U8.9 · Henderson-Hasselbalch EquationPractice · Henderson-Hasselbalch 方程
Lessons covered:U8.4 Acid-Base Reactions and BuffersU8.9 Henderson-Hasselbalch EquationU8.10 Buffer Capacity
- 1Numeric
A buffer contains 0.15 M CH₃COOH and 0.25 M CH₃COONa; acetic acid has pKa = 4.74. What is the pH?
ExplanationpH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.25/0.15) = 4.74 + 0.22 = 4.96. Flipping the ratio gives 4.52.
- 2Numeric
A buffer at pH 9.00 is made from NH₃ and NH₄Cl. NH₄⁺ has pKa = 9.26. What should [NH₃] ÷ [NH₄⁺] be?
ExplanationNH₄⁺ is the acid form and NH₃ the base form. 9.00 = 9.26 + log([NH₃]/[NH₄⁺]), so log of the ratio = −0.26 and the ratio = 10−0.26 = 0.55. The target pH is below the pKa, so a bit more of the acid form NH₄⁺ is needed.
- 3Multiple choice
To make a buffer at pH 7.40, which weak acid and its sodium salt is the best choice? (pKa: HCOOH 3.74, CH₃COOH 4.74, HClO 7.54, HCN 9.31)
ExplanationA buffer works best near pH ≈ pKa, where the two components are present in similar amounts. HClO has pKa = 7.54, closest to 7.40, needing only [ClO⁻]/[HClO] = 10−0.14 ≈ 0.72. The others would need very lopsided ratios and would buffer poorly.
Review: U8.9 Henderson-Hasselbalch EquationU8.10 Buffer Capacity
- 4Numeric
60.0 mL of 0.20 M NH₃ is mixed with 20.0 mL of 0.20 M HCl. What is the pH of the mixture (pKa of NH₄⁺ = 9.26)?
ExplanationNH₃ is 12.0 mmol and H⁺ 4.0 mmol. NH₃ + H⁺ → NH₄⁺ leaves 8.0 mmol NH₃ and makes 4.0 mmol NH₄⁺: a buffer. The ratio can use moles directly (same volume): pH = 9.26 + log(8.0/4.0) = 9.56.
Review: U8.9 Henderson-Hasselbalch EquationU8.4 Acid-Base Reactions and Buffers
- 5Multiple choice
When does the Henderson–Hasselbalch equation give an accurate buffer pH?
ExplanationThe equation assumes [HA] and [A⁻] at equilibrium equal the amounts added, meaning the acid’s ionization and the base’s reaction with water are negligible. That holds when the concentrations are large enough and the ratio isn’t extreme. With only the weak acid (no conjugate base) the ratio is zero and the equation fails; use an ICE table instead.
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