Calculate pH of 10-8 M NaOH
This interactive calculator solves the pH of extremely dilute sodium hydroxide solutions. For concentrations such as 1.0 × 10-8 M, a simple pOH = -log[OH–] shortcut is not accurate because water itself contributes hydroxide and hydrogen ions. This tool applies the proper equilibrium correction.
NaOH pH Calculator
Results
Enter values and click Calculate to compute the pH of a very dilute NaOH solution.
Expert Guide: How to Calculate the pH of 10-8 M NaOH Correctly
If you are trying to calculate the pH of 10-8 M NaOH, the first thing to know is that this is not a standard strong-base homework problem where you can directly set hydroxide concentration equal to the listed molarity and stop. At moderate concentrations, sodium hydroxide behaves as a strong base and dissociates essentially completely, so many students learn the fast method: [OH–] = C, then pOH = -log[OH–], then pH = 14 – pOH at 25 C. However, that shortcut breaks down when the concentration is extremely low, especially near 10-7 M or below.
The reason is simple but important: pure water already contains ions because of autoionization. At 25 C, pure water has [H+] = 1.0 × 10-7 M and [OH–] = 1.0 × 10-7 M. So if you prepare a 10-8 M NaOH solution, the hydroxide coming from the base is actually smaller than the hydroxide already present from water. That means the solution is basic, but only slightly basic, not strongly basic.
Why the simple shortcut gives the wrong answer
Let us start with the shortcut many people try first:
- Assume NaOH fully dissociates, so [OH–] = 1.0 × 10-8 M
- Compute pOH = -log(1.0 × 10-8) = 8
- Compute pH = 14 – 8 = 6
This predicts an acidic solution, which cannot be correct for sodium hydroxide, a base. The contradiction reveals the flaw: the method ignored water autoionization.
In a very dilute strong base, total hydroxide in solution is not just the hydroxide supplied by the solute. Instead, the equilibrium between hydrogen ions and hydroxide ions in water must also be considered. At 25 C, the ion product of water is:
Kw = [H+][OH–] = 1.0 × 10-14
The correct equilibrium setup
For a strong monohydroxide base such as NaOH at concentration C, the chemically added hydroxide is C. But because water also ionizes, the total hydroxide concentration is slightly larger than C, while hydrogen ion concentration is slightly smaller than 10-7 M.
Let [H+] = x. Then by water equilibrium:
[OH–] = Kw / x
Charge balance for a dilute NaOH solution gives:
[Na+] + [H+] = [OH–]
Since NaOH dissociates completely, [Na+] = C = 1.0 × 10-8 M. Therefore:
C + x = [OH–]
Now combine both expressions:
x(C + x) = Kw
x2 + Cx – Kw = 0
Using C = 1.0 × 10-8 and Kw = 1.0 × 10-14 at 25 C:
x = [-C + √(C2 + 4Kw)] / 2
Substituting values:
x = [-(1.0 × 10-8) + √((1.0 × 10-8)2 + 4.0 × 10-14)] / 2
This gives approximately:
- [H+] ≈ 9.512 × 10-8 M
- [OH–] ≈ 1.051 × 10-7 M
- pH ≈ 7.022
- pOH ≈ 6.978
So the correct pH of 10-8 M NaOH at 25 C is about 7.02, not 6.00 and not 8.00. The solution is basic, but only very slightly.
Interpretation of the result
This result makes chemical sense. Adding a tiny amount of a strong base to pure water should shift the pH above 7, but not by much if the concentration is only 10-8 M. Because the concentration is below the natural 10-7 M ion level of water, autoionization strongly influences the final answer.
This is one of the classic examples used in general chemistry to show why approximations have limits. A strong electrolyte does fully dissociate, but that does not mean the listed concentration is automatically the final equilibrium ion concentration when the solvent contributes comparable amounts of ions.
Step by step method you can use every time
- Identify that the strong base concentration is very small, especially near or below 10-6 M.
- Recognize that water autoionization cannot be neglected.
- Write the charge balance: [OH–] = C + [H+] for NaOH.
- Apply Kw = [H+][OH–].
- Solve the quadratic: x2 + Cx – Kw = 0.
- Use the positive physical root for x = [H+].
- Calculate pH = -log[H+].
Comparison table: ideal shortcut versus corrected equilibrium
| NaOH concentration (M) | Ideal shortcut pH | Corrected pH at 25 C | Difference |
|---|---|---|---|
| 1.0 × 10-2 | 12.00 | 12.00 | Negligible |
| 1.0 × 10-5 | 9.00 | 9.00 | Very small |
| 1.0 × 10-7 | 7.00 | 7.21 | About 0.21 pH unit |
| 1.0 × 10-8 | 6.00 | 7.02 | About 1.02 pH units |
| 1.0 × 10-9 | 5.00 | 7.00 | About 2.00 pH units |
The table shows a useful trend. At high concentration, the shortcut is fine. At low concentration, it becomes seriously misleading. At 10-8 M, the shortcut even predicts the wrong chemical behavior, calling a basic solution acidic.
What changes with temperature?
Many textbooks teach pH and pOH with the statement pH + pOH = 14, but this is only exactly true at 25 C. More generally:
pH + pOH = pKw
Because Kw changes with temperature, the neutral pH also changes. As temperature rises, Kw increases, meaning pure water contains more H+ and OH– ions. A neutral solution at elevated temperature can have a pH below 7 while still remaining neutral because [H+] equals [OH–].
| Temperature | Kw | pKw | Approximate neutral pH |
|---|---|---|---|
| 0 C | 1.15 × 10-15 | 14.94 | 7.47 |
| 25 C | 1.00 × 10-14 | 14.00 | 7.00 |
| 35 C | 2.92 × 10-14 | 13.53 | 6.77 |
| 50 C | 5.48 × 10-14 | 13.26 | 6.63 |
For this reason, any serious pH calculator should allow the user to account for Kw at the chosen temperature. The calculator above does exactly that.
Common mistakes when solving this problem
- Ignoring water autoionization. This is the biggest error and leads to impossible predictions like acidic NaOH.
- Using pH + pOH = 14 at every temperature. The correct relation is pH + pOH = pKw.
- Forgetting stoichiometry for bases that release more than one OH–. Ba(OH)2 and Ca(OH)2 contribute two hydroxides per formula unit.
- Rounding too early. With dilute solutions, small concentration differences matter, so carry extra digits through the calculation.
- Assuming neutral always means pH 7. Neutral means [H+] = [OH–], not necessarily pH 7 at all temperatures.
When is the water correction necessary?
A good practical rule is to think carefully any time the acid or base concentration drops into the 10-6 to 10-8 M range. Above that level, the solute contribution usually dominates strongly enough that the shortcut is acceptable for introductory work. Below that level, the solvent contribution is no longer negligible. In very dilute systems, the correct equilibrium treatment is the only reliable method.
In analytical chemistry, environmental chemistry, and high-purity water systems, this distinction matters. Even a tiny error in ion accounting can produce a noticeable pH shift. The physics of the solvent starts to matter as much as the chemistry of the dissolved reagent.
Final answer for 10-8 M NaOH at 25 C
Using the full equilibrium calculation with water autoionization included:
- [H+] ≈ 9.512 × 10-8 M
- [OH–] ≈ 1.051 × 10-7 M
- pH ≈ 7.02
- pOH ≈ 6.98
That is the correct chemistry result and the one expected in a rigorous textbook or exam solution.
Authoritative references for pH, water ionization, and aqueous equilibria
- LibreTexts Chemistry for detailed derivations of acid-base equilibrium and water autoionization.
- USGS.gov: pH and Water for reliable background on pH and aqueous systems.
- EPA.gov: pH Overview for environmental context and accepted pH definitions.