Acids, Bases & Titration — Problem set, Grade 11

Separate strength from concentration and connect proton transfer with pH, buffers, and titration.

Review the key ideas

Acids & Bases

Acid–base roles depend on the chosen model and partner. Proton-transfer strength, formal concentration and measured pH are distinct; polyprotic steps and base reactions with water require the appropriate equilibrium account.

pH & Neutralization

pH is a logarithmic activity measure, approximated by dilute hydronium concentration. Neutrality and pH+pOH depend on temperature. Neutralization requires reacting-amount bookkeeping and the final equilibrium; equivalence is not universally pH 7.

Acid Strength, Conjugate Pairs & Ka

Strength and concentration are independent descriptors; acid equilibria determine hydronium concentration.

Titration, Buffers & Neutralization Curves

Equivalence follows reaction ratios; buffers resist small changes through a conjugate pair.

Problem 1

In the reaction NH₃ + H₂O → NH₄⁺ + OH⁻, identify the Brønsted-Lowry acid and the Brønsted-Lowry base.
Answer and explanation

Water is the acid (proton donor); ammonia is the base (proton acceptor).

A Brønsted-Lowry acid donates a proton (H⁺); a base accepts one. Ammonia (NH₃) gains a proton to become NH₄⁺, so it is the base. Water (H₂O) loses a proton to become OH⁻, so here it acts as the acid.

Problem 2

A dilute strong acid and a concentrated weak acid are described without numerical concentrations. Can you determine which has more hydronium? Explain.
Answer and explanation

Not from those labels alone. Strength concerns ionization fraction and concentration concerns amount per volume; numerical concentration and ionization are needed.

Compare actual hydronium using concentration and the relevant ionization model, not strength alone.

Problem 3

A solution has a pH of 3. Is it acidic, neutral, or basic?
Answer and explanation

Acidic.

On the pH scale, any value below 7 is acidic. A pH of 3 is well below 7, so the solution is acidic.

Problem 4

Write the products of the neutralization reaction when hydrochloric acid (HCl) reacts with potassium hydroxide (KOH).
Answer and explanation

HCl + KOH → KCl + H₂O (the salt potassium chloride plus water).

Neutralization produces water and a salt. The H⁺ from HCl combines with the OH⁻ from KOH to form water, H₂O. The remaining ions K⁺ and Cl⁻ combine to form the salt potassium chloride, KCl. Counting atoms confirms the equation is balanced.

Problem 5

A solution has a hydrogen ion concentration [H⁺] of 1 × 10⁻⁴ M. What is its pH, and how many times more acidic is it than a neutral solution of pH 7?
Answer and explanation

pH = 4; it is 1000 times more acidic.

pH = −log(1 × 10⁻⁴) = 4. The difference from neutral is 7 − 4 = 3 pH units, and each unit is a factor of 10, so the solution has 10³ = 1000 times more hydrogen ions than a neutral solution.

Problem 6

Explain why NH₃ is a base without containing OH in its formula.
Answer and explanation

It accepts a proton from water, forming NH₄⁺ and OH⁻; Brønsted bases need not contain hydroxide initially.

Response criteria

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

Problem 7

Chapter review: In NH₃+H₂O⇌NH₄⁺+OH⁻, identify the acid, base and both conjugate pairs. Explain why NH₃ can act as a base without OH in its formula.
Answer and explanation

H₂O donates a proton and NH₃ accepts it. Pairs are H₂O/OH⁻ and NH₄⁺/NH₃. NH₃ produces hydroxide through proton transfer with water rather than having to release pre-existing OH⁻ groups.

H₂O donates a proton and NH₃ accepts it. Pairs are H₂O/OH⁻ and NH₄⁺/NH₃. NH₃ produces hydroxide through proton transfer with water rather than having to release pre-existing OH⁻ groups.

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.

Problem 8

Compare pH3 andpH5 hydronium concentrations.
Answer and explanation

pH3 has 10^(5−3)=100 times the hydronium concentration in the simple concentration model.

Response criteria

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

Problem 9

Chapter review: 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.

Problem 10

Give the conjugate base of H₂CO₃ after losing one proton.
Answer and explanation

HCO₃⁻, not CO₃²⁻ after just one proton loss.

Response criteria

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

Problem 11

A monoprotic strong acid has concentration .010 M. Estimate pH.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

2

−log(.010)=2.

Problem 12

Chapter review: For a hypothetical monoprotic acid with C=0.200 M and Ka=2.0×10⁻⁵, estimate hydronium and pH, then test the small-ionization approximation.
Answer and explanation

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

x≈√(KaC)=√(4.0×10⁻⁶)=0.0020 M. pH≈2.70. x/C≈0.010=1%, supporting the approximation for ordinary classroom accuracy under the stated species model.

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.

Problem 13

30.0 mL .100 M NaOH neutralize 15.0 mL HCl. Find HCl molarity.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

0.2 M

.003 mol/.015L=.200M.

Problem 14

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 15

Chapter review: 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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