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Apply Proton Transfer Reactions - IB Chemistry

  1. Which pair of statements correctly compares 0.1 mol dm⁻³ NaOH(aq) with 0.1 mol dm⁻³ NH₃(aq)?
    Answer: NaOH has a higher pH and is a stronger base. Strength of a base refers to its degree of dissociation. NaOH is a strong base, meaning it dissociates completely in water to produce a high [OH⁻]. NH₃ is a weak base, meaning it only partially reacts with water, resulting in a lower [OH⁻]. A higher [OH⁻] corresponds to a higher pH. Therefore, the NaOH solution has a higher pH and NaOH is classified as a stronger base than NH₃.
  2. A titration is performed by adding a strong base to a strong acid. Which combination of observations is most consistent with this type of titration?
    Answer: The initial pH is around 1 and the pH at the equivalence point is 7. A strong acid will have a very low initial pH, typically around 1 for a concentration like 0.1 mol dm⁻³. When a strong acid is neutralized by a strong base, the salt formed (e.g., NaCl from HCl + NaOH) does not hydrolyze water, so the pH at the equivalence point is neutral, which is 7 at 298 K. The other options describe titrations involving weak acids/bases or are reversed (base in flask).
  3. What is the approximate pH of a 0.010 mol dm⁻³ solution of hydrocyanic acid, HCN? (Ka for HCN = 6.2 × 10⁻¹⁰)
    Answer: 5.60 For a weak acid, Ka = [H⁺][A⁻]/[HA]. Assuming [H⁺] = [A⁻] and that the dissociation is small so [HA] ≈ 0.010, we have Ka ≈ [H⁺]² / [HA]. Therefore, [H⁺]² ≈ Ka × [HA] = (6.2 × 10⁻¹⁰) × 0.010 = 6.2 × 10⁻¹². Taking the square root, [H⁺] ≈ 2.5 × 10⁻⁶ mol dm⁻³. Then, pH = -log([H⁺]) = -log(2.5 × 10⁻⁶) ≈ 5.60. A pH of 2.00 would be for a strong acid.
  4. A solution is prepared by dissolving sodium ethanoate (CH₃COONa) in water. What is the net ionic equation for the reaction that occurs and how does it affect the pH?
    Answer: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq); pH increases. Sodium ethanoate is the salt of a strong base (NaOH) and a weak acid (CH₃COOH). The Na⁺ ion is a spectator ion. The ethanoate ion, CH₃COO⁻, is the conjugate base of a weak acid and will act as a Brønsted-Lowry base, accepting a proton from water. This process is called hydrolysis. The reaction is CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). The production of hydroxide ions (OH⁻) makes the solution basic, thus increasing the pH above 7.
  5. Phosphoric acid, H₃PO₄, is a triprotic acid. When titrated with NaOH, which species will be present in the highest concentration when exactly 1.5 equivalents of NaOH have been added?
    Answer: H₂PO₄⁻ and HPO₄²⁻ in equal concentration. Adding 1.0 equivalent of NaOH converts all H₃PO₄ to H₂PO₄⁻ (the first equivalence point). Adding another 0.5 equivalent of NaOH (for a total of 1.5) will convert half of the H₂PO₄⁻ into its conjugate base, HPO₄²⁻. This point is the second half-equivalence point. At this stage, the solution contains equal concentrations of the weak acid H₂PO₄⁻ and its conjugate base HPO₄²⁻.
  6. Two buffer solutions, X and Y, are prepared using ethanoic acid and sodium ethanoate. Buffer X contains 0.10 mol dm⁻³ of each component. Buffer Y contains 0.50 mol dm⁻³ of each component. A small amount of strong acid is added to both buffers. Which statement correctly describes the result?
    Answer: The pH of buffer X changes more than the pH of buffer Y. Both buffers have a [A⁻]/[HA] ratio of 1, so their initial pH will be the same (pH = pKa). However, their buffer capacities are different. Buffer Y has higher concentrations of the acid and conjugate base, meaning it can neutralize more added acid or base before its pH changes significantly. Buffer X has a lower capacity. Therefore, upon adding the same amount of strong acid, the pH of the more dilute buffer (X) will change more than the pH of the more concentrated buffer (Y).
  7. Solutions of 0.1 mol dm⁻³ hydrochloric acid, HCl(aq), and 0.1 mol dm⁻³ ethanoic acid, CH₃COOH(aq), are compared at 298 K. Which property will be approximately the same for both solutions?
    Answer: The volume of 0.1 mol dm⁻³ NaOH required for neutralization. Neutralization depends on the total moles of acid present, not the initial concentration of H⁺ ions. Since both solutions have the same concentration (0.1 mol dm⁻³) and implicitly the same volume is being considered for neutralization, they contain the same moles of acid molecules (HCl or CH₃COOH). Therefore, they will require the same volume of NaOH for complete neutralization. The other properties (pH, conductivity, reaction rate) depend on the concentration of H⁺ ions, which is much lower for the weak acid (ethanoic acid) than the strong acid (hydrochloric acid).
  8. A student prepares a buffer solution by mixing 0.10 mol of ammonia (NH3NH_3) with 0.080 mol of ammonium chloride (NH4ClNH_4Cl) in 1.0 L of solution. If 0.020 mol of solid sodium hydroxide (NaOHNaOH) is then added to this buffer, what is the approximate pH of the resulting solution? (KbK_b for NH3NH_3 = 1.8×10−51.8 \times 10^{-5})
    Answer: 9.7 Initially: [NH₃] = 0.10 M, [NH₄⁺] = 0.080 M. When NaOH is added, it reacts with NH₄⁺: NH₄⁺ + OH⁻ → NH₃ + H₂O. After reaction: [NH₃] = 0.10 + 0.020 = 0.12 M, [NH₄⁺] = 0.080 - 0.020 = 0.060 M. Using Henderson-Hasselbalch: pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.060/0.12) = 4.74 - 0.30 = 4.44. Therefore pH = 14 - 4.44 = 9.56 ≈ 9.7.
  9. Which mixture will produce a buffer solution with a pH greater than 7?
    Answer: 50 cm³ of 0.1 mol dm⁻³ NH₃ and 25 cm³ of 0.1 mol dm⁻³ HCl A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Option C involves mixing a weak base (NH₃) with a strong acid (HCl) in a 2:1 mole ratio. The HCl will react with half of the NH₃ to form its conjugate acid, NH₄⁺. The resulting solution contains significant amounts of both the weak base (NH₃) and its conjugate acid (NH₄⁺), forming a buffer. Since it is based on the NH₃/NH₄⁺ system (pKb of NH₃ is 4.75, so pKa of NH₄⁺ is 9.25), the buffer will have a pH in the basic region (>7). Option B creates an acidic buffer. Option A is a neutralization reaction resulting in a neutral salt. Option D is a reaction between a weak acid and weak base.
  10. In different reactions, a species can act as a Brønsted-Lowry acid or a Brønsted-Lowry base. Which species is amphiprotic?
    Answer: HSO₃⁻ An amphiprotic species can both donate and accept a proton. HSO₃⁻ can donate a proton to form SO₃²⁻, acting as an acid. It can also accept a proton to form H₂SO₃, acting as a base. CO₃²⁻ can only accept a proton (base). H₂SO₄ is a strong acid that can only donate protons. NH₄⁺ can only donate a proton (acid).
  11. An aqueous solution of ammonium nitrate, NH₄NO₃, is tested with litmus paper. Which of the following correctly describes the observation and the ion responsible?
    Answer: The solution is acidic because the NH₄⁺ ion hydrolyzes water. Ammonium nitrate is the salt of a weak base (NH₃) and a strong acid (HNO₃). The nitrate ion (NO₃⁻) is the conjugate base of a strong acid and does not hydrolyze. The ammonium ion (NH₄⁺) is the conjugate acid of a weak base and will hydrolyze water according to the equation: NH₄⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NH₃(aq). The production of H₃O⁺ ions makes the solution acidic.
  12. A weak acid is titrated with a strong base, and the pH is monitored. The equivalence point is found to occur at pH = 8.8. Which indicator would be the most suitable to determine the end point of this titration?
    Answer: Phenolphthalein (pKa = 9.6) A suitable indicator for a titration has a pKa value close to the pH at the equivalence point. The indicator's color change range (approximately pKa ± 1) should bracket the equivalence point pH. The equivalence point is at pH 8.8. Of the choices, phenolphthalein (pKa = 9.6) is the closest, and its color change range (approx. 8.6-10.6) effectively covers the equivalence point. The other indicators would change color too early or too late.
  13. In a titration of 20.0 mL of 0.100 M CH3NH2CH_3NH_2 (methylamine, Kb=4.4×10−4K_b = 4.4 \times 10^{-4}) with 0.100 M HClHCl, at what volume of added HClHCl will the pH equal the pKapK_a of the CH3NH3+CH_3NH_3^+ ion?
    Answer: 10.0 mL The pH equals pKa when [base] = [conjugate acid], which occurs at the half-equivalence point. Initial moles of CH₃NH₂ = 0.020 L × 0.100 M = 2.0 × 10⁻³ mol. At half-equivalence, 1.0 × 10⁻³ mol of HCl is needed: Volume = (1.0 × 10⁻³ mol)/(0.100 M) = 0.010 L = 10.0 mL. At this point, [CH₃NH₂] = [CH₃NH₃⁺] and pH = pKa.
  14. In aqueous solution, the diprotic acid H2AH_2A undergoes the following equilibria: H2A⇌H++HA−H_2A \rightleftharpoons H^+ + HA^- (Ka1=1.0×10−3K_{a1} = 1.0 \times 10^{-3}) and HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-} (Ka2=1.0×10−8K_{a2} = 1.0 \times 10^{-8}). At what pH will the concentration of HA−HA^- be maximized?
    Answer: 5.5 The concentration of HA⁻ is maximized when d[HA⁻]/d[H⁺] = 0. This occurs at the geometric mean of the two Ka values: pH = ½(pKa1 + pKa2) = ½(3.0 + 8.0) = 5.5. At this pH, the rates of formation (from H₂A) and consumption (to A²⁻) of HA⁻ are balanced optimally.
  15. If the hydrogen ion concentration, [H⁺], in a solution is increased by a factor of 1000, how does the pH value of the solution change?
    Answer: It decreases by 3. The pH scale is logarithmic. pH = -log₁₀[H⁺]. Let the initial concentration be [H⁺]₁ and the final be [H⁺]₂ = 1000 × [H⁺]₁. The change in pH is ΔpH = pH₂ - pH₁ = -log([H⁺]₂) - (-log([H⁺]₁)) = -log(1000 × [H⁺]₁) + log([H⁺]₁) = -(log(1000) + log([H⁺]₁)) + log([H⁺]₁) = -log(1000) - log([H⁺]₁) + log([H⁺]₁) = -log(10³) = -3. Therefore, the pH decreases by 3.
  16. The base dissociation constant, Kb, for ammonia (NH₃) is 1.8 × 10⁻⁵. What is the acid dissociation constant, Ka, for the ammonium ion (NH₄⁺) at 298 K? (Kw = 1.0 × 10⁻¹⁴)
    Answer: 5.6 × 10⁻¹⁰ For a conjugate acid-base pair, the relationship Ka × Kb = Kw holds. The ammonium ion, NH₄⁺, is the conjugate acid of the base ammonia, NH₃. Therefore, Ka(NH₄⁺) = Kw / Kb(NH₃) = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) ≈ 5.6 × 10⁻¹⁰.
  17. In a titration, 20.00 cm³ of a sulfuric acid (H₂SO₄) solution completely neutralizes 30.00 cm³ of 0.150 mol dm⁻³ potassium hydroxide (KOH). What is the concentration of the sulfuric acid solution?
    Answer: 0.113 mol dm⁻³ The balanced equation is H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. The mole ratio of H₂SO₄ to KOH is 1:2. First, find the moles of KOH used: n(KOH) = C × V = 0.150 mol dm⁻³ × 0.03000 dm³ = 0.00450 mol. Using the mole ratio, find the moles of H₂SO₄: n(H₂SO₄) = 0.00450 mol KOH × (1 mol H₂SO₄ / 2 mol KOH) = 0.00225 mol H₂SO₄. Finally, calculate the concentration of H₂SO₄: C = n / V = 0.00225 mol / 0.02000 dm³ = 0.1125 mol dm⁻³, which is 0.113 mol dm⁻³.
  18. Consider the titration of a weak base with a strong acid. At which point on the titration curve is the concentration of the weak base equal to the concentration of its conjugate acid?
    Answer: At the half-equivalence point. The half-equivalence point is the point at which exactly half of the initial weak base has been neutralized by the strong acid, converting it into its conjugate acid. At this specific point, the concentrations of the remaining weak base and the newly formed conjugate acid are equal. This is also the point where pH = pKa (or pOH = pKb) for the conjugate pair.
  19. Which of the following represents the balanced net ionic equation for the reaction between aqueous nitric acid and solid calcium carbonate?
    Answer: 2H⁺(aq) + CaCO₃(s) → Ca²⁺(aq) + H₂O(l) + CO₂(g) The full molecular equation is 2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g). To get the net ionic equation, we write the ions for the strong acid and the soluble salt: 2H⁺(aq) + 2NO₃⁻(aq) + CaCO₃(s) → Ca²⁺(aq) + 2NO₃⁻(aq) + H₂O(l) + CO₂(g). The nitrate ions (NO₃⁻) are spectator ions and are cancelled out. The remaining species give the net ionic equation: 2H⁺(aq) + CaCO₃(s) → Ca²⁺(aq) + H₂O(l) + CO₂(g).
  20. A biochemist studies the amino acid glycine, which has two ionizable groups: −COOH-COOH (pKa1=2.3pK_{a1} = 2.3) and −NH3+-NH_3^+ (pKa2=9.6pK_{a2} = 9.6). At what pH will glycine exist predominantly as a zwitterion (+NH3CH2COO−^+NH_3CH_2COO^-)?
    Answer: pH 4.5-7.5 When you encounter amino acid ionization problems, focus on the relationship between pH and pK_a values to determine which functional groups are protonated or deprotonated. A zwitterion exists when the amino acid carries both a positive and negative charge simultaneously. For glycine, this means the carboxyl group (−COOH-COOH) must be deprotonated to −COO−-COO^- while the amino group remains protonated as −NH3+-NH_3^+. The carboxyl group deprotonates when pH > pKa1pK_{a1} (2.3), and the amino group remains protonated when pH < pKa2pK_{a2} (9.6). Therefore, glycine exists as a zwitterion when the pH falls between these two pK_a values: 2.3 < pH < 9.6. Answer choice D (pH 4.5-7.5) is the only range that falls entirely within this window. Answer A (pH 1.0-3.0) is incorrect because at these low pH values, both groups remain protonated, giving glycine a net positive charge (+NH3CH2COOH^+NH_3CH_2COOH). Answer B (pH 11.5-13.0) is wrong because at such high pH, both groups are deprotonated, creating a net negative charge (NH2CH2COO−NH_2CH_2COO^-). Answer C (pH 8.0-11.0) is incorrect because while it includes some zwitterionic range, it extends beyond pKa2pK_{a2} = 9.6, where the amino group begins deprotonating. Remember: zwitterions exist between the two pK_a values of diprotic amino acids. Always identify which functional groups should be charged at the given pH by comparing to their respective pK_a values.

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