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.

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.

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].

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.

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.

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.

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.

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.

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.

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.