pKa and pKb: Converting Ka to pKa and Ranking Acid Strength
In school and entrance exams, acid strength is given in two ways: as a small number like 1.8 × 10−5 (Ka) or as a simple number like 4.74 (pKa). Questions often ask you to move between the two, to rank acids, or to find a pH from pKa. This page shows the conversion, the ranking rule and the pH shortcut, with every number worked out.
The definition
pKb = −log10 Kb pKa + pKb = pKw = 14 (for a conjugate pair at 25 °C)
- Ka is the acid dissociation constant: Ka = [H+][A−] / [HA]. It has no unit in most school problems.
- pKa is just the “p” (negative log) of Ka, the same idea as pH from [H+].
- The direction flips. A bigger Ka means a stronger acid. A bigger pKa means a weaker acid. Remember: “small pKa, strong acid”.
- Kw = 1.0 × 10−14 at 25 °C, so pKw = 14. At other temperatures pKw is different, so use 14 only when the question says 25 °C or gives no temperature.
Typical textbook values
These values are typical figures from school textbooks at 25 °C. Different books round them slightly, so use the value your question gives.
| Acid | Ka | pKa | Strength rank |
|---|---|---|---|
| HF (hydrofluoric acid) | 6.8 × 10−4 | 3.17 | Strongest of the three |
| CH3COOH (acetic acid) | 1.8 × 10−5 | 4.74 | Middle |
| HCN (hydrocyanic acid) | 4.9 × 10−10 | 9.31 | Weakest of the three |
Notice that the pKa column is in reverse order to the Ka column. The acid with the smallest pKa is the strongest.
Worked examples
Example 1: Ka to pKa. Acetic acid has Ka = 1.8 × 10−5. Find pKa.
log10(1.8 × 10−5) = log10(1.8) + (−5) = 0.2553 − 5 = −4.7447
pKa = −(−4.7447) = 4.74
Check: 10−4.74 = 100.26 × 10−5 = 1.82 × 10−5. ✓
Example 2: pKa to Ka. An acid has pKa = 9.25. Find Ka.
Ka = 10−9.25. Write −9.25 as −10 + 0.75. Then Ka = 100.75 × 10−10 = 5.62 × 10−10.
So Ka = 5.6 × 10−10. The trick is to split the decimal so that the power of 10 is a whole number.
Example 3: conjugate base. For acetic acid (pKa = 4.74), find pKb and Kb of the acetate ion at 25 °C.
pKb = 14 − 4.74 = 9.26
Kb = 10−9.26 = 100.74 × 10−10 = 5.5 × 10−10
Second route: Kb = Kw / Ka = 1.0 × 10−14 / 1.8 × 10−5 = 5.6 × 10−10. The tiny difference comes from rounding pKa to two decimals. ✓
Example 4: pH of a weak acid from pKa. Find the pH of 0.10 M acetic acid (pKa = 4.74).
For a weak acid that ionises only a little, [H+] = √(Ka × C). Taking −log of both sides gives a shortcut:
pH = ½ (pKa − log C) = ½ (4.74 − log 0.10) = ½ (4.74 + 1) = 2.87
Check with Ka: [H+] = √(1.8 × 10−5 × 0.10) = √(1.8 × 10−6) = 1.34 × 10−3 M, so pH = −log(1.34 × 10−3) = 2.87. ✓
Is the approximation fair? The ionised part is 1.34 × 10−3 / 0.10 = 1.3% of the acid, which is small, so ignoring the change in C is acceptable.
Example 5: how much stronger? How many times larger is Ka of acetic acid than that of HCN?
Ratio = 1.8 × 10−5 / 4.9 × 10−10 = 3.7 × 104. By pKa: difference = 9.31 − 4.74 = 4.57, and 104.57 = 3.7 × 104. ✓
A difference of 1 in pKa means a factor of 10 in Ka.
Common mistakes
- Saying a higher pKa means a stronger acid. It is the opposite.
- Forgetting the minus sign in pKa = −log Ka, which gives a negative pKa for a weak acid.
- Writing 10−9.25 as 9.25 × 10−10. The decimal part of the power is not the coefficient; use 100.75 × 10−10.
- Using pKa + pKb = 14 for two different acids. It works only for an acid and its own conjugate base.
- Using the shortcut pH = ½(pKa − log C) for a very dilute solution or a fairly strong acid, where ionisation is large. Then solve the quadratic equation.
Where this is used in exams
| Topic | How pKa is used |
|---|---|
| Ionic equilibrium (Class 11) | Ranking acids, finding pH of weak acids and bases. |
| Buffers | pH = pKa + log([salt]/[acid]) uses pKa directly. |
| Titrations | At half-equivalence of a weak acid with a strong base, pH = pKa. |
| Organic chemistry | Comparing acidity of carboxylic acids and phenols. |
Always follow the values and rounding given in your own textbook or question paper.
Use the log and 10x keys of the Scientific Calculator to check your pKa and Ka conversions.
Open the Chemistry Calculator Suite →Trends and exceptions are asked every year: practise previous-year CSIR-NET questions and quizzes.
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