Acid-Base Titrations - AP Chemistry

1. A student titrates $50.0\,\text{mL}$ of $0.100\,\text{M}$ CH$_3$NH$_2$ ($K_b=4.4\times 10^{-4}$) with $0.100\,\text{M}$ HCl. At the equivalence point, which statement is correct about the pH of the solution?
   Answer: The pH is less than 7.00 because CH$_3$NH$_3^+$ is a weak acid. This question tests the skill of acid–base titrations. Stoichiometry shows that equivalence requires 50 mL HCl for 50 mL 0.1 M CH3NH2, so at equivalence, all base is converted to CH3NH3+, with Cl-. The species is CH3NH3+, a weak acid conjugate of the weak base. The pH is less than 7, determined by the hydrolysis of CH3NH3+. A tempting distractor is that pH is 7.00, but that's for strong, not for weak base strong acid salt, which is acidic. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.
2. A student titrates $50.0\ \text{mL}$ of $0.100\ \text{M}$ HCl with $0.200\ \text{M}$ NaOH. What volume of NaOH is required to reach the equivalence point?
   Answer: 25.0 mL This question involves acid–base titrations. Stoichiometry calculates the equivalence volume as moles of HCl (5.0 mmol) divided by NaOH concentration (0.200 M), requiring 25.0 mL of NaOH. At equivalence, all acid is neutralized, leaving neutral ions, but the question focuses on volume, which determines when species shift from excess acid to neutral. This point controls pH to 7.00 for strong-strong titrations; Henderson–Hasselbalch is not relevant. A tempting distractor is 50.0 mL, possibly from forgetting to account for the higher NaOH concentration. Determine the titration stage first, then choose the appropriate method like mole equality for equivalence volumes.
3. A student titrates $50.0\,\text{mL}$ of $0.100\,\text{M}$ benzoic acid, HC$_7$H$_5$O$_2$ ($K_a=6.3\times 10^{-5}$), with $0.100\,\text{M}$ NaOH. After $10.0\,\text{mL}$ of NaOH has been added (before equivalence), which species is present in the greatest amount (ignoring water)?
   Answer: HC$_7$H$_5$O$_2$ This question tests the skill of acid–base titrations. Stoichiometry shows that 10 mL of 0.100 M NaOH adds 1 mmol OH-, which reacts with 1 mmol of the 5 mmol HC7H5O2, leaving 4 mmol HA and producing 1 mmol A-. The species remaining are HA, A-, and Na+, with HA in the greatest amount. The pH is controlled by the buffer of HA and A- using Henderson-Hasselbalch. A tempting distractor is OH-, thinking of the added base, but the OH- is consumed in the reaction. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.
4. A student titrates $25.0\,\text{mL}$ of $0.200\,\text{M}$ ammonia, $\text{NH}_3$ ($K_b = 1.8\times10^{-5}$; $pK_b\approx 4.74$), with $0.100\,\text{M}$ HCl. At the half-equivalence point, which statement best describes the solution?
   Answer: The pH equals $pK_a$ of $\text{NH}_4^+$ and $[\text{NH}_3]=[\text{NH}_4^+]$. This problem examines acid-base titrations of a weak base with a strong acid. At the half-equivalence point, exactly half the NH₃ has been converted to NH₄⁺, creating a buffer with equal concentrations of both species. The Henderson-Hasselbalch equation for this conjugate acid-base pair gives: pH = pKₐ(NH₄⁺) + log([NH₃]/[NH₄⁺]) = pKₐ + log(1) = pKₐ. Since pKₐ + pKᵦ = 14, we have pKₐ(NH₄⁺) = 14 - 4.74 = 9.26, so pH = 9.26. A common mistake is thinking the pH equals 7.00 because equal moles have reacted, but buffers don't necessarily have neutral pH. Always identify whether you have a buffer system, then apply Henderson-Hasselbalch when [conjugate base] = [conjugate acid].
5. A student titrates $25.0\ \text{mL}$ of $0.100\ \text{M}$ HCl with $0.100\ \text{M}$ NaOH. What is the pH after $30.0\ \text{mL}$ of NaOH has been added? (Assume additive volumes.)
   Answer: 11.96 This question involves acid–base titrations. Stoichiometry shows that adding 30.0 mL of 0.100 M NaOH provides 3.0 mmol of OH-, exceeding the initial 2.5 mmol of HCl by 0.5 mmol. Past the equivalence point in a strong acid-strong base titration, excess OH- remains, controlling the pH directly. The total volume is 55.0 mL, so [OH-] = 0.0005 mol / 0.055 L ≈ 0.00909 M, pOH ≈ 2.04, pH ≈ 11.96; Henderson–Hasselbalch is inapplicable without a buffer. A tempting distractor is pH = 12.00, perhaps from rounding [OH-] to 0.01 M without precise calculation. Determine the titration stage first, then choose the appropriate method such as -log[OH-] for excess strong base.
6. A student titrates $30.0\ \text{mL}$ of $0.100\ \text{M}$ HF ($pK_a\approx 3.17$) with $0.100\ \text{M}$ NaOH. After $15.0\ \text{mL}$ of NaOH is added, what is the pH? (Use Henderson–Hasselbalch.)
   Answer: 3.17 This question involves acid–base titrations. Stoichiometry shows that 15.0 mL of 0.100 M NaOH neutralizes 1.5 mmol of HF, leaving 1.5 mmol HF and producing 1.5 mmol F- from initial 3.0 mmol. At the half-equivalence point in a weak acid-strong base titration, equal [HF] and [F-] buffer the solution, controlling pH. Henderson–Hasselbalch is appropriate: pH = 3.17 + log(1) = 3.17. A tempting distractor is pH = 7.00, from assuming neutrality at half-neutralization. Determine the titration stage first, then choose the appropriate method such as Henderson–Hasselbalch for equal conjugate pairs.
7. A student titrates $25.0\,\text{mL}$ of $0.100\,\text{M}$ HNO$_3$ with $0.100\,\text{M}$ KOH. What is the pH after $30.0\,\text{mL}$ of KOH has been added?
   Answer: 11.96 This problem involves acid-base titrations with excess base. Starting with 25.0 mL × 0.100 M = 2.50 mmol HNO₃ and adding 30.0 mL × 0.100 M = 3.00 mmol KOH, we have 0.50 mmol excess OH⁻ in 55.0 mL total volume. The [OH⁻] = 0.50 mmol / 55.0 mL = 0.00909 M, giving pOH = -log(0.00909) = 2.04, so pH = 14 - 2.04 = 11.96. A common mistake is calculating pH directly from excess base moles without accounting for the total volume change. For strong acid-strong base titrations past equivalence, calculate excess [OH⁻], find pOH, then convert to pH.
8. A student titrates $50.0\ \text{mL}$ of $0.100\ \text{M}$ CH$_3$COOH (acetic acid, $K_a=1.8\times10^{-5}$) with $0.100\ \text{M}$ NaOH. At the half-equivalence point, what is the pH? (You may use $pK_a\approx 4.74$.)
   Answer: 4.74 This question involves acid–base titrations. Stoichiometry indicates that the half-equivalence point occurs when half the initial moles of CH3COOH (2.5 mmol) have been neutralized by NaOH, leaving equal amounts of CH3COOH and CH3COO-. At this buffer stage in a weak acid-strong base titration, the species present are CH3COOH and its conjugate base CH3COO- in equal concentrations, controlling the pH via the buffer system. Henderson–Hasselbalch is appropriate here, giving pH = pKa + log([CH3COO-]/[CH3COOH]) = 4.74 + log(1) = 4.74. A tempting distractor is pH = 7.00, which might come from confusing the half-equivalence with the equivalence point where pH is not neutral for weak acids. Determine the titration stage first, then choose the appropriate method like Henderson–Hasselbalch for buffer regions.
9. A student titrates $20.0\,\text{mL}$ of $0.100\,\text{M}$ NH$_3$ ($K_b=1.8\times 10^{-5}$) with $0.100\,\text{M}$ HCl. After $10.0\,\text{mL}$ of HCl has been added (half-equivalence), which statement best describes the solution?
   Answer: The pH equals $pK_a$ of NH$_4^+$, and NH$_3$ and NH$_4^+$ are present in equal amounts. This question tests the skill of acid–base titrations. Stoichiometry shows that 10 mL of 0.100 M HCl adds 1 mmol H+, which reacts with half of the 2 mmol NH3, leaving 1 mmol NH3 and producing 1 mmol NH4+. At this half-equivalence point for a weak base, the concentrations of B and BH+ are equal. The pH is equal to pKa of the conjugate acid NH4+, using Henderson–Hasselbalch for the buffer. A tempting distractor is that the pH is 7.00, but that would be for strong acid-strong base at half-equivalence, not for weak base where it's pKa. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.
10. A student titrates $50.0\,\text{mL}$ of $0.100\,\text{M}$ HBr with $0.200\,\text{M}$ NaOH. What is the pH at the equivalence point?
    Answer: 7.00 This problem tests acid-base titrations of strong acid with strong base. Starting with 50.0 mL × 0.100 M = 5.00 mmol HBr, we need 5.00 mmol NaOH to reach equivalence. With 0.200 M NaOH, this requires 25.0 mL. At equivalence, all H⁺ from HBr neutralizes all OH⁻ from NaOH, leaving only water and spectator ions (Na⁺ and Br⁻). Since neither ion hydrolyzes, the solution is neutral with pH = 7.00. A common error is thinking all equivalence points have pH = 7.00, but this is only true for strong acid-strong base titrations. For strong-strong titrations at equivalence, the pH is always 7.00 because no hydrolysis occurs.
11. A student titrates $40.0\ \text{mL}$ of $0.100\ \text{M}$ HCOOH (formic acid, $K_a=1.8\times10^{-4}$; $pK_a\approx 3.74$) with $0.100\ \text{M}$ NaOH. After $20.0\ \text{mL}$ of NaOH is added, what is the pH? (Use Henderson–Hasselbalch.)
    Answer: 3.74 This question involves acid–base titrations. Stoichiometry reveals that 20.0 mL of 0.100 M NaOH neutralizes 2.0 mmol of HCOOH, leaving 2.0 mmol HCOOH and producing 2.0 mmol HCOO- from initial 4.0 mmol. At this half-equivalence point in a weak acid-strong base titration, equal [HCOOH] and [HCOO-] form a buffer controlling pH. Henderson–Hasselbalch is suitable: pH = pKa + log([HCOO-]/[HCOOH]) = 3.74 + log(1) = 3.74. A tempting distractor is pH = 7.00, mistakenly equating half-equivalence to neutrality. Determine the titration stage first, then choose the appropriate method like Henderson–Hasselbalch for buffer calculations.
12. A student titrates $25.0\,\text{mL}$ of $0.100\,\text{M}$ acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$ ($K_a = 1.8\times10^{-5}$; $pK_a\approx 4.74$), with $0.100\,\text{M}$ NaOH. What is the pH of the solution after $12.5\,\text{mL}$ of NaOH has been added?
    Answer: 4.74 This problem tests understanding of acid-base titrations. At the halfway point (12.5 mL added to reach 25.0 mL equivalence), exactly half the acetic acid has been neutralized, creating a buffer with equal concentrations of HC₂H₃O₂ and C₂H₃O₂⁻. The Henderson-Hasselbalch equation applies: pH = pKₐ + log([A⁻]/[HA]) = 4.74 + log(1) = 4.74. A common misconception is thinking the pH should be 7.00 at the halfway point, but this only occurs at the equivalence point of strong acid-strong base titrations. To solve titration problems, first identify the titration stage (before, at half-equivalence, at equivalence, or after), then apply the appropriate calculation method.
13. A student titrates $30.0\,\text{mL}$ of $0.100\,\text{M}$ HF ($K_a=6.8\times 10^{-4}$) with $0.100\,\text{M}$ NaOH. At the equivalence point, which species is primarily responsible for determining the pH of the solution?
    Answer: F$^-$ This question tests the skill of acid–base titrations. Stoichiometry shows that the equivalence point requires 30 mL NaOH for 30 mL 0.1 M HF, so at equivalence, all HF is converted to F-, with Na+. The species remaining are Na+ and F- in 60 mL. The pH is determined by the hydrolysis of F-, the weak base conjugate of HF, leading to pH >7. A tempting distractor is H3O+ from the strong acid, but there is no strong acid at equivalence, it's the salt of weak acid strong base. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.
14. A student titrates $50.0\,\text{mL}$ of $0.100\,\text{M}$ HNO$_3$ with $0.100\,\text{M}$ KOH. What is the pH at the equivalence point?
    Answer: 7.00 This question tests the skill of acid–base titrations. Stoichiometry shows that equivalence requires 50 mL KOH for 50 mL 0.1 M HNO3, so at equivalence, all acid and base are neutralized to KNO3. The species are K+ and NO3-, both from strong, so no hydrolysis. The pH is 7.00, as it's neutral. A tempting distractor is 1.00, perhaps thinking of initial pH of the acid, but at equivalence, it's neutral for strong strong. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.
15. A student titrates $25.0\,\text{mL}$ of $0.100\,\text{M}$ propionic acid, HC$_2$H$_5$COO ($K_a = 1.3\times10^{-5}$; $pK_a\approx 4.89$), with $0.100\,\text{M}$ NaOH. Which statement is correct about the pH at the equivalence point compared with 7.00?
    Answer: pH > 7.00 because the conjugate base hydrolyzes to form OH$^-$. This problem involves acid-base titrations examining pH at equivalence. When propionic acid (weak acid) is completely neutralized by NaOH (strong base), all acid converts to propionate ion (C₂H₅COO⁻). This conjugate base hydrolyzes: C₂H₅COO⁻ + H₂O ⇌ HC₂H₅COO + OH⁻, producing OH⁻ and making pH > 7.00. The extent of hydrolysis depends on Kᵦ = Kw/Kₐ = 1.0×10⁻¹⁴/1.3×10⁻⁵ = 7.7×10⁻¹⁰. A common misconception is that all titrations are neutral at equivalence, but only strong acid-strong base titrations have pH = 7.00. For weak acid-strong base titrations, the conjugate base formed at equivalence always hydrolyzes, making pH > 7.00.
16. A student titrates $25.0\ \text{mL}$ of $0.100\ \text{M}$ CH$_3$COOH ($pK_a\approx 4.74$) with $0.100\ \text{M}$ NaOH. After $12.5\ \text{mL}$ of NaOH is added, what is the pH? (Use Henderson–Hasselbalch.)
    Answer: 4.74 This question involves acid–base titrations. Stoichiometry determines that 12.5 mL of 0.100 M NaOH neutralizes 1.25 mmol of CH3COOH, leaving 1.25 mmol CH3COOH and 1.25 mmol CH3COO- from initial 2.5 mmol. At half-equivalence in a weak acid-strong base titration, the equal acid and conjugate base control pH as a buffer. Henderson–Hasselbalch gives pH = 4.74 + log(1) = 4.74. A tempting distractor is pH = 7.00, mistakenly thinking half-equivalence means neutral like strong acids. Determine the titration stage first, then choose the appropriate method for buffer pH calculations.
17. A student titrates $30.0\,\text{mL}$ of $0.100\,\text{M}$ methylamine, CH$_3$NH$_2$ ($K_b = 4.4\times10^{-4}$; $pK_b\approx 3.36$), with $0.100\,\text{M}$ HCl. At the half-equivalence point, what is the pH? (Use $pK_a = 14.00 - pK_b$ for CH$_3$NH$_3^+$.)
    Answer: 10.64 This problem involves acid-base titrations of a weak base with strong acid. At the half-equivalence point, half the methylamine (CH₃NH₂) is converted to its conjugate acid (CH₃NH₃⁺), creating a buffer with equal concentrations. For this buffer, pH = pKₐ(CH₃NH₃⁺) + log([CH₃NH₂]/[CH₃NH₃⁺]) = pKₐ + log(1) = pKₐ. Given pKᵦ = 3.36 for methylamine, pKₐ(CH₃NH₃⁺) = 14.00 - 3.36 = 10.64, so pH = 10.64. A common mistake is using pKᵦ directly instead of converting to pKₐ for the conjugate acid. For weak base titrations, always convert to the conjugate acid's pKₐ when applying Henderson-Hasselbalch.
18. A student titrates $25.0\,\text{mL}$ of $0.100\,\text{M}$ hydrofluoric acid, HF ($K_a = 6.8\times10^{-4}$; $pK_a\approx 3.17$), with $0.100\,\text{M}$ NaOH. Which species is primarily responsible for determining the pH at the equivalence point?
    Answer: F$^-$, because it hydrolyzes to produce OH$^-$. This problem examines acid-base titrations at the equivalence point. When HF is completely neutralized by NaOH, all HF is converted to F⁻ (fluoride ion), with no excess acid or base remaining. The F⁻ ion undergoes hydrolysis: F⁻ + H₂O ⇌ HF + OH⁻, producing OH⁻ and making the solution basic. This hydrolysis reaction determines the pH at equivalence. A common misconception is that Na⁺ affects pH, but it's a spectator ion from a strong base and doesn't hydrolyze. At the equivalence point of weak acid-strong base titrations, identify the conjugate base formed and recognize it will hydrolyze to control pH.
19. A student titrates $30.0\ \text{mL}$ of $0.200\ \text{M}$ HNO$_3$ with $0.100\ \text{M}$ KOH. What is the pH at the equivalence point?
    Answer: 7.00 This question involves acid–base titrations. Stoichiometry determines that the equivalence point requires 60.0 mL of 0.100 M KOH to neutralize 6.0 mmol of HNO3 from 30.0 mL of 0.200 M HNO3. At the equivalence point in a strong acid-strong base titration, all acid and base are neutralized, leaving only spectator ions K+ and NO3-, resulting in a neutral solution. The pH is controlled by water autoionization, so pH = 7.00; Henderson–Hasselbalch is not appropriate as no buffer is present. A tempting distractor is pH = 13.00, possibly from mistakenly calculating excess base instead of recognizing neutralization. Determine the titration stage first, then choose the appropriate method such as checking for neutral salt in strong-strong titrations.
20. A student titrates $25.0\ \text{mL}$ of $0.100\ \text{M}$ CH$_3$COOH ($pK_a\approx 4.74$) with $0.100\ \text{M}$ NaOH. Which statement is true at the half-equivalence point?
    Answer: The pH equals $pK_a$ and $[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]$. This question involves acid–base titrations. Stoichiometry shows that at half-equivalence (12.5 mL NaOH), half the CH3COOH is neutralized, leaving equal [CH3COOH] and [CH3COO-]. This buffer stage controls pH such that pH = pKa = 4.74. Henderson–Hasselbalch confirms pH = pKa + log(1) = 4.74. A tempting distractor is pH = 7.00 because amounts are equal, mistakenly applying strong acid logic. Determine the titration stage first, then choose the appropriate method like recognizing pH = pKa at half-equivalence for weak acids.

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