AP Chemistry Quiz: Acid Base Titrations

What this quiz covers

This quiz focuses on Acid Base Titrations, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Na$^+$
2. F$^-$ (correct answer)
3. H$_3$O$^+$ from the strong acid
4. HF
5. OH$^-$ from the strong base

Explanation: 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.

Question 2

1. 1.00
2. 9.00
3. 13.00
4. 7.00 (correct answer)
5. 5.00

Explanation: 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.

Question 3

1. The pH equals $pK_a$ of $\text{NH}_4^+$ and $[\text{NH}_3]=[\text{NH}_4^+]$. (correct answer)
2. The pH equals 7.00 because equal moles of acid and base have reacted.
3. The pH equals $pK_b$ of $\text{NH}_3$ and $[\text{H}^+]=[\text{OH}^-]$.
4. Only $\text{NH}_4^+$ is present in significant amount; the solution is not a buffer.
5. Only excess HCl controls the pH; the solution is strongly acidic.

Explanation: 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].

Question 4

1. 7.00
2. 3.74 (correct answer)
3. 8.26
4. 11.26
5. 2.74

Explanation: 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.

Question 5

1. 2.87
2. 4.74 (correct answer)
3. 7.00
4. 9.26
5. 12.00

Explanation: 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.

Question 6

1. 1.59
2. 3.17 (correct answer)
3. 4.76
4. 7.00
5. 10.83

Explanation: 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.

Question 7

1. 1.83
2. 3.17 (correct answer)
3. 7.00
4. 10.83
5. 12.17

Explanation: This question involves acid–base titrations. Stoichiometry indicates that the half-equivalence point is reached with 10.0 mL of 0.100 M NaOH, neutralizing half of the 2.0 mmol of HF, leaving equal HF and F-. In this weak acid-strong base titration, the buffer region has equal [HF] and [F-], controlling pH through the conjugate pair. Henderson–Hasselbalch applies: pH = pKa + log([F-]/[HF]) = 3.17 + log(1) = 3.17. A tempting distractor is pH = 7.00, which could arise from assuming half-neutralization leads to neutrality like in strong acids. Determine the titration stage first, then choose the appropriate method like Henderson–Hasselbalch for buffers.

Question 8

1. HF, because weak acids always remain mostly undissociated.
2. F$^-$, because it hydrolyzes to produce OH$^-$. (correct answer)
3. Na$^+$, because it is the conjugate acid of NaOH.
4. OH$^-$ from excess NaOH, because equivalence means excess base.
5. H$_3$O$^+$ from excess HF, because equivalence means excess acid.

Explanation: 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.

Question 9

1. 4.74
2. 7.00
3. 9.26 (correct answer)
4. 12.48
5. 2.00

Explanation: This question involves acid–base titrations. Stoichiometry determines that the half-equivalence point uses 12.5 mL of 0.100 M HCl to protonate half of the 2.5 mmol NH3, yielding equal NH3 and NH4+. In this weak base-strong acid titration, the buffer of NH3 and NH4+ controls pH at this stage. Henderson–Hasselbalch for the conjugate acid gives pH = pKa + log([NH3]/[NH4+]) = (14 - 4.74) + log(1) = 9.26. A tempting distractor is pH = 7.00, from confusing half-equivalence with the neutral equivalence point. Determine the titration stage first, then choose the appropriate method such as converting pKb to pKa for weak base buffers.

Question 10

1. OH$^-$
2. H$_3$O$^+$
3. HC$_7$H$_5$O$_2$ (correct answer)
4. C$_7$H$_5$O$_2^-$
5. Na$^+$

Explanation: 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.

Question 11

1. 7.00
2. 4.74 (correct answer)
3. 11.26
4. 9.26
5. 2.74

Explanation: This question tests the skill of acid–base titrations. Stoichiometry shows that 25.0 mL of 0.100 M NaOH adds 2.5 mmol of OH-, which reacts with half of the 5 mmol of acetic acid, leaving 2.5 mmol HA and producing 2.5 mmol A-. At this half-equivalence point, the concentrations of HA and A- are equal in the total volume of 75 mL. Since [HA] = [A-], the pH is equal to pKa according to the Henderson–Hasselbalch equation, pH = -log($1.8×10^{-5}$) = 4.74. A tempting distractor is 9.26, which is the pH at equivalence for this titration, but at half-equivalence, it's pKa, not the equivalence pH. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.

Question 12

1. 10.0 mL
2. 25.0 mL (correct answer)
3. 50.0 mL
4. 75.0 mL
5. 100.0 mL

Explanation: 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.

Question 13

1. The pH equals $pK_a$ and $[\text{CH}_3\text{COOH}] = [\text{CH}_3\text{COO}^-]$. (correct answer)
2. The pH is 7.00 because the amounts of acid and base are equal.
3. Excess OH$^-$ is present and controls the pH.
4. All CH$_3$COOH has been converted to CH$_3$COO$^-$.
5. The solution contains only CH$_3$COOH and water (no conjugate base).

Explanation: 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.

Question 14

1. 4.74 (correct answer)
2. 7.00
3. 9.26
4. 2.37
5. 11.63

Explanation: 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.

Question 15

1. 1.30
2. 7.00
3. 12.00
4. 11.96 (correct answer)
5. 2.04

Explanation: 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.

Question 16

1. 2.74
2. 4.74
3. 7.00
4. 11.96 (correct answer)
5. 13.00

Explanation: This question tests the skill of acid–base titrations. Stoichiometry shows that 30 mL of 0.100 M NaOH adds 3 mmol OH-, which reacts with the 2.5 mmol HCOOH, leaving 0.5 mmol OH- excess. The excess OH- is in total volume of 55 mL, so [OH-] = 0.5/55 ≈ 0.00909 M. Since it's after equivalence, pH = 14 - (-log 0.00909) = 14 - 2.04 = 11.96. A tempting distractor is 7.00, thinking it's equivalence, but we are past equivalence with excess base, so pH >7. To solve titration pH problems, determine the titration stage first, then choose the appropriate method.

Question 17

1. 3.00
2. 7.00 (correct answer)
3. 1.00
4. 9.00
5. 5.00

Explanation: 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.

Question 18

1. The pH equals $pK_a$ of NH$_4^+$, and NH$_3$ and NH$_4^+$ are present in equal amounts. (correct answer)
2. The pH is 7.00 because the moles of acid added equal half the initial moles of base.
3. The pH is determined primarily by excess strong acid, HCl.
4. Only NH$_4^+$ remains in solution, so the pH is controlled by hydrolysis of NH$_4^+$ alone.
5. Only NH$_3$ remains in solution, so the pH is controlled by $K_b$ of NH$_3$ alone.

Explanation: 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.

Question 19

1. The pH is less than 7.00 because CH$_3$NH$_3^+$ is a weak acid. (correct answer)
2. The pH is determined only by the concentration of Cl$^-$.
3. The pH is greater than 7.00 because excess weak base remains.
4. The pH is determined only by the concentration of HCl remaining in excess.
5. The pH is 7.00 because equal moles of acid and base have reacted.

Explanation: 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.

Question 20

1. 1.00
2. 2.00
3. 7.00
4. 11.96 (correct answer)
5. 13.00

Explanation: 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.