Calculate The Ph Of A 1.0 10 8 M Hbr

Use this interactive calculator to find the exact pH of an extremely dilute hydrobromic acid solution. Because 1.0 × 10^-8 M is close to the hydrogen ion concentration produced by pure water, the correct calculation must include water autoionization rather than relying only on the simple strong-acid shortcut.

pH Calculator

How to Calculate the pH of a 1.0 × 10^-8 M HBr Solution

If you are asked to calculate the pH of a 1.0 × 10^-8 M HBr solution, the problem looks simple at first. HBr is a strong acid, so many students immediately assume that the hydrogen ion concentration is exactly the same as the acid concentration. That shortcut often works for stronger solutions, but here it gives the wrong answer. The reason is that 1.0 × 10^-8 M is extremely dilute, and it is actually lower than the hydrogen ion concentration contributed by pure water at 25 °C, which is 1.0 × 10^-7 M.

This means water itself is no longer negligible. To solve the problem correctly, you must include the autoionization of water. Once that is accounted for, the pH comes out slightly less than 7, not 8. That result makes chemical sense: adding a strong acid to water should produce an acidic solution, even if only very slightly acidic.

Key result: for 1.0 × 10^-8 M HBr at 25 °C, the exact pH is about 6.9783, not 8.0000.

Why the Simple Strong-Acid Shortcut Fails

In general chemistry, strong acids are usually treated as fully dissociated. For HBr, that means:

HBr → H⁺ + Br^–

From this, many people write [H⁺] = 1.0 × 10^-8 M and then compute:

pH = -log(1.0 × 10^-8) = 8

But pH 8 is basic, and that cannot be right for a solution made by adding a strong acid. The contradiction tells you the approximation is invalid. The issue is that pure water already generates hydrogen ions and hydroxide ions through autoionization:

H₂O ⇌ H⁺ + OH^–

At 25 °C, the ion-product constant for water is:

K_w = [H⁺][OH^–] = 1.0 × 10^-14

In pure water, [H⁺] = [OH^–] = 1.0 × 10^-7 M. Since the acid concentration here is only 1.0 × 10^-8 M, it is too small to dominate the equilibrium completely. Instead, the final hydrogen ion concentration comes from both the added acid and water.

The Correct Equation

Let the formal concentration of HBr be C. Since HBr is a strong acid, it contributes approximately C mol/L of bromide ion and extra hydrogen ion. The total hydrogen ion concentration in solution is not just C because water also contributes some H⁺.

Use these two relationships:

• Charge balance: [H⁺] = [OH^–] + C
• Water equilibrium: [H⁺][OH^–] = K_w

Substitute [OH^–] = K_w/[H⁺] into the charge balance:

[H⁺] = K_w/[H⁺] + C

Multiply both sides by [H⁺]:

[H⁺]² – C[H⁺] – K_w = 0

This is a quadratic equation. Solving for the physically meaningful positive root gives:

[H⁺] = (C + √(C² + 4K_w)) / 2

Now plug in the values:

• C = 1.0 × 10^-8 M
• K_w = 1.0 × 10^-14

[H⁺] = (1.0 × 10^-8 + √((1.0 × 10^-8)² + 4.0 × 10^-14)) / 2

[H⁺] = (1.0 × 10^-8 + √(1.0 × 10^-16 + 4.0 × 10^-14)) / 2

[H⁺] = (1.0 × 10^-8 + √(4.01 × 10^-14)) / 2

[H⁺] ≈ (1.0 × 10^-8 + 2.0025 × 10^-7) / 2

[H⁺] ≈ 1.05125 × 10^-7 M

Then:

pH = -log(1.05125 × 10^-7) ≈ 6.9783

Step-by-Step Method You Can Reuse

1. Identify that HBr is a strong acid and fully dissociates.
2. Check whether the acid concentration is much larger than 1.0 × 10^-7 M.
3. If it is not much larger, include water autoionization.
4. Use the equation [H⁺] = (C + √(C² + 4K_w))/2.
5. Compute pH using pH = -log[H⁺].

This procedure works for any extremely dilute strong monoprotic acid such as HCl, HBr, or HI, as long as the idealized assumptions are appropriate and the temperature is close to the value used for K_w.

Comparison of Approximate and Exact Results

The biggest conceptual mistake in this problem is trusting the naive strong-acid approximation outside its valid range. The table below shows why the exact method matters more and more as concentration approaches the intrinsic [H⁺] of water.

HBr concentration (M)	Naive pH using pH = -log C	Exact pH including Kw	Difference
1.0 × 10^-2	2.0000	2.0000	Negligible
1.0 × 10^-4	4.0000	4.0000	Negligible
1.0 × 10^-6	6.0000	5.9957	0.0043 pH unit
1.0 × 10^-8	8.0000	6.9783	1.0217 pH units
1.0 × 10^-9	9.0000	6.9978	2.0022 pH units

Notice the trend: once the formal acid concentration falls into the 10^-8 to 10^-9 M range, the shortcut becomes physically misleading. The exact pH remains slightly below 7 because the solution is still acidic, but only weakly so.

What Is Really Happening Chemically?

At higher concentrations, HBr overwhelmingly controls the hydrogen ion concentration. Water’s own contribution is tiny in comparison and can safely be ignored. At 1.0 × 10^-8 M, however, the situation changes. Water contributes roughly 1.0 × 10^-7 M H⁺ in pure water, and the acid only nudges the equilibrium a little. The final [H⁺] is therefore just above 1.0 × 10^-7 M, giving a pH just below 7.

Another way to think about it is through charge balance. Every bromide ion introduced by HBr must be balanced by positive charge somewhere. The final hydrogen ion concentration increases enough to satisfy charge neutrality while still obeying the K_w relationship.

Common Errors Students Make

• Ignoring water autoionization: this is the number one mistake in very dilute acid and base problems.
• Assuming pH 8 for a strong acid: if your answer says a strong acid solution is basic, revisit your assumptions.
• Using the wrong K_w value: K_w depends on temperature, so 1.0 × 10^-14 is tied to 25 °C.
• Forgetting the quadratic solution: when concentration is close to 10^-7 M, the exact expression matters.

Practical Significance in Analytical Chemistry

Although this may seem like a classroom detail, the distinction matters in analytical chemistry, environmental chemistry, and any system involving ultra-dilute solutions. pH measurements near neutrality can be strongly influenced by dissolved gases, ionic strength, electrode calibration, and temperature. In real laboratory work, preparing and verifying a true 1.0 × 10^-8 M strong acid solution is more difficult than simply calculating it on paper. Carbon dioxide from the air can dissolve into water and form carbonic acid, shifting the measured pH enough to compete with such a tiny amount of added HBr.

Reference point	[H^+] (M)	pH	Interpretation
Pure water at 25 °C	1.0 × 10^-7	7.0000	Neutral benchmark
1.0 × 10^-8 M HBr, acid-only assumption	1.0 × 10^-8	8.0000	Incorrect because it predicts a basic solution
1.0 × 10^-8 M HBr, exact calculation	1.05125 × 10^-7	6.9783	Correct slightly acidic result

When Can You Ignore Water Autoionization?

A good rule of thumb is that if the concentration of a strong acid is at least 100 times larger than 1.0 × 10^-7 M, then the contribution from water is usually negligible for introductory calculations. That means around 1.0 × 10^-5 M and above, the shortcut often works reasonably well. Below that range, especially around 10^-7 to 10^-9 M, you should be cautious.

For 1.0 × 10^-8 M HBr, the exact equation is not optional. It is the chemically correct approach.

Summary Answer

To calculate the pH of a 1.0 × 10^-8 M HBr solution, do not simply use pH = 8. Because the acid is so dilute, you must include water autoionization. Using the quadratic expression

[H⁺] = (C + √(C² + 4K_w)) / 2

with C = 1.0 × 10^-8 M and K_w = 1.0 × 10^-14, you obtain:

[H⁺] ≈ 1.05125 × 10^-7 M

pH ≈ 6.9783

That final answer is slightly acidic, which is exactly what chemistry predicts for a very dilute strong acid solution.

Authoritative Sources for Further Reading

• LibreTexts Chemistry for detailed acid-base equilibrium explanations
• U.S. Environmental Protection Agency for pH fundamentals and water quality context
• U.S. Geological Survey for pH concepts in aqueous systems

Note: The numerical result shown by this calculator assumes ideal behavior at 25 °C with K_w = 1.0 × 10^-14. In real experimental systems, dissolved gases and non-ideal effects can shift observed pH slightly.