pH & Neutralization — Practice · Grade 11 Science

Chapter: Acids, Bases & Titration

What this practice covers

By the end of pH & Neutralization you should be able to:

9 questions, multiple choice, numeric and written answer, each with a worked explanation.

Vocabulary used: pH scale, pH, neutralization, salt, indicator, titration.

Problem 1

A solution has a pH of 12. It is…
  1. impossible, since pH cannot exceed 7
  2. strongly acidic
  3. neutral
  4. basic
Answer and explanation

basic

A pH above 7 is basic; 12 is well above 7.

Problem 2

In the introductory dilute model at 25°C, the pH of pure water is approximately ___.
Answer and explanation

7

At 25°C, Kw≈1.0×10⁻¹⁴ and [H₃O⁺]=[OH⁻]≈1.0×10⁻⁷ M, giving pH≈7. Neutral pH varies with temperature.

Problem 3

A solution has a hydrogen ion concentration [H⁺] of 1 × 10⁻⁵ M. What is its pH?
Answer and explanation

5

pH = −log(1 × 10⁻⁵) = 5.

Problem 4

Describe the net reaction when dilute strong acid and hydroxide react. Does the final mixture necessarily have pH 7?
Answer and explanation

H⁺(aq)+OH⁻(aq)→H₂O(l); spectator ions remain dissolved. Final pH depends on reacting moles and chemical identities. Near 25°C an ideal strong-acid/strong-base mixture is near pH 7 at equivalence, but excess acid or base changes the final pH, and weak-species hydrolysis can change equivalence pH.

Problem 5

How many times greater is hydrogen-ion activity at pH 3 than at pH 6?
Answer and explanation

1000 times

The activity ratio is 10^(6−3)=1000. This is not automatically a ratio of total acid amounts or titratable acidity.

Problem 6

Find pH for [H₃O⁺]=1.0×10⁻⁴ M using the dilute approximation. Show the reasoning, not just the conclusion.
Hints

A concentration between 10⁻² and 10⁻³ M gives pH between 2 and 3.

Answer and explanation

pH=4.00.

Problem 7

At 25°C, a dilute solution has pOH=3.00. Find pH using pH+pOH=14.00.
Hints

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

Answer and explanation

pH=11.00, under the stated temperature and dilute approximation.

Problem 8

Reaction stoichiometry determines excess before a pH calculationReaction stoichiometry determines excess before a pH calculationMix 25.0 mL 0.100 M acid + 30.0 mL 0.100 M baseAmountInitial strong monoprotic acid0.00250 molInitial strong hydroxide base0.00300 molExcess OH⁻ after reaction0.00050 molFinal volume, additive model0.0550 L[OH⁻]≈0.00909 M → pOH≈2.04 → pH≈11.96
Original complete strong-acid/strong-base mixture calculation at 25°C with approximate additive volumes.
Interpret the supplied visual. Why is the excess hydroxide divided by 55.0 mL rather than the 30.0 mL base portion alone?
Answer and explanation

It is distributed through the final mixed solution under the additive-volume model, not confined to its original portion.

For a selected strong monoprotic acid and strong hydroxide base, first subtract reacting mole amounts using the one-to-one net ratio. Then divide the excess amount by the final solution volume and use the water equilibrium to connect pH and pOH. Averaging the two initial pH values is not a neutralization calculation. Weak acids, polyprotic species and buffering can require additional equilibrium steps. State temperature, extent and volume assumptions before presenting the final number.

It is distributed through the final mixed solution under the additive-volume model, not confined to its original portion.

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 9

Mix 20.0 mL 0.100 M strong monoprotic acid with 25.0 mL 0.100 M strong base at 25°C. Assume additive volumes and complete neutralization. Find approximate pH.
Answer and explanation

Acid=0.00200 mol and base=0.00250 mol, leaving 0.00050 mol OH⁻ in 0.0450 L. [OH⁻]≈0.0111 M, pOH≈1.95 and pH≈12.05. Do stoichiometry before logarithms.

Acid=0.00200 mol and base=0.00250 mol, leaving 0.00050 mol OH⁻ in 0.0450 L. [OH⁻]≈0.0111 M, pOH≈1.95 and pH≈12.05. Do stoichiometry before logarithms.

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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