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 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 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 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 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.

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.

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.