Titration, Buffers & Neutralization Curves — Practice · Grade 11 Science

Chapter: Acids, Bases & Titration

What this practice covers

By the end of Titration, Buffers & Neutralization Curves you should be able to:

8 questions, numeric and written answer, each with a worked explanation.

Vocabulary used: endpoint, equivalence point, buffer capacity.

Problem 1

A buffer has equal conjugate species with pKa=4.5. Find pH.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

4.5

log 1=0.

Problem 2

Why is a weak-acid titration’s equivalence often above pH 7?
Answer and explanation

The remaining conjugate base accepts protons from water, producing OH⁻.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

A buffer has pKa=6 and conjugate-base/acid ratio 100. Estimate pH.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

8

6+log 100=8.

Problem 4

Why does a buffer fail after a large acid addition?
Answer and explanation

The conjugate-base component is depleted; the small-addition approximation no longer applies.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

For pK_a=5.0 and [A⁻]/[HA]=10, find approximate pH. Show the reasoning, not just the conclusion.
Hints

A buffer resists limited additions; it does not have unlimited capacity.

Answer and explanation

5.0+log10=6.0.

Problem 6

Why can a weak-acid/strong-base equivalence solution have pH above 7 at 25°C?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

The conjugate base produced can react with water to generate OH⁻; equivalence is stoichiometric, not synonymous with neutral pH.

Problem 7

The Henderson–Hasselbalch relation is a conditional equilibrium rearrangementThe Henderson–Hasselbalch relation is a conditional equilibrium rearrangementpH ≈ pKa + log₁₀([A⁻]/[HA])pKa=5.00; postreaction ratio=0.110/0.090pH ≈ 5.00 + 0.087 = 5.09Use a common volume for a mole ratio; validate the approximation.
Original selected pKa=5.00 buffer calculation using rounded postreaction amounts and ideal concentration behavior.
Interpret the supplied visual. Why can the postreaction mole amounts be used in this ratio without separately dividing each by volume?
Answer and explanation

They occupy the same solution volume, so that common denominator cancels in [A⁻]/[HA] under the stated approximation.

The Henderson–Hasselbalch form rearranges the acid equilibrium into a logarithmic conjugate-base/acid ratio. The familiar concentration version approximates activities and often substitutes postreaction component amounts when both species are appreciable. A common solution volume cancels from their mole ratio. Near exhaustion of one component, at very low concentration or under strong nonideality, the simple approximation may fail and a fuller equilibrium is needed. It is not a formula for pure acid before any conjugate base exists.

They occupy the same solution volume, so that common denominator cancels in [A⁻]/[HA] under the stated approximation.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

A buffer initially has 0.080 mol HA and 0.120 mol A⁻, with pKa=4.80. Add 0.020 mol strong acid without exhausting either component. Estimate pH after reaction and state an approximation.
Answer and explanation

H⁺ converts A⁻ to HA, leaving 0.100 mol of each. The ratio is one, so pH≈4.80. Assume the usual dilute activity/concentration approximation with both conjugate components appreciable and a common final volume.

H⁺ converts A⁻ to HA, leaving 0.100 mol of each. The ratio is one, so pH≈4.80. Assume the usual dilute activity/concentration approximation with both conjugate components appreciable and a common final volume.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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