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Mathematical Skills in Chemistry

Logarithms, pH and pKa

  • 3 things to know
  • Component 3

Overview

Recall questions on what a logarithm is and the relationship between a number and its logarithm, converting between hydrogen ion concentration and pH in both directions, between Ka and pKa, why a change of one pH unit is a factor of ten, and worked conversions.

What you need to know

Answer each one before you reveal it.

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Key term

What is plotted against what to obtain a straight line from the Arrhenius equation, and what does the gradient give?

Key term

Why is acidity expressed on a logarithmic scale rather than as a concentration?

Numbers

Calculate the pH of a solution in which the hydrogen ion concentration is 0.0200 mol dm⁻³.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is acidity expressed on a logarithmic scale rather than as a concentration?

    Model answer

    Hydrogen ion concentrations in ordinary solutions span more than fourteen orders of magnitude, from about 1 to 10⁻¹⁴ mol dm⁻³, and taking the negative logarithm compresses that enormous range onto a scale of small numbers that can be compared at a glance.

  2. 2

    Calculate the pH of a solution in which the hydrogen ion concentration is 0.0200 mol dm⁻³.

    Model answer

    The pH is −log(2.00 × 10⁻²), which is 1.70.

  3. 3

    A solution has a pH of 5.80. Calculate its hydrogen ion concentration and give the answer to two significant figures.

    Model answer

    [H⁺] is 10 raised to the power −5.80, which is 1.6 × 10⁻⁶ mol dm⁻³. Taking the negative logarithm of that value returns 5.80, which confirms it.

  4. 4

    Calculate the pH of 0.0500 mol dm⁻³ sodium hydroxide solution at 298 K, taking Kw as 1.00 × 10⁻¹⁴ mol² dm⁻⁶.

    Model answer

    Dividing Kw by the hydroxide ion concentration gives [H⁺] = 1.00 × 10⁻¹⁴ ÷ 0.0500 = 2.00 × 10⁻¹³ mol dm⁻³, so the pH is 12.70.

  5. 5

    An acid has Ka = 1.8 × 10⁻⁵ mol dm⁻³. Calculate its pKa.

    Model answer

    Taking the negative logarithm gives pKa = −log(1.8 × 10⁻⁵) = 4.74.

  6. 6

    An acid has a pKa of 3.75. Calculate its acid dissociation constant.

    Model answer

    Taking the inverse logarithm gives Ka = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴ mol dm⁻³.

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