Questions & explanations
1. Which of the following statements correctly defines a buffer solution and its types?
- A buffer solution is one that resists change in pH on addition of small amounts of acid or base; acidic buffers are mixtures of weak acid and its salt with strong base, basic buffers are mixtures of weak base and its salt with strong acid.
- A buffer solution is one that maintains a constant pH regardless of any addition; acidic buffers contain strong acid and its salt, basic buffers contain strong base and its salt.
- A buffer solution is one that has a pH of 7; acidic buffers have pH < 7, basic buffers have pH > 7.
- A buffer solution is one that changes pH drastically upon addition of acid or base; acidic buffers are made of strong acid and strong base, basic buffers of weak acid and weak base.
Answer: A buffer solution is one that resists change in pH on addition of small amounts of acid or base; acidic buffers are mixtures of weak acid and its salt with strong base, basic buffers are mixtures of weak base and its salt with strong acid.
A buffer solution resists pH change when small amounts of acid or base are added. An acidic buffer consists of a weak acid and its salt with a strong base (e.g., acetic acid and sodium acetate). A basic buffer consists of a weak base and its salt with a strong acid (e.g., ammonium hydroxide and ammonium chloride). The other options incorrectly describe buffer properties or composition.
2. In the reaction: HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻, which pair is a conjugate acid-base pair?
- A. HCO₃⁻ and H₂O
- B. HCO₃⁻ and CO₃²⁻
- C. H₂O and H₃O⁺
- D. H₃O⁺ and CO₃²⁻
Answer: B. HCO₃⁻ and CO₃²⁻
A conjugate acid-base pair differs by one proton. HCO₃⁻ donates a proton to become CO₃²⁻; thus HCO₃⁻ (acid) and CO₃²⁻ (conjugate base) form a conjugate pair. H₂O/H₃O⁺ is also a pair but the question asks for the pair in this reaction; however, HCO₃⁻/CO₃²⁻ is the correct answer as it is the acid–base pair involved in the proton transfer.
3. The solubility product (Ksp) of a sparingly soluble salt is defined as:
- The product of the molar concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient.
- The product of the solubility of the salt in moles per liter.
- The product of the molar concentrations of the ions in any solution.
- The product of the solubility and the molar mass.
Answer: The product of the molar concentrations of the ions in a saturated solution, each raised to the power of its stoichiometric coefficient.
Ksp is the equilibrium constant for the dissolution of a sparingly soluble salt. It equals the product of the concentrations of the constituent ions (in a saturated solution) with each concentration raised to the power of the stoichiometric coefficient in the balanced dissolution equation. The other options are incorrect definitions.
4. Buffer capacity is defined as:
- The amount of acid or base that can be added to a buffer without changing its pH at all.
- The number of moles of acid or base required to change the pH of one liter of buffer solution by one unit.
- The ratio of salt to acid in the buffer.
- The pH range over which a buffer is effective.
Answer: The number of moles of acid or base required to change the pH of one liter of buffer solution by one unit.
Buffer capacity is a quantitative measure: it is the number of moles of strong acid or strong base needed to change the pH of one liter of buffer by exactly one unit. Option A is unrealistic because some pH change always occurs. Option C describes the buffer ratio, and option D describes the buffer range, not capacity.
5. A solution contains 0.01 M Ba²⁺ and 0.01 M SO₄²⁻. Ksp of BaSO₄ is 1.0 × 10⁻¹⁰. Which of the following is correct?
- Precipitation will occur because ionic product (0.0001) > Ksp.
- Precipitation will not occur because ionic product = 1.0 × 10⁻⁴ < Ksp.
- Precipitation will occur because ionic product < Ksp.
- The solution is saturated and no further precipitation occurs.
Answer: Precipitation will occur because ionic product (0.0001) > Ksp.
The ionic product = [Ba²⁺][SO₄²⁻] = (0.01)(0.01) = 1 × 10⁻⁴. Since 1 × 10⁻⁴ > 1 × 10⁻¹⁰ (Ksp), the solution is supersaturated, so precipitation will occur. Option B incorrectly states ionic product is less; option C reverses the condition; option D incorrectly describes a saturated solution (saturated when IP = Ksp).
6. For a conjugate acid-base pair, the relationship between their ionization constants Ka (for the acid) and Kb (for the base) is given by:
- Ka × Kb = Kw
- Ka / Kb = Kw
- Ka + Kb = Kw
- Ka = Kb
Answer: Ka × Kb = Kw
For any conjugate acid-base pair, the product of the acid ionization constant (Ka) and the base ionization constant (Kb) equals the ion-product constant of water (Kw) at a given temperature. This relationship is derived from the equilibrium expressions for the acid dissociation and the conjugate base hydrolysis.
7. For the reaction N2(g) + 3H2(g) ⇌ 2NH3(g), what is the correct equilibrium constant expression in terms of concentration (Kc)?
- Kc = [NH3]² / [N2][H2]³
- Kc = [NH3]² / [N2][H2]
- Kc = [N2][H2]³ / [NH3]²
- Kc = [NH3] / [N2][H2]
Answer: Kc = [NH3]² / [N2][H2]³
According to the law of mass action, the equilibrium constant is the product of product concentrations raised to their stoichiometric coefficients divided by the product of reactant concentrations raised to their coefficients. For NH3 the coefficient is 2, for N2 it is 1, and for H2 it is 3.
8. The common ion effect refers to:
- The increase in dissociation of a weak electrolyte when a strong electrolyte with a common ion is added.
- The suppression of dissociation of a weak electrolyte by adding a strong electrolyte that shares a common ion.
- The increase in solubility of a sparingly soluble salt when a common ion is added.
- The decrease in pH of a buffer solution when a small amount of acid is added.
Answer: The suppression of dissociation of a weak electrolyte by adding a strong electrolyte that shares a common ion.
The common ion effect describes the shift in equilibrium that occurs when a common ion (from a strong electrolyte) is added to a solution of a weak electrolyte, resulting in decreased dissociation of the weak electrolyte. For example, adding NH₄Cl to NH₄OH suppresses the ionization of NH₄OH.
9. The solubility of calcium fluoride (CaF2) in pure water is 2.1 × 10^{-4} mol/L. Its solubility in a 0.01 M NaF solution is approximately: (Ksp of CaF2 = 3.9 × 10^{-11})
- 2.1 × 10^{-4} M
- 3.9 × 10^{-7} M
- 3.9 × 10^{-9} M
- 1.0 × 10^{-4} M
Answer: 3.9 × 10^{-7} M
In presence of common ion F⁻ (0.01 M from NaF), the solubility of CaF₂ is greatly reduced. Using Ksp = [Ca²⁺][F⁻]², with [F⁻] ≈ 0.01 M, [Ca²⁺] = Ksp/(0.01)² = 3.9×10⁻¹¹/1×10⁻⁴ = 3.9×10⁻⁷ M. The small amount of F⁻ from dissolution is negligible, so solubility equals [Ca²⁺].
10. For the reaction: 2SO2(g) + O2(g) ⇌ 2SO3(g), what is the effect of decreasing the volume of the container at constant temperature?
- Equilibrium shifts to left (backward direction)
- Equilibrium shifts to right (forward direction)
- No change in equilibrium
- The reaction stops
Answer: Equilibrium shifts to right (forward direction)
Decreasing volume increases pressure. According to Le Chatelier's principle, the equilibrium shifts towards the side with fewer moles of gas. Here, reactants side has 3 moles (2+1) and products side has 2 moles, so the shift occurs to the right to reduce pressure.
11. Which of the following statements is correct regarding the effect of a catalyst on a chemical equilibrium?
- A catalyst shifts the equilibrium to the right
- A catalyst increases the equilibrium constant
- A catalyst increases the rate of both forward and reverse reactions equally
- A catalyst increases the yield of products
Answer: A catalyst increases the rate of both forward and reverse reactions equally
A catalyst lowers the activation energy for both forward and reverse reactions by the same amount, thereby increasing their rates equally. It does not affect the equilibrium position or the equilibrium constant; it only helps the system reach equilibrium faster.
12. In the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) (red complex), what will happen if a few drops of concentrated KSCN solution are added?
- The equilibrium will shift to the left, and the red color will fade
- The equilibrium will shift to the right, and the red color will deepen
- No change will occur because the system is already at equilibrium
- The equilibrium constant will increase
Answer: The equilibrium will shift to the right, and the red color will deepen
Adding SCN⁻ (a reactant) increases its concentration. According to Le Chatelier’s principle, the equilibrium shifts to the right to consume the added SCN⁻, producing more FeSCN²⁺ (red complex), thus the color deepens. K remains unchanged at a given temperature.