State the approximate expression used to find [H⁺] in a solution of a weak acid.

[H⁺] = √(Kₐ × [HA]), where [HA] is the concentration of acid made up.

Which two approximations does that expression rest on?

That the hydrogen ions from water are negligible, so [H⁺] = [A⁻], and that so little acid ionises that [HA] at equilibrium equals the concentration made up.

Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Kₐ = 1.74 × 10⁻⁵ mol dm⁻³.

[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ mol dm⁻³, so pH = 2.88.

Calculate the pH of 0.150 mol dm⁻³ propanoic acid, Kₐ = 1.3 × 10⁻⁵ mol dm⁻³.

[H⁺] = √(1.3 × 10⁻⁵ × 0.150) = 1.4 × 10⁻³ mol dm⁻³, so pH = 2.85.

A 0.250 mol dm⁻³ solution of a weak acid has pH 2.72. Calculate its Kₐ.

[H⁺] = 10⁻²·⁷² = 1.91 × 10⁻³ mol dm⁻³, so Kₐ = (1.91 × 10⁻³)²/0.250 = 1.45 × 10⁻⁵ mol dm⁻³.

A weak acid with Kₐ = 6.3 × 10⁻⁵ mol dm⁻³ gives a solution of pH 2.70. Calculate its concentration.

[H⁺] = 10⁻²·⁷⁰ = 2.00 × 10⁻³ mol dm⁻³, so [HA] = [H⁺]²/Kₐ = 4.0 × 10⁻⁶/6.3 × 10⁻⁵ = 0.063 mol dm⁻³.

OCR A-Level Chemistry B H433

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pH of weak acids, dilution and the limits of the approximation

1 thing to know
Components 01, 02 and 03

Overview

Recall questions on the weak acid expression and its two approximations, the effect of tenfold dilution, and when the approximation fails.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Numbers

What expression gives [H⁺] for a weak acid, which two approximations does it rest on, and what are three strong acids?

Questions from this subtopic

Each one with the model answer underneath.

1
State the approximate expression used to find [H⁺] in a solution of a weak acid.
Model answer
[H⁺] = √(Kₐ × [HA]), where [HA] is the concentration of acid made up.
2
Which two approximations does that expression rest on?
Model answer
That the hydrogen ions from water are negligible, so [H⁺] = [A⁻], and that so little acid ionises that [HA] at equilibrium equals the concentration made up.
3
Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, Kₐ = 1.74 × 10⁻⁵ mol dm⁻³.
Model answer
[H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ mol dm⁻³, so pH = 2.88.
4
Calculate the pH of 0.150 mol dm⁻³ propanoic acid, Kₐ = 1.3 × 10⁻⁵ mol dm⁻³.
Model answer
[H⁺] = √(1.3 × 10⁻⁵ × 0.150) = 1.4 × 10⁻³ mol dm⁻³, so pH = 2.85.
5
A 0.250 mol dm⁻³ solution of a weak acid has pH 2.72. Calculate its Kₐ.
Model answer
[H⁺] = 10⁻²·⁷² = 1.91 × 10⁻³ mol dm⁻³, so Kₐ = (1.91 × 10⁻³)²/0.250 = 1.45 × 10⁻⁵ mol dm⁻³.
6
A weak acid with Kₐ = 6.3 × 10⁻⁵ mol dm⁻³ gives a solution of pH 2.70. Calculate its concentration.
Model answer
[H⁺] = 10⁻²·⁷⁰ = 2.00 × 10⁻³ mol dm⁻³, so [HA] = [H⁺]²/Kₐ = 4.0 × 10⁻⁶/6.3 × 10⁻⁵ = 0.063 mol dm⁻³.

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