NO2(g) has ΔH = +57.2 kJ mol–1 (endothermic) because the reverse reaction (formation of N2O4 from 2NO2) is exothermic at –57.2 kJ mol–1. At equilibrium, the brown colour of NO2 is at a constant intensity.
On immersing in hot water (363 K): the temperature increases. Since the forward reaction is endothermic, Le Chatelier's principle predicts that increasing temperature shifts the equilibrium to the right (forward), increasing the concentration of NO2 (brown) and decreasing N2O4 (colourless). The intensity of the brown colour increases, and Kc increases.
On placing in a freezing mixture (270 K): the temperature decreases. The exothermic direction (reverse, forming N2O4) is favoured; equilibrium shifts left, decreasing [NO2] and increasing [N2O4]. The brown colour fades (decreases in intensity), confirming conversion of NO2 to colourless N2O4. Kc decreases.
This experiment provides direct visual evidence that temperature change alters both the equilibrium position and the value of Kc, unlike concentration and pressure changes which only shift Qc relative to the unchanged Kc.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

The Ksp of AgCl is 1.8 × 10–10 and that of AgBr is 5.0 × 10–13. A solution contains both Cl– and Br– ions each at concentration 0.1 M. When AgNO3 solution is added slowly, which salt precipitates first? Determine the [Ag+] at which each begins to precipitate.

For precipitation to begin, the ionic product must just exceed Ksp. AgCl starts precipitating when [Ag+] > Ksp(AgCl)/[Cl–] = 1.8 × 10–10/0.1 = 1.8 × 10–9 M.
AgBr starts precipitating when [Ag+] > Ksp(AgBr)/[Br–] = 5.0 × 10–13/0.1 = 5.0 × 10–12 M.
Since AgBr requires a lower [Ag+] to begin precipitating (5.0 × 10–12 < 1.8 × 10–9), AgBr precipitates first; the less soluble salt precipitates preferentially when the common cation is added gradually.
When AgCl starts to precipitate at [Ag+] = 1.8 × 10–9 M, the remaining [Br–] = Ksp(AgBr)/[Ag+] = 5.0 × 10–13/1.8 × 10–9 = 2.78 × 10–4 M — nearly all Br– has been removed. This selective precipitation is used analytically to separate and identify halide ions.

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Chapter 6: Equilibrium | Application Questions

A soda water bottle is kept in a refrigerator and then opened at room temperature. With reference to Henry's law and gas-liquid equilibrium, explain why it fizzes much more vigorously when opened at room temperature than when opened directly from the refrigerator.

In the manufacture of sulphuric acid by the contact process, the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is carried out using V2O5 catalyst at about 450°C. Explain why a catalyst is necessary even though Kc = 1.7 × 1026 at room temperature, and why the yield decreases at very high temperatures.

A student prepares a blood sample for analysis and notes that human blood has a pH of 7.4. Calculate the concentration of H+ ions in blood. What type of buffer system maintains this pH?

A chemist studying the formation of iron thiocyanate complex [Fe(SCN)]2+ in a solution observes that the red colour intensifies upon addition of KSCN but fades when oxalic acid is added. Explain these observations using Le Chatelier's principle and the concept of ionic equilibrium.

A geologist finds that limestone (CaCO3) deposits dissolve more rapidly in acidic rainwater. With reference to the equilibrium CaCO3(s) ⇌ Ca2+(aq) + CO32–(aq), explain the effect of acid rain (H+ ions) on solubility, using Le Chatelier's principle.

A pharmaceutical company needs to prepare a buffer solution of pH 9.25 for a drug formulation. Using your knowledge of the ammonia-ammonium chloride buffer system, explain how this buffer is prepared and how it resists pH change.

In a gravimetric analysis, a chemist wants to precipitate all Ag+ ions from solution as AgCl by adding HCl. After adding excess HCl, will the precipitate re-dissolve? Explain using common ion effect and solubility product (Ksp of AgCl = 1.8 × 10–10).

A student dissolves 0.1 mol of acetic acid in water to make 1 L solution. The pH is found to be 2.87. Calculate the degree of ionization and Ka. Then predict what happens to pH and degree of ionization on adding 0.1 mol of sodium acetate to this solution.

With reference to the equilibrium N2O4(g) ⇌ 2NO2(g), explain what will be observed if a sealed tube containing the equilibrium mixture at room temperature is (a) immersed in hot water and (b) placed in a freezing mixture.

The Ksp of AgCl is 1.8 × 10–10 and that of AgBr is 5.0 × 10–13. A solution contains both Cl– and Br– ions each at concentration 0.1 M. When AgNO3 solution is added slowly, which salt precipitates first? Determine the [Ag+] at which each begins to precipitate.

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