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Chemistry · 07 / 10

Acids and bases: proton transfer

Final-year concepts · Worked example and 4 exercises with solutions, at your pace.

Need a foundation?Atoms, ions and moleculesBalance a reaction equation
UnderstandSee an exampleGuided exercise3 progressive challenges

The principle, simply

A Brønsted acid donates a proton; a base accepts one. A conjugate pair differs by one proton. pH measures hydronium activity; dilute-solution exercises often approximate it as −log₁₀([H₃O⁺]/c°), with c° = 1 mol/L. Check reaction and dilution assumptions.

p=-log(C)
THE PICTURE TO REMEMBERAcids and bases: proton transfer
  1. 01Acid donates H⁺
  2. 02Base accepts H⁺
  3. 03System equilibrium
  4. 04pH and assumptions

WORKED EXAMPLE

From principle to reasoning

In the dilute model, 10⁻³ mol/L hydronium gives pH ≈ 3. A tenfold concentration increase lowers pH by one.

Solve a problem in this subject

  1. Identify species

    Write formulas and useful states. Distinguish atoms, ions, molecules and amount of substance. Total charge is conserved.

  2. Connect quantities

    Use m = nM and C = n/V with consistent units. V is the final solution volume.

  3. Use coefficients

    Balance before building an extent table. Compare initial amounts divided by coefficients to identify the limiting reactant.

  4. Check context

    Remaining amounts cannot be negative. Check conservation, units and assumptions: dilution, medium, completion or equilibrium.

YOUR TURN

01 · Try before checking

What happens when hydronium falls from 10⁻³ to 10⁻⁴ mol/L?

One hint

Evaluate −log₁₀(10⁻⁴).

Reveal and understand the solution

pH increases from 3 to 4 as hydronium concentration falls tenfold.

Compare method, conditions and answer; identify the exact part to revisit.

IMPROVE THROUGH PRACTICE

Three additional challenges with solutions

Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.

Reflect after practising
02 · Recognise and understandLink to this exercise

A solution has [H₃O⁺] = 10⁻³ mol/L. Find pH using the usual dilute-solution model.

A hint to get started

pH = −log₁₀([H₃O⁺]/c°), with c° = 1 mol/L.

Step-by-step solution
  1. log₁₀(10⁻³) = −3.
  2. The minus sign gives pH = 3.
  3. A tenfold concentration decrease would raise this pH by one.

Which step did you miss? Retry without help, then explain why the method works.

03 · Apply the methodLink to this exercise

What is the conjugate acid of NH₃?

A hint to get started

Add a proton and adjust charge.

Step-by-step solution
  1. NH₃ accepts H⁺.
  2. The product is NH₄⁺.
  3. The conjugate pair NH₄⁺/NH₃ differs by one proton.

Which step did you miss? Retry without help, then explain why the method works.

04 · Go furtherLink to this exercise

For HA/A⁻ with pKa = 4, which form predominates at pH = 5?

A hint to get started

Compare pH and pKa in Henderson–Hasselbalch.

Step-by-step solution
  1. pH − pKa = log₁₀([A⁻]/[HA]) = 1.
  2. The concentration ratio is 10.
  3. A⁻ predominates, while both forms remain present in the model.

Which step did you miss? Retry without help, then explain why the method works.

The trap to avoid

A weak acid concentration is not automatically the hydronium concentration.

Without notes, can you…

  • Explain the principle in your own words?
  • Rebuild the diagram and its connections?
  • Solve the exercise and justify the method?

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