Calculate the pH of 2.5 x 10-8 HBr
This calculator solves the pH of a very dilute hydrobromic acid solution using the correct equilibrium treatment. Because 2.5 x 10-8 M is below the natural 1.0 x 10-7 M hydrogen ion scale associated with pure water at 25 C, water autoionization must be included for an accurate answer.
HBr pH Calculator
Use the exact strong acid plus water equilibrium model for ultradilute solutions.
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Expert guide: how to calculate the pH of 2.5 x 10-8 HBr correctly
If you are trying to calculate the pH of 2.5 x 10-8 M HBr, the biggest trap is using the usual strong acid shortcut without thinking about the concentration range. In most classroom problems involving HBr, hydrobromic acid is treated as a strong acid that dissociates completely, so students often write [H+] = 2.5 x 10-8 M and then calculate pH as minus log of that number. That shortcut would give a pH of about 7.60, which suggests the solution is basic. That cannot be right for a solution containing a strong acid.
The reason the shortcut fails is that the acid concentration is extremely low. Pure water at 25 C already contributes hydrogen ions and hydroxide ions through autoionization. In pure water, [H+] and [OH–] are each close to 1.0 x 10-7 M, with Kw near 1.0 x 10-14. When the strong acid concentration drops below or near that 10-7 M scale, the water contribution is no longer negligible. For a problem like 2.5 x 10-8 M HBr, the correct treatment must include both complete acid dissociation and the water equilibrium.
Why HBr is usually easy, but not here
HBr is a strong acid, so it dissociates essentially completely in dilute aqueous solution:
For common concentrations such as 0.1 M, 0.01 M, or even 1.0 x 10-5 M, the acid itself overwhelmingly controls hydrogen ion concentration, so the water contribution can be ignored. But at 2.5 x 10-8 M, the acid concentration is smaller than the hydrogen ion concentration in neutral pure water. That is exactly the regime where the standard approximation breaks down.
The correct equilibrium setup
To solve the problem properly, use two core ideas:
- HBr is a strong acid, so it contributes a formal concentration Ca of bromide ions and acidic charge.
- Water still obeys Kw = [H+][OH–].
The electrical neutrality condition is the most convenient way to start. In this solution, the positive charge from hydrogen ions must balance the negative charge from bromide and hydroxide:
Since HBr dissociates completely, [Br–] = Ca = 2.5 x 10-8 M. Also, from water equilibrium:
Substitute [OH–] = Kw / [H+] into the charge balance:
Multiply through by [H+] to obtain a quadratic equation:
The physically meaningful solution is:
At 25 C, using Ca = 2.5 x 10-8 M and Kw = 1.01 x 10-14 gives:
- Ca2 = 6.25 x 10-16
- 4Kw = 4.04 x 10-14
- Ca2 + 4Kw = 4.1025 x 10-14
- Square root = 2.0255 x 10-7
- [H+] = (2.5 x 10-8 + 2.0255 x 10-7) / 2 = 1.1378 x 10-7 M
Now calculate pH:
Why the naive answer is wrong
The naive strong acid method says [H+] = 2.5 x 10-8 M, which leads to pH = 7.60. That answer is chemically inconsistent because adding a strong acid should not make water more basic than neutral. The problem is that the shortcut silently assumes water contributes almost no hydrogen ions. At this concentration, that assumption is false.
In reality, pure water already sits near pH 7 at 25 C. When you add a tiny amount of strong acid, you do not simply replace the water contribution with the acid concentration. Instead, the acid shifts the water equilibrium and slightly increases total [H+] above the neutral water level. That is why the exact pH is just below 7, not far above it.
Comparison table: exact vs naive pH for dilute strong acid solutions
The table below shows how the exact approach and the shortcut compare as strong acid concentration decreases at 25 C. The exact values come from the quadratic relation above with Kw = 1.01 x 10-14.
| HBr concentration (M) | Naive pH using [H+] = Ca | Exact [H+] (M) | Exact pH | Practical takeaway |
|---|---|---|---|---|
| 1.0 x 10^-2 | 2.000 | 1.000 x 10^-2 | 2.000 | Water contribution is negligible. |
| 1.0 x 10^-5 | 5.000 | 1.001 x 10^-5 | 4.9996 | Shortcut still works very well. |
| 1.0 x 10^-7 | 7.000 | 1.621 x 10^-7 | 6.790 | Water autoionization matters a lot. |
| 2.5 x 10^-8 | 7.602 | 1.138 x 10^-7 | 6.944 | Shortcut gives a chemically misleading result. |
| 1.0 x 10^-9 | 9.000 | 1.005 x 10^-7 | 6.998 | Solution is still slightly acidic at 25 C. |
Step by step method you can use on exams
If you encounter any ultradilute strong acid problem, this is the fastest reliable workflow:
- Identify whether the acid is strong and fully dissociated. HBr is.
- Check whether the acid concentration is near or below 1 x 10-6 to 1 x 10-7 M. If yes, think about water autoionization.
- Write charge balance: [H+] = Ca + [OH–].
- Use Kw = [H+][OH–].
- Substitute and solve the quadratic: [H+]2 – Ca[H+] – Kw = 0.
- Take the positive root and compute pH.
This method is robust, physically meaningful, and works for HCl, HBr, and HI in the ultradilute limit. It also explains why the answer remains acidic even when the formal strong acid concentration is below 1 x 10-7 M.
Temperature also affects the result
Neutral pH equals 7.00 only at about 25 C when pKw is close to 14.00. As temperature changes, Kw changes too, which shifts the pH of neutral water. That means the exact pH of a very dilute HBr solution also depends somewhat on temperature. The effect is not huge for routine calculations, but it is real and measurable.
| Temperature | Approximate Kw | pKw | Neutral pH | Exact pH for 2.5 x 10^-8 M HBr |
|---|---|---|---|---|
| 20 C | 6.81 x 10^-15 | 14.167 | 7.083 | 7.019 |
| 25 C | 1.01 x 10^-14 | 13.996 | 6.998 | 6.944 |
| 37 C | 2.40 x 10^-14 | 13.620 | 6.810 | 6.773 |
This temperature sensitivity is one reason advanced chemistry courses emphasize that pH 7 is not a universal definition of neutrality. Neutrality means [H+] = [OH–], not necessarily pH 7. In a highly dilute strong acid solution, that distinction matters.
Common mistakes students make
- Ignoring water autoionization: This is the main mistake for concentrations near 10-7 M.
- Assuming pH above 7 means basic in every context: A wrong calculation can produce that result for a strong acid, but chemistry should alert you that something is off.
- Using the wrong equilibrium equation: The correct relation for an ultradilute strong acid is the quadratic based on charge balance and Kw.
- Forgetting temperature effects: Kw changes with temperature, so neutral pH changes too.
- Over-rounding intermediate values: Since the numbers are small and close together, keep enough significant figures until the final step.
Interpreting the chemistry behind the number
A pH of about 6.94 means the solution is only slightly acidic. That makes sense because 2.5 x 10-8 M is a tiny concentration. The acid does not overwhelm water. Instead, it nudges the hydrogen ion concentration above the neutral-water value. This is a subtle but important conceptual point: pH is determined by the total equilibrium hydrogen ion concentration, not just the formal amount of acid added.
In laboratory work, very dilute pH measurements can also be affected by dissolved carbon dioxide, ionic strength, electrode calibration, and contamination. So while the theoretical calculation is clear, actual measured pH in an open container may differ because atmospheric CO2 can acidify water significantly more than a 2.5 x 10-8 M strong acid addition. That practical reality is another reason to understand the equilibrium framework rather than relying on a memorized shortcut.
Useful references and authoritative sources
For additional reading on pH, water chemistry, and reference data, consult these authoritative sources:
Bottom line
To calculate the pH of 2.5 x 10-8 HBr correctly, you must include water autoionization. The exact strong acid expression gives [H+] around 1.14 x 10-7 M at 25 C, corresponding to a pH of about 6.94. The common shortcut pH = -log(2.5 x 10-8) fails because it ignores the fact that water itself already contributes hydrogen ions at about the 10-7 M level. For ultradilute strong acid problems, the quadratic equilibrium approach is the correct method and the one you should trust.