Calculate the pH of 3.8 × 10-8 M HNO3
Use this premium calculator to find the exact pH of a very dilute nitric acid solution. For ultra-dilute strong acids such as 3.8 × 10-8 M HNO3, the correct calculation should include water autoionization at 25°C, not just the simple shortcut pH = -log[H+].
Enter the concentration values or use the default example of 3.8 × 10-8 M HNO3, then click the button to see the exact pH, hydrogen ion concentration, hydroxide ion concentration, and a comparison chart.
How to calculate the pH of 3.8 × 10-8 M HNO3 correctly
When students first learn pH, they usually memorize the shortcut formula pH = -log[H+]. That shortcut is useful in many standard chemistry problems, but it becomes unreliable for very dilute strong acid solutions. The concentration 3.8 × 10-8 M HNO3 is a perfect example. At this level, the acid concentration is actually lower than the hydrogen ion concentration already present from pure water at 25°C, which is 1.0 × 10-7 M. Because of that, you cannot simply set [H+] equal to 3.8 × 10-8 M and stop there.
Nitric acid, HNO3, is a strong monoprotic acid. In ordinary textbook problems, we assume it dissociates completely:
HNO3 → H+ + NO3–
If this were a more concentrated solution, that would be enough to estimate [H+] directly from the acid molarity. But at 3.8 × 10-8 M, the water itself contributes a significant amount of hydrogen ions and hydroxide ions through autoionization:
H2O ⇌ H+ + OH–
At 25°C, the ion product of water is:
Kw = [H+][OH–] = 1.0 × 10-14
The exact setup
Let the total hydrogen ion concentration be x. Since HNO3 is a strong acid, it contributes 3.8 × 10-8 M nitrate, and charge balance gives:
x = 3.8 × 10-8 + [OH–]
From water equilibrium:
[OH–] = Kw / x
Substitute that into the charge balance equation:
x = 3.8 × 10-8 + (1.0 × 10-14 / x)
Multiply both sides by x:
x2 = (3.8 × 10-8)x + 1.0 × 10-14
Rearrange into quadratic form:
x2 – (3.8 × 10-8)x – 1.0 × 10-14 = 0
Now solve with the quadratic formula:
x = [(3.8 × 10-8) + √((3.8 × 10-8)2 + 4(1.0 × 10-14))] / 2
This gives:
[H+] ≈ 1.21 × 10-7 M
Then:
pH = -log(1.21 × 10-7) ≈ 6.92
Why the shortcut pH = 7.42 is wrong
If someone ignores water autoionization and simply computes pH from the formal acid concentration, they would do this:
pH = -log(3.8 × 10-8) ≈ 7.42
That result suggests the acidic solution is basic because the pH is above 7. This is chemically impossible for a solution made by adding a strong acid to pure water. The mistake happens because the shortcut treats the acid as the only source of H+, which fails when the acid concentration is extremely small. In reality, pure water already contains 1.0 × 10-7 M H+ at 25°C. Adding any strong acid increases the hydrogen ion concentration above that level, so the pH must fall below 7.
Step by step interpretation of the chemistry
- Recognize that HNO3 is a strong acid and dissociates essentially completely.
- Notice the concentration is only 3.8 × 10-8 M, which is less than 1.0 × 10-7 M.
- Realize water autoionization is no longer negligible.
- Use charge balance together with Kw to write the exact quadratic.
- Solve for the physically meaningful positive root.
- Convert the resulting hydrogen ion concentration into pH.
Comparison table: naive versus exact pH for dilute strong acid solutions
The table below shows why the exact method matters. The “naive pH” column assumes [H+] equals the acid molarity. The “exact pH” column includes water autoionization at 25°C.
| HNO3 Concentration (M) | Naive pH | Exact [H+] (M) | Exact pH | Absolute pH Error |
|---|---|---|---|---|
| 1.0 × 10-3 | 3.00 | 1.0000001 × 10-3 | 3.00 | 0.00 |
| 1.0 × 10-5 | 5.00 | 1.0001 × 10-5 | 5.00 | 0.00 |
| 1.0 × 10-7 | 7.00 | 1.6180 × 10-7 | 6.79 | 0.21 |
| 3.8 × 10-8 | 7.42 | 1.2099 × 10-7 | 6.92 | 0.50 |
| 1.0 × 10-8 | 8.00 | 1.0512 × 10-7 | 6.98 | 1.02 |
What this result means physically
The exact pH of about 6.92 tells you the solution is slightly acidic. That makes sense. The acid is extremely dilute, so it only nudges the hydrogen ion concentration above the level of pure water. The effect is small, but it is still enough to make the pH less than 7. This is a subtle but important concept in analytical chemistry, general chemistry, environmental chemistry, and any laboratory situation involving trace concentrations.
You can think of the final hydrogen ion concentration as having two contributors:
- The strong acid itself, contributing nitrate and pushing the equilibrium toward more H+.
- The water equilibrium, which always supplies some H+ and OH–.
At moderate acid concentrations, the water contribution is negligible. At very low acid concentrations like 3.8 × 10-8 M, it is not negligible at all.
When should you include water autoionization?
A practical rule is this: if the acid or base concentration is near 1.0 × 10-6 M or smaller, you should at least check whether water autoionization matters. Once you get close to 1.0 × 10-7 M, the correction becomes especially important. For the current problem, 3.8 × 10-8 M is below that threshold, so the exact method is absolutely the correct one.
| Quantity | Value for 3.8 × 10-8 M HNO3 | Meaning |
|---|---|---|
| Formal acid concentration | 3.8 × 10-8 M | The amount of HNO3 added |
| Kw at 25°C | 1.0 × 10-14 | Water autoionization constant |
| Exact [H+] | 1.2099 × 10-7 M | Total hydrogen ion concentration after equilibrium |
| Exact [OH–] | 8.2648 × 10-8 M | Hydroxide concentration from Kw / [H+] |
| Exact pH | 6.92 | Correct acidity of the solution |
Common mistakes students make
- Using only pH = -log C for every strong acid problem, even when C is below 10-6 M.
- Forgetting that pure water already has ions at 25°C.
- Reporting pH above 7 for a strong acid solution, which should trigger an immediate reality check.
- Ignoring temperature. The neutral pH is 7.00 only at 25°C because Kw changes with temperature.
- Dropping the wrong quadratic root or making algebra mistakes while solving the equilibrium equation.
Fast exam strategy for ultra-dilute acid questions
If you see a strong acid concentration around 10-8 M, pause before calculating. Ask whether the concentration is comparable to 10-7 M, the hydrogen ion concentration from water. If yes, use the exact relation:
[H+] = (C + √(C2 + 4Kw)) / 2
For this specific problem, plug in C = 3.8 × 10-8 and Kw = 1.0 × 10-14. This gives the exact answer rapidly without needing a long derivation during an exam.
Why this matters outside the classroom
These calculations are not just academic exercises. Ultra-dilute acid and base systems appear in environmental monitoring, atmospheric chemistry, water treatment, and trace-level analytical work. When concentrations approach the natural ion background of water, simplistic formulas can produce misleading answers. Using the exact equilibrium approach avoids conceptual errors and gives results that match physical reality.
For example, pH measurements in very pure water can be challenging because dissolved carbon dioxide from air, trace contaminants, electrode limitations, and temperature shifts all influence the reading. This makes the distinction between formal concentration and actual equilibrium ion concentrations especially important.
Authoritative chemistry references
For more background on water chemistry, pH, and equilibrium data, consult these authoritative sources:
- U.S. Geological Survey: pH and Water
- LibreTexts Chemistry hosted by educational institutions
- U.S. Environmental Protection Agency: Water Quality Criteria
Bottom line
To calculate the pH of 3.8 × 10-8 M HNO3, do not use the oversimplified shortcut by itself. Because the acid concentration is lower than the hydrogen ion concentration in pure water, water autoionization must be included. Solving the exact equilibrium expression gives [H+] ≈ 1.21 × 10-7 M and pH ≈ 6.92 at 25°C. That answer is chemically reasonable, mathematically correct, and the one you should report in serious chemistry work.