What concentration of sodium ethanoate must be added to 0.100 mol dm⁻³ ethanoic acid to give a buffer of pH 5.00, if Kₐ = 1.74 × 10⁻⁵ mol dm⁻³?

[H⁺] must be 1.00 × 10⁻⁵ mol dm⁻³, so [salt] = Kₐ[acid]/[H⁺] = 1.74 × 10⁻⁵ × 0.100/1.00 × 10⁻⁵ = 0.174 mol dm⁻³.

25.0 cm³ of 0.100 mol dm⁻³ propanoic acid is mixed with 25.0 cm³ of 0.100 mol dm⁻³ sodium propanoate. Calculate the pH, given Kₐ = 1.3 × 10⁻⁵ mol dm⁻³.

Both are diluted equally so the ratio stays at one, giving [H⁺] = 1.3 × 10⁻⁵ mol dm⁻³ and pH = 4.89.

0.010 mol of hydrogen chloride is added to 1.00 dm³ of a buffer containing 0.100 mol of ethanoic acid and 0.100 mol of sodium ethanoate. Calculate the new pH, given Kₐ = 1.74 × 10⁻⁵ mol dm⁻³.

The acid rises to 0.110 mol and the salt falls to 0.090 mol, so [H⁺] = 1.74 × 10⁻⁵ × 0.110/0.090 = 2.13 × 10⁻⁵ mol dm⁻³ and pH = 4.67.

Which two approximations are made in buffer pH calculations?

That the concentration of the weak acid at equilibrium equals the concentration added, since its ionisation is suppressed by the common ion, and that the concentration of the conjugate base equals the concentration of salt added.

What limits the buffering capacity of a solution?

It fails once so much acid or alkali has been added that one of the two components is largely used up and the concentration ratio changes sharply.

OCR A-Level Chemistry B H433

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Oceans

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Oceans

Buffer calculations, capacity and the natural buffers

1 thing to know
Components 01, 02 and 03

Overview

Recall questions on calculating buffer pH from Ka and the two concentrations, why dilution barely changes it, and the blood and seawater buffers.

What you need to know

Answer each one before you reveal it.

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Numbers

Why does diluting a buffer leave its pH almost unchanged?

Questions from this subtopic

Each one with the model answer underneath.

1
What concentration of sodium ethanoate must be added to 0.100 mol dm⁻³ ethanoic acid to give a buffer of pH 5.00, if Kₐ = 1.74 × 10⁻⁵ mol dm⁻³?
Model answer
[H⁺] must be 1.00 × 10⁻⁵ mol dm⁻³, so [salt] = Kₐ[acid]/[H⁺] = 1.74 × 10⁻⁵ × 0.100/1.00 × 10⁻⁵ = 0.174 mol dm⁻³.
2
25.0 cm³ of 0.100 mol dm⁻³ propanoic acid is mixed with 25.0 cm³ of 0.100 mol dm⁻³ sodium propanoate. Calculate the pH, given Kₐ = 1.3 × 10⁻⁵ mol dm⁻³.
Model answer
Both are diluted equally so the ratio stays at one, giving [H⁺] = 1.3 × 10⁻⁵ mol dm⁻³ and pH = 4.89.
3
Why does diluting a buffer leave its pH almost unchanged?
Model answer
Dilution reduces the acid and salt concentrations by the same factor, so their ratio and therefore [H⁺] are unaffected.
4
0.010 mol of hydrogen chloride is added to 1.00 dm³ of a buffer containing 0.100 mol of ethanoic acid and 0.100 mol of sodium ethanoate. Calculate the new pH, given Kₐ = 1.74 × 10⁻⁵ mol dm⁻³.
Model answer
The acid rises to 0.110 mol and the salt falls to 0.090 mol, so [H⁺] = 1.74 × 10⁻⁵ × 0.110/0.090 = 2.13 × 10⁻⁵ mol dm⁻³ and pH = 4.67.
5
Which two approximations are made in buffer pH calculations?
Model answer
That the concentration of the weak acid at equilibrium equals the concentration added, since its ionisation is suppressed by the common ion, and that the concentration of the conjugate base equals the concentration of salt added.
6
What limits the buffering capacity of a solution?
Model answer
It fails once so much acid or alkali has been added that one of the two components is largely used up and the concentration ratio changes sharply.

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