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pKa is −log₁₀ Ka, a way of expressing very small Ka values on a manageable scale, just as pH does for [H⁺]. Free AQA notes and 11 questions answered.
Calculations with Ka and pKa is part of Acids and bases (A-level only) in A Level AQA Chemistry.
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pKa is −log₁₀ Ka, a way of expressing very small Ka values on a manageable scale, just as pH does for [H⁺]. Free AQA notes and 11 questions answered.
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Use this lesson to stay inside Acids and bases (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Read the core explanations from Calculations with Ka and pKa before testing yourself from memory.
Strong acids dissociate completely — you already know how to find their pH from concentration alone. Weak acids are different: they only partially dissociate, so you need an extra piece of information to find [H+]. That piece of information is Ka, the acid dissociation constant.
Ka tells you exactly how far the dissociation equilibrium lies. A large Ka means the equilibrium lies further to the right — more H+ ions in solution, lower pH. A small Ka means the acid barely dissociates at all. Once you can work with Ka, you can calculate the pH of any weak acid and compare acid strengths quantitatively.
Consider a general weak acid HA dissociating in water: $\text{HA}(aq) \rightleftharpoons \text{H}^+(aq) + \text{A}^-(aq)$. The equilibrium constant expression for this dissociation is written with products over reactants, but water is omitted because it is the solvent and its concentration is effectively constant.
Notice that Ka has units of mol dm-3 for a monoprotic acid, because the numerator has two concentration terms and the denominator has one. You will sometimes see Ka quoted as a dimensionless number in data books, but AQA expects you to include the units mol dm-3 unless told otherwise.
Here is where the key simplification comes in. When a weak acid HA of initial concentration c dissociates, it produces x mol dm-3 of H+ and x mol dm-3 of A-, while the concentration of HA falls to (c - x). Because x is very small compared with c, we approximate (c - x) as c. This means $[\text{HA}]_{\text{eqm}} \approx c$, the initial concentration.
Applying this approximation and noting that $[\text{H}^+] = [\text{A}^-] = x$ gives a much simpler expression: $K_a \approx \frac{x^2}{c}$, so $x = \sqrt{K_a \times c}$. Since $x = [\text{H}^+]$, you can then find pH directly.
The calculation always follows the same three-stage sequence: find [H+] using the Ka expression, then convert to pH. Work through the example below step by step before trying the practice question that follows.
Worked example: Calculating the pH of a weak acid from Ka and concentration
Calculate the pH of a 0.100 mol dm-3 solution of ethanoic acid. Ka(CH3COOH) = 1.74 x 10-5 mol dm-3 at 298 K.
Answer: pH = 2.88 (to 2 decimal places)
The calculation can also run in reverse: if you are given the pH and the initial concentration of a weak acid, you can work back to find Ka. This is a common exam question format, so it is worth practising the reverse route explicitly.
Worked example: Calculating Ka from pH and concentration
A 0.200 mol dm-3 solution of a weak acid HX has a pH of 3.40 at 298 K. Calculate Ka for HX.
Answer: Ka = 7.92 x 10-7 mol dm-3
Notice the pattern: pH to [H+] uses the antilog ($10^{-\text{pH}}$), then you square [H+] and divide by c. Keeping extra significant figures until the final step prevents rounding errors from distorting Ka, which is often a very small number.
Ka values for weak acids are typically very small numbers — ethanoic acid has Ka = 1.74 x 10-5 mol dm-3. Working with numbers like this is cumbersome, so chemists use pKa in exactly the same way that pH compresses [H+] into a manageable scale.
The relationship mirrors pH exactly: a larger Ka gives a smaller pKa, just as a higher [H+] gives a lower pH. This means that when comparing two weak acids, the one with the lower pKa is the stronger acid — it has a larger Ka and dissociates to a greater extent.
Worked example: Converting Ka to pKa and pKa to Ka
Part A: Ka for methanoic acid is 1.77 x 10-4 mol dm-3. Calculate pKa. Part B: A weak acid has pKa = 4.75. Calculate Ka.
Answer: Part A: pKa = 3.75 | Part B: Ka = 1.78 x 10-5 mol dm-3
One of the most useful applications of pKa is comparing the relative strengths of weak acids without doing a full pH calculation. The table below shows three common weak acids — notice how pKa and Ka move in opposite directions.
| Acid | Ka / mol dm-3 | pKa | Relative strength |
|---|---|---|---|
| Methanoic acid (HCOOH) | 1.77 x 10-4 | 3.75 | Strongest of the three |
| Ethanoic acid (CH3COOH) | 1.74 x 10-5 | 4.76 | Intermediate |
| Propanoic acid (CH3CH2COOH) | 1.35 x 10-5 | 4.87 | Weakest of the three |
Methanoic acid has the largest Ka and the smallest pKa, confirming it is the strongest of the three. At the same concentration, it will produce the highest [H+] and therefore the lowest pH. This is the kind of comparison an AQA examiner may ask you to make using the command word 'compare' — you need to state both the direction of the difference and the reason.
Ka and pKa calculations are reliable mark-earners if you follow a consistent method. The most common source of lost marks is not the maths — it is failing to state the assumption, misreading the direction of the log relationship, or rounding too early.
AQA mark schemes for 'calculate' questions award marks at each stage: one method mark for the correct Ka expression or rearrangement, one mark for correct substitution, and one accuracy mark for the final answer. Even if you make an arithmetic slip, you can still earn the method marks by showing clear, logical working.
These three spec requirements — calculating pH from Ka and c, calculating Ka from pH and c, and interconverting Ka and pKa — form a tightly connected set of skills. Mastering the direction of each conversion and the one key approximation gives you the tools to answer every AQA question in this area with confidence.
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Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
pKa is −log₁₀ Ka, a way of expressing very small Ka values on a manageable scale, just as pH does for [H⁺]. A lower pKa means a larger Ka and a stronger acid: methanoic acid, pKa 3.75, is stronger than ethanoic acid, pKa 4.76.
Ka is calculated from pH by first finding [H⁺] as 10 to the power −pH, then using Ka ≈ [H⁺]² ÷ c. For a 0.200 mol dm⁻³ weak acid with pH 3.40, [H⁺] = 3.981 × 10⁻⁴ mol dm⁻³, so Ka = (3.981 × 10⁻⁴)² ÷ 0.200 = 7.92 × 10⁻⁷ mol dm⁻³.
To calculate the pH of a weak acid, assume [HA] at equilibrium equals the initial concentration c and [H⁺] = [A⁻], so [H⁺] = √(Ka × c), then pH = −log₁₀[H⁺]. For 0.100 mol dm⁻³ ethanoic acid with Ka = 1.74 × 10⁻⁵ mol dm⁻³, [H⁺] = 1.319 × 10⁻³ and pH = 2.88.
The degree of dissociation is the fraction of acid molecules that have dissociated at equilibrium. For a weak acid it is small, typically less than 5%, which is why [HA] at equilibrium can be approximated as the initial concentration in Ka calculations.
pKa is converted to Ka using Ka = 10 to the power −pKa; for pKa = 4.75, Ka = 1.78 × 10⁻⁵ mol dm⁻³. Ka is converted to pKa using pKa = −log₁₀(Ka); for methanoic acid, Ka = 1.77 × 10⁻⁴ gives pKa = 3.75.
In weak acid calculations, [HA] at equilibrium is approximated as the initial concentration c because the acid is weak and dissociation is small, so x, the amount dissociated, is very small compared with c. The approximation is valid when the degree of dissociation is small, typically less than 5%.
The pH of 0.100 mol dm⁻³ ethanoic acid is 2.88. With Ka = 1.74 × 10⁻⁵ mol dm⁻³, [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.319 × 10⁻³ mol dm⁻³, and pH = −log₁₀(1.319 × 10⁻³) = 2.88, higher than the pH of 1.00 for a strong acid at the same concentration.
A lower pKa means a stronger weak acid, because pKa = −log₁₀ Ka, so a larger Ka gives a smaller pKa. The acid with the lower pKa dissociates to a greater extent and gives the higher [H⁺] and lower pH at the same concentration.
Methanoic acid has pKa 3.75, ethanoic acid 4.76 and propanoic acid 4.87. Methanoic acid has the largest Ka and smallest pKa, so it is the strongest of the three, while propanoic acid, with the highest pKa, is the weakest.
[H⁺] = √(Ka × c) for a weak acid because [H⁺] = [A⁻] = x and the approximation (c − x) ≈ c simplifies Ka = x² ÷ (c − x) to Ka ≈ x² ÷ c. Rearranging gives x = √(Ka × c), and since x = [H⁺], pH follows directly.
The most common errors in Ka and pKa calculations are failing to state the weak acid assumption, misreading the direction of the log relationship, and rounding too early. Keeping extra significant figures until the final step stops rounding errors distorting Ka, which is often a very small number.