Understand Proton Transfer Reactions - IB Chemistry

1. Which of the following species is amphiprotic?
   Answer: HSO₄⁻ An amphiprotic species can both donate and accept a proton (H⁺). The hydrogen sulfate ion, HSO₄⁻, can donate a proton to form SO₄²⁻ (acting as an acid) and can accept a proton to form H₂SO₄ (acting as a base). H₃O⁺ can only act as an acid. SO₄²⁻ can only act as a base. CH₃COOH is a weak acid and does not typically accept a proton in aqueous solution.
2. What are the major species present, other than water, when equimolar amounts of aqueous ethanoic acid (CH₃COOH) and aqueous sodium hydroxide (NaOH) are mixed?
   Answer: CH₃COO⁻(aq) and Na⁺(aq) The neutralization reaction is CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l). When equimolar amounts are mixed, the acid and base are completely consumed, forming the salt sodium ethanoate (CH₃COONa). This salt is soluble and fully dissociates in water into sodium ions (Na⁺) and ethanoate ions (CH₃COO⁻). Therefore, these two ions are the major species present in the solution, apart from water.
3. A buffer solution contains $\text{HClO}$ and $\text{ClO}^-$ with a pH of 7.8. When a small amount of strong base is added, the primary proton transfer reaction that maintains the buffer capacity is $\text{HClO} + \text{OH}^- \rightarrow \text{ClO}^- + \text{H}_2\text{O}$. However, a secondary reaction $\text{H}_2\text{O} + \text{ClO}^- \rightarrow \text{HClO} + \text{OH}^-$ also occurs simultaneously. Which factor most significantly determines the relative rates of these competing proton transfer processes?
   Answer: The difference in activation energies between the forward neutralization reaction and the reverse hydrolysis reaction of the conjugate base. When analyzing competing reactions in buffer systems, you need to distinguish between thermodynamic favorability (which reaction is more favorable at equilibrium) and kinetic factors (which reaction proceeds faster). This question tests your understanding of what controls reaction rates when multiple pathways are possible. The correct answer is D because reaction rates are fundamentally determined by activation energies, not equilibrium positions. The neutralization reaction $\text{HClO} + \text{OH}^- \rightarrow \text{ClO}^- + \text{H}_2\text{O}$ involves a direct proton transfer with typically low activation energy. The competing hydrolysis reaction $\text{H}_2\text{O} + \text{ClO}^- \rightarrow \text{HClO} + \text{OH}^-$ requires breaking the strong O-H bond in water, which has a significantly higher activation energy. This difference in activation barriers determines which reaction dominates kinetically. Option A incorrectly focuses on concentration ratios. While concentrations affect reaction rates through the rate law, they don't determine the relative rates of competing mechanisms. Option B confuses equilibrium composition with reaction kinetics - pH tells you what species are present, not which reactions are faster. Option C represents a common misconception that thermodynamically favorable reactions are automatically faster. Equilibrium constants indicate final positions, not reaction rates. Remember that kinetics and thermodynamics are separate concepts. When analyzing competing reactions, always consider activation energies for rate determination. Reactions with lower activation barriers will dominate kinetically, regardless of their thermodynamic favorability. This principle applies broadly to mechanism problems in organic and physical chemistry.
4. Consider the equilibrium system: $\text{H}_3\text{PO}_4 + 3\text{NH}_3 \rightleftharpoons \text{PO}_4^{3-} + 3\text{NH}_4^+$. A student calculates that this overall reaction has an equilibrium constant of $K = 2.4 \times 10^{-10}$. Based on this information and the principles of proton transfer reactions, which statement about the relative basicity is most accurate?
   Answer: $\text{PO}_4^{3-}$ is a much stronger base than $\text{NH}_3$ since the equilibrium lies far to the left, favoring the stronger base in its protonated form. Since K = 2.4 × 10⁻¹⁰ << 1, the equilibrium lies far to the left, meaning H₃PO₄ + 3NH₄⁺ is strongly favored over PO₄³⁻ + 3NH₃. This indicates that PO₄³⁻ is a much stronger base than NH₃ because the stronger base (PO₄³⁻) preferentially exists in its protonated form (H₃PO₄) when in competition with the weaker base (NH₃). The equilibrium favors the weaker base (NH₃) being deprotonated and the stronger base (PO₄³⁻) being protonated. (A) reverses the base strength relationship. (B) incorrectly interprets what 'retaining protons' means. (D) is wrong because equilibrium constants for overall reactions do provide information about relative strengths.
5. What is the pH of a solution formed by mixing 10.0 cm³ of 0.100 mol dm⁻³ HCl(aq) with 90.0 cm³ of 0.010 mol dm⁻³ HCl(aq)?
   Answer: 1.72 First, calculate the moles of H⁺ from each solution. Moles from first solution = 0.100 mol dm⁻³ × 0.0100 dm³ = 0.00100 mol. Moles from second solution = 0.010 mol dm⁻³ × 0.0900 dm³ = 0.00090 mol. The total moles of H⁺ = 0.00100 + 0.00090 = 0.00190 mol. The total volume is 10.0 + 90.0 = 100.0 cm³ = 0.1000 dm³. The final concentration of H⁺ is [H⁺] = total moles / total volume = 0.00190 mol / 0.1000 dm³ = 0.0190 mol dm⁻³. The pH is -log₁₀(0.0190) ≈ 1.72.
6. What is the concentration of H⁺(aq) ions, in mol dm⁻³, in a 0.050 mol dm⁻³ solution of potassium hydroxide, KOH(aq), at 298 K? (Kₗ = 1.0 × 10⁻¹⁴ mol² dm⁻⁶)
   Answer: 2.0 × 10⁻¹³ Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water. Therefore, [OH⁻] = 0.050 mol dm⁻³. The ion product constant for water, Kₗ, is given by Kₗ = [H⁺][OH⁻]. Rearranging for [H⁺] gives [H⁺] = Kₗ / [OH⁻]. Substituting the values: [H⁺] = (1.0 × 10⁻¹⁴ mol² dm⁻⁶) / (0.050 mol dm⁻³) = 2.0 × 10⁻¹³ mol dm⁻³.
7. 100.0 cm³ of 0.100 mol dm⁻³ Ba(OH)₂(aq) is mixed with 100.0 cm³ of 0.100 mol dm⁻³ HCl(aq). What is the pH of the resulting solution at 298 K?
   Answer: 12.70 The reaction is Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O. First, calculate the moles of H⁺ and OH⁻. Moles of H⁺ = 0.100 dm³ × 0.100 mol dm⁻³ = 0.0100 mol. Moles of Ba(OH)₂ = 0.100 dm³ × 0.100 mol dm⁻³ = 0.0100 mol, which provides 2 × 0.0100 = 0.0200 mol of OH⁻ ions. The H⁺ is the limiting reactant. Moles of excess OH⁻ = 0.0200 - 0.0100 = 0.0100 mol. The final volume is 100.0 + 100.0 = 200.0 cm³ = 0.2000 dm³. The final [OH⁻] = 0.0100 mol / 0.2000 dm³ = 0.0500 mol dm⁻³. Using Kₗ = 1.0 × 10⁻¹⁴, [H⁺] = Kₗ / [OH⁻] = 1.0 × 10⁻¹⁴ / 0.0500 = 2.0 × 10⁻¹³ mol dm⁻³. The pH = -log(2.0 × 10⁻¹³) = 12.70. Alternatively, pOH = -log(0.0500) = 1.30, and pH = 14.00 - pOH = 12.70.
8. A researcher studies the proton transfer reaction $\text{HA} + \text{B} \rightleftharpoons \text{A}^- + \text{BH}^+$ and finds that when [HA] = [B] = 0.10 M initially, the equilibrium concentrations are [$\text{A}^-$] = [$\text{BH}^+$] = 0.040 M. If the researcher then prepares a new solution with [HA] = 0.20 M and [B] = 0.050 M, what will be the equilibrium concentration of $\text{BH}^+$?
   Answer: 0.035 M, because the equilibrium constant remains constant but the stoichiometry requires adjustment for the different initial concentrations. First, calculate K from the initial experiment: K = [A⁻][BH⁺]/[HA][B] = (0.040)(0.040)/[(0.10-0.040)(0.10-0.040)] = 0.0016/0.0036 = 0.444. For the new solution, let x = [BH⁺] at equilibrium. Then [A⁻] = x, [HA] = 0.20-x, [B] = 0.050-x. Setting up: 0.444 = x²/[(0.20-x)(0.050-x)]. Solving this quadratic gives x ≈ 0.035 M. (A) incorrectly assumes simple proportionality. (C) wrongly assumes product concentrations are independent of initial reactant concentrations. (D) overestimates by not properly accounting for the limiting effect of the lower B concentration.
9. A student observes that when equal molar amounts of $\text{CH}_3\text{NH}_2$ (methylamine) and $\text{CH}_3\text{COOH}$ (acetic acid) are mixed in water, the resulting solution has a pH of 7.4. Based on this experimental observation, which conclusion about the relative strengths of these species as a base and acid respectively is most justified?
   Answer: Methylamine is a stronger base than acetic acid is an acid, since the basic hydrolysis of $\text{CH}_3\text{NH}_3^+$ predominates over acidic hydrolysis of $\text{CH}_3\text{COO}^-$. When equal molar amounts react: CH₃NH₂ + CH₃COOH → CH₃NH₃⁺ + CH₃COO⁻. The resulting solution contains equal concentrations of CH₃NH₃⁺ (conjugate acid) and CH₃COO⁻ (conjugate base). Since pH = 7.4 > 7, the solution is basic, meaning CH₃NH₃⁺ is a weaker acid than CH₃COO⁻ is a base. This occurs when the original base (CH₃NH₂) is stronger than the original acid (CH₃COOH). (B) is incorrect - there's complete neutralization, not excess acid. (C) is wrong because equal reaction stoichiometry doesn't mean equal acid/base strengths. (D) incorrectly describes the hydrolysis - CH₃COO⁻ undergoes basic hydrolysis, not acidic.
10. Which species can act as a Brønsted-Lowry base but NOT as a Brønsted-Lowry acid in aqueous solution?
    Answer: CO₃²⁻ A Brønsted-Lowry acid is a proton (H⁺) donor, and a Brønsted-Lowry base is a proton acceptor. To act as an acid, a species must have a proton to donate. To act as a base, it must be able to accept a proton. The carbonate ion, CO₃²⁻, can accept a proton to form the hydrogen carbonate ion, HCO₃⁻, so it can act as a base. However, it has no protons to donate, so it cannot act as an acid. H₂PO₄⁻ and H₂O are amphiprotic (can act as both). NH₄⁺ can donate a proton to form NH₃ (acting as an acid) but cannot accept another proton.
11. In studying proton transfer reactions, a researcher observes that the reaction $\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$ has a rate constant for the forward reaction that is independent of pH, while the reverse reaction rate constant increases significantly as pH increases. Which molecular-level explanation best accounts for this pH-dependent kinetic behavior?
    Answer: The forward reaction has a pH-independent activation energy since it involves breaking the same H-A bond regardless of conditions, while the reverse reaction activation energy decreases with increasing pH. When analyzing pH-dependent kinetics in acid-base reactions, focus on how pH affects the energy barriers for forward and reverse processes differently. The key insight is that activation energy can be pH-dependent. For the forward reaction ($\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-$), you're breaking the same H-A bond regardless of solution conditions, so the activation energy remains constant and the rate constant is pH-independent. However, the reverse reaction involves $\text{H}_3\text{O}^+$ transferring a proton to $\text{A}^-$. At higher pH (lower $[\text{H}_3\text{O}^+]$), the remaining hydronium ions are more reactive and the hydroxide-rich environment facilitates proton abstraction, effectively lowering the activation energy and increasing the rate constant. Answer A incorrectly focuses on concentration effects rather than the intrinsic rate constants. While concentrations do change with pH, the question specifically asks about rate constants, which are temperature-dependent properties independent of concentration. Answer B mischaracterizes the mechanism. Both forward and reverse reactions are intermolecular - the forward involves HA colliding with water molecules, not intramolecular rearrangement. Answer C confuses rate constants with reaction rates. Rate constants are intrinsic properties that don't depend on concentrations, while reaction rates do. The question explicitly states that rate constants (not rates) show this pH dependence. Remember: distinguish between rate constants (intrinsic molecular properties affected by activation energy) and reaction rates (which depend on both rate constants and concentrations). IB Chemistry often tests this distinction.
12. A strong acid is titrated with a strong base, and the pH is monitored. Which statement correctly describes the composition of the solution after the equivalence point has been reached?
    Answer: It contains the excess strong base, the salt of the strong acid, and water. At the equivalence point, the moles of acid and base are stoichiometrically equal, and they react to form a salt and water. After the equivalence point, all of the initial acid has been consumed, and there is an excess of the titrant (the strong base). Therefore, the solution contains the salt formed from the neutralization reaction, the excess strong base that has been added, and water.
13. What is the concentration of a sulfuric acid solution (H₂SO₄) if 15.00 cm³ of the acid is completely neutralized by 25.00 cm³ of 0.300 mol dm⁻³ potassium hydroxide (KOH)?
    Answer: 0.250 mol dm⁻³ The balanced neutralization equation is H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l). The mole ratio of H₂SO₄ to KOH is 1:2. First, calculate the moles of KOH used: n(KOH) = C × V = 0.300 mol dm⁻³ × 0.02500 dm³ = 0.00750 mol. Next, use the mole ratio to find the moles of H₂SO₄: n(H₂SO₄) = n(KOH) / 2 = 0.00750 mol / 2 = 0.00375 mol. Finally, calculate the concentration of the H₂SO₄ solution: [H₂SO₄] = n / V = 0.00375 mol / 0.01500 dm³ = 0.250 mol dm⁻³.
14. In the reversible reaction: H₂S(aq) + CN⁻(aq) ⇌ HS⁻(aq) + HCN(aq), which two species are acting as Brønsted-Lowry bases?
    Answer: CN⁻ and HS⁻ A Brønsted-Lowry base is a proton acceptor. In the forward reaction, H₂S donates a proton and CN⁻ accepts it, so CN⁻ is the base. In the reverse reaction, HCN donates a proton and HS⁻ accepts it, so HS⁻ is the base. Therefore, the two species acting as bases are CN⁻ and HS⁻.
15. Which pair represents a Brønsted-Lowry conjugate acid-base pair?
    Answer: H₂CO₃ and HCO₃⁻ A conjugate acid-base pair consists of two species that differ from each other by a single proton (H⁺). Carbonic acid (H₂CO₃) is an acid, and when it donates a proton, it forms its conjugate base, the hydrogen carbonate ion (HCO₃⁻). These two species differ by one H⁺. H₂SO₄ and SO₄²⁻ differ by two protons. HCl and NaOH are an acid and a base, but not a conjugate pair. H₃O⁺ and OH⁻ differ by two protons (or H₂O), not one.
16. Consider the amphiprotic species $\text{HPO}_4^{2-}$ in aqueous solution. When analyzing its behavior in a buffer system, which statement most accurately describes the conditions under which $\text{HPO}_4^{2-}$ will predominantly act as a proton donor rather than a proton acceptor?
    Answer: When the solution pH is significantly above the $pK_a$ of the $\text{HPO}_4^{2-}/\text{PO}_4^{3-}$ conjugate acid-base pair, favoring deprotonation reactions. When you encounter amphiprotic species like $\text{HPO}_4^{2-}$, remember that these molecules can both donate and accept protons. The key to predicting their behavior lies in understanding how solution pH relates to relevant $pK_a$ values and applying Le Châtelier's principle. For $\text{HPO}_4^{2-}$ to act as a proton donor, it must give up a proton: $\text{HPO}_4^{2-} + \text{H}_2\text{O} \rightleftharpoons \text{PO}_4^{3-} + \text{H}_3\text{O}^+$. This deprotonation is favored when the solution pH is significantly above the $pK_a$ of the $\text{HPO}_4^{2-}/\text{PO}_4^{3-}$ pair. Under basic conditions, the equilibrium shifts right to produce more hydronium ions, making $\text{HPO}_4^{2-}$ predominantly act as an acid. Answer A correctly identifies this relationship. Answer B is backwards—when pH is below the $pK_a$ of $\text{H}_2\text{PO}_4^-/\text{HPO}_4^{2-}$, the protonated form ($\text{H}_2\text{PO}_4^-$) is favored, meaning $\text{HPO}_4^{2-}$ would act as a proton acceptor, not donor. Answer C contradicts chemical logic. Adding strong acids increases $\text{H}_3\text{O}^+$ concentration, which would drive $\text{HPO}_4^{2-}$ to accept protons (act as a base), not donate them. Answer D focuses on concentration ratios rather than the pH-$pK_a$ relationship that actually determines acid-base behavior. Study tip: For amphiprotic species, always compare solution pH to the relevant $pK_a$. When pH > $pK_a$, the species acts as a proton donor; when pH < $pK_a$, it acts as a proton acceptor.
17. Two solutions of a strong monoprotic acid, Solution X and Solution Y, have pH values of 3 and 5 respectively. Which statement correctly compares the hydrogen ion concentration in the two solutions?
    Answer: [H⁺] in X is 100 times greater than in Y. The pH scale is logarithmic. A difference of 1 pH unit corresponds to a 10-fold difference in [H⁺]. A difference of 2 pH units corresponds to a 10² or 100-fold difference. Specifically, for Solution X, [H⁺] = 10⁻³ mol dm⁻³. For Solution Y, [H⁺] = 10⁻⁵ mol dm⁻³. The ratio of the concentrations is [H⁺]ₓ / [H⁺]ᵧ = 10⁻³ / 10⁻⁵ = 10². Therefore, the hydrogen ion concentration in Solution X is 100 times greater than in Solution Y.
18. A student has two beakers, one containing 100 cm³ of 0.10 mol dm⁻³ HCl(aq) and the other containing 100 cm³ of 0.10 mol dm⁻³ CH₃COOH(aq) at 298 K. Which statement correctly compares the two solutions?
    Answer: Both solutions will require the same volume of 0.10 mol dm⁻³ NaOH(aq) for complete neutralization. HCl is a strong acid and CH₃COOH is a weak acid. Neutralization depends on the total moles of acid present. Since both solutions have the same concentration and volume (n = C × V), they contain the same moles of acid (0.010 mol). Therefore, they will require the same volume of NaOH for neutralization. A is incorrect because the strong acid (HCl) will have a lower pH. B is incorrect because HCl fully dissociates, creating a higher concentration of ions and thus higher conductivity. D is incorrect because the rate of reaction depends on [H⁺], which is higher in the HCl solution, leading to a faster initial rate.
19. In a solution containing $\text{HF}$, $\text{F}^-$, $\text{HSO}_4^-$, and $\text{SO}_4^{2-}$, the following equilibrium is established: $\text{HF} + \text{SO}_4^{2-} \rightleftharpoons \text{F}^- + \text{HSO}_4^-$. Given that $K_a(\text{HF}) = 6.8 \times 10^{-4}$ and $K_a(\text{HSO}_4^-) = 1.2 \times 10^{-2}$, which prediction about this proton transfer equilibrium is most accurate?
    Answer: The equilibrium lies significantly to the left because $\text{HSO}_4^-$ is a stronger acid than HF and the reverse reaction is favored. The equilibrium constant for this reaction is K = Ka(HF)/Ka(HSO₄⁻) = (6.8 × 10⁻⁴)/(1.2 × 10⁻²) ≈ 0.057. Since K << 1, the equilibrium lies significantly to the left. This makes sense because HSO₄⁻ (Ka = 1.2 × 10⁻²) is a much stronger acid than HF (Ka = 6.8 × 10⁻⁴), so the reverse reaction (HSO₄⁻ donating a proton to F⁻) is strongly favored. (A) incorrectly identifies HF as the stronger acid. (C) is wrong because K = 0.057 indicates the equilibrium is not close to center. (D) is incorrect because the equilibrium position (K value) is determined solely by the relative acid strengths.
20. A chemist studies the proton transfer kinetics of the reaction $\text{HCN} + \text{OH}^- \rightarrow \text{CN}^- + \text{H}_2\text{O}$ and finds that the reaction rate is limited by the encounter frequency between reactant molecules rather than by the activation energy for proton transfer. What does this observation suggest about the mechanism of this particular proton transfer reaction?
    Answer: The reaction is diffusion-controlled, with essentially every collision between HCN and $\text{OH}^-$ resulting in immediate and successful proton transfer. When a reaction rate is limited by encounter frequency rather than activation energy, it means the reaction is diffusion-controlled. This occurs when the activation barrier for the chemical step (proton transfer) is so low that essentially every collision between reactants leads to successful reaction. The rate is then determined by how quickly the molecules can diffuse through solution to encounter each other. (A) is incorrect because thermodynamically unfavorable reactions would have high activation barriers, not diffusion-limited kinetics. (B) describes a different mechanism with pre-equilibrium steps. (D) would still involve activation energy considerations for proper orientation, not pure diffusion control.

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