The degree of ionisation and Ostwald’s dilution law
Chemistry Sec 3 Term 1, Chapter 3: pp. 66–71
GoalCalculate the degree of ionisation α from Ka and the concentration, and explain why dilution raises α while Ka stays the same.
New words
degree of ionisation α · the fraction of the dissolved molecules that ionise: ionised moles ÷ moles before ionisationOstwald’s dilution law · Ka = α²C ÷ (1 − α), which for a weak acid becomes Ka ≈ α²C and α = √(Ka ÷ C)dilution · adding solvent so that the concentration falls
Explanation
The degree of ionisation α is the fraction of the dissolved molecules that have ionised: α = ionised moles ÷ moles before ionisation. For a percentage, multiply by 100. Take C mol/L of a weak acid HA. At equilibrium αC mol/L has ionised, so [H⁺] = [A⁻] = αC and [HA] = C(1 − α). Then Ka = (αC)(αC) ÷ (C(1 − α)) = α²C ÷ (1 − α). This is Ostwald’s dilution law, after the scientist who found it. For a weak acid α is small, so 1 − α is nearly 1 and Ka ≈ α²C, which gives α = √(Ka ÷ C). The formula shows that a more dilute solution has a larger α: when the concentration is four times smaller, α is twice as large. Dilution raises the degree of ionisation but not Ka, because Ka depends only on the acid and the temperature. Be careful with two things. The solution still has fewer ions in each litre, because C has fallen. And a larger α does not make the acid stronger: its strength is Ka. The square-root rule is a good approximation while α is small, below about 5 %.
ƒOstwald’s dilution law
Ka = α²C ÷ (1 − α) ≈ α²C ; α = √(Ka ÷ C)
α is a fraction; multiply by 100 for a percentage
Worked examples
ƒα of the invented acid
α = √(2.0 × 10⁻⁵ ÷ 0.050) = √(4.0 × 10⁻⁴) = 0.020
α from Ka. An invented weak acid has Ka = 2.0 × 10⁻⁵ and C = 0.050 mol/L. α = √(Ka ÷ C) = √(2.0 × 10⁻⁵ ÷ 0.050) = √(4.0 × 10⁻⁴) = 0.020, which is 2 %. Check the approximation: 1 − α = 0.98, close to 1, and the exact formula gives Ka = 0.020² × 0.050 ÷ 0.98 = 2.04 × 10⁻⁵, which is the same to two figures.
Dilution. The same acid is diluted four times, to C = 0.0125 mol/L. α = √(2.0 × 10⁻⁵ ÷ 0.0125) = √(1.6 × 10⁻³) = 0.040, which is 4 %: α has doubled. But the ions in each litre have fallen: [H⁺] = αC = 0.040 × 0.0125 = 5.0 × 10⁻⁴ mol/L, against 0.020 × 0.050 = 1.0 × 10⁻³ mol/L before. Ka has not changed.
Ka from α. For an invented acid at C = 0.10 mol/L a test shows that α = 0.010 (1 %). Since α is small, Ka ≈ α²C = 0.010² × 0.10 = 1.0 × 10⁻⁴ × 0.10 = 1.0 × 10⁻⁵. Read in the other direction, Ka = 1.0 × 10⁻⁵ and C = 0.10 give α = √(1.0 × 10⁻⁴) = 0.010, so the two directions agree.
Class activity
Fill in a table, by hand, with a grown-up. An invented acid has Ka = 4.0 × 10⁻⁶. For each concentration C = 0.16, 0.040, 0.010 and 0.0025 mol/L, divide Ka by C and take the square root to find α, then write the percentage. You should find 0.0050, 0.010, 0.020 and 0.040. Spot the pattern: each time C falls to a quarter, α doubles. Then draw the four values as bars on squared paper. No lab chemicals are used.
Practice
1
An invented acid has Ka = 9.0 × 10⁻⁸ and C = 0.010 mol/L. Find α and the percentage ionised.
A student says: “When a weak acid is diluted, α rises, so the acid has become stronger.” Reply in two sentences.
Strength is measured by Ka, and Ka does not change on dilution. A larger α only means a larger fraction of the molecules have ionised, while the ions in each litre are fewer.
