pH of Very Dilute Acids and Mixtures of Strong Acids and Bases
Questions on the pH of a strong acid look easy: write the concentration, take −log. But exam setters like two twists. One is the very dilute acid, such as 10−8 M HCl, where water can no longer be ignored. The other is a mixture of acids, or of an acid and a base, where you must add moles first and divide by the final volume. This page shows both, step by step.
The formulas
Mixing: [H+] = (M1V1 + M2V2) / (V1 + V2)
Acid + base: excess moles / total volume
Very dilute strong acid: [H+] = C + Kw/[H+], so [H+]2 − C[H+] − Kw = 0
C is the molar concentration of the strong acid. Kw = 1.0 × 10−14 at 25 °C. V is volume (use mL for both volumes, the units cancel in the ratio). For H2SO4 in school problems, take both hydrogens as released, so 1 mol of acid gives 2 mol of H+; note that your book may treat the second step as incomplete in dilute solution.
Why 10−8 M acid is not pH 8
An acid can never make a solution basic. If you write pH = −log(10−8) = 8, you get a basic answer from an acid, which is impossible. Pure water already holds 10−7 M H+. When the acid gives only 10−8 M, water’s own contribution is larger and cannot be ignored. The correct approach adds the two sources: [H+] = C + [OH−] (charge balance), and [OH−] = Kw/[H+]. That gives the quadratic in the formula box.
As a rule of thumb, if C is about 10−6 M or larger, the simple formula is accurate enough. Below that, use the quadratic.
Worked example 1: 1.0 × 10−8 M HCl
[H+]2 − (1.0 × 10−8)[H+] − 1.0 × 10−14 = 0.
[H+] = [C + √(C2 + 4Kw)] / 2 = [1.0 × 10−8 + √(1.0 × 10−16 + 4.0 × 10−14)] / 2.
√(4.0001 × 10−14) = 2.0000 × 10−7, so [H+] = (0.1 × 10−7 + 2.0 × 10−7)/2 = 1.05 × 10−7 M.
pH = −log(1.05 × 10−7) = 6.98. The answer is just below 7, as an acid should give.
Worked example 2: mixing two strong acids
100 mL of 0.100 M HCl is mixed with 400 mL of 0.0500 M HNO3. Find the pH.
Moles of H+: 0.100 × 100 = 10.0 mmol and 0.0500 × 400 = 20.0 mmol. Total = 30.0 mmol.
Total volume = 500 mL, so [H+] = 30.0 / 500 = 0.0600 M. pH = −log(0.0600) = 1.22.
Wrong shortcut: averaging pH values (1.00 and 1.30) gives 1.15, which is not correct. Always average concentrations by moles, never pH.
Worked example 3: strong acid with a smaller amount of strong base
50 mL of 0.200 M HCl is mixed with 30 mL of 0.200 M NaOH.
H+ = 0.200 × 50 = 10.0 mmol. OH− = 0.200 × 30 = 6.0 mmol. Acid is in excess: 10.0 − 6.0 = 4.0 mmol H+ left.
Total volume = 80 mL, so [H+] = 4.0 / 80 = 0.0500 M. pH = 1.30.
Worked example 4: base in excess, with H2SO4
25 mL of 0.100 M NaOH is mixed with 25 mL of 0.0400 M H2SO4. Treat both H+ of H2SO4 as released.
OH− = 0.100 × 25 = 2.50 mmol. H+ = 2 × 0.0400 × 25 = 2.00 mmol. Base is in excess by 0.50 mmol.
[OH−] = 0.50 / 50 = 0.0100 M. pOH = 2.00, so pH = 14.00 − 2.00 = 12.00.
Worked example 5: mixing solutions of known pH
100 mL of a solution with pH 2 is mixed with 400 mL of a solution with pH 4 (both strong acids).
[H+] = 10−2 M and 10−4 M. Moles: 10−2 × 100 = 1.000 mmol and 10−4 × 400 = 0.040 mmol. Total = 1.040 mmol in 500 mL.
[H+] = 2.08 × 10−3 M, pH = 2.68. The stronger acid dominates, so the result is close to pH 2 and not to the middle value.
Common mistakes
- Giving pH 8 for 10−8 M HCl. Always check that an acid solution has pH below 7.
- Averaging the pH values of the two solutions. pH is a logarithm; add moles first.
- Using the old volume instead of the total volume after mixing.
- Forgetting to subtract the neutralised part when an acid and a base are mixed.
- Stopping at pOH. If the excess is a base, convert with pH = 14 − pOH.
- Counting only one H+ from H2SO4 when your book says to count two.
Exam relevance
| Situation | Method |
|---|---|
| Acid concentration 10−6 M or more | pH = −log C |
| Acid concentration below about 10−6 M | Use the quadratic with Kw |
| Two acids mixed | Add mmol of H+, divide by total mL |
| Acid with base | Subtract mmol, divide the excess by total mL |
| Excess is base | Find pOH first, then pH = 14 − pOH |
| Exact equivalence (strong/strong) | pH = 7 at 25 °C |
Practice with the numbers above, then change the volumes and check your answer. Your school board and your teacher decide whether the quadratic step is needed, so follow your own syllabus.
After you work out each [H+] by hand, check the final pH with the pH / pOH calculator.
Open the pH / pOH Calculator →Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
More Equilibrium articles