Calculate The Ph Of 2.8X10-8 M Hi Solution

Use this premium calculator to find the pH of an extremely dilute hydroiodic acid solution. Because 2.8×10^-8 M is close to the hydrogen ion level contributed by pure water, the correct calculation must include water autoionization rather than relying on the simple pH = -log[H⁺] shortcut alone.

pH Calculator

For very dilute strong acids like 2.8×10^-8 M HI, water contributes a significant amount of H+ on its own. This calculator uses the quadratic expression: [H+] = (C + sqrt(C² + 4Kw)) / 2

Quick Concept Summary

• HI is a strong acid and is treated as fully dissociated in dilute aqueous solution.
• At ordinary classroom concentrations, pH is often estimated using pH = -log(C).
• At 2.8×10^-8 M, that shortcut predicts pH 7.55, which is chemically impossible for an acid solution.
• Pure water already contains about 1.0×10^-7 M H+ and 1.0×10^-7 M OH- at 25 degrees C.
• The proper equilibrium treatment gives a slightly acidic answer, close to pH 6.94.

Exact setup
Let the acid concentration be C. If water contributes x M additional H+, then:
[H+] = C + x
[OH-] = x
Kw = (C + x)(x)
Therefore, total [H+] = (C + sqrt(C² + 4Kw)) / 2

How to Calculate the pH of 2.8×10^-8 M HI Solution Correctly

If you are trying to calculate the pH of 2.8×10^-8 M HI solution, you are dealing with one of the most common “advanced intro chemistry” traps. Hydroiodic acid, HI, is a strong acid. In many ordinary problems, students are taught that a strong acid dissociates completely, so the hydrogen ion concentration is equal to the acid concentration. Using that simple shortcut, you might write [H⁺] = 2.8×10^-8 M and then compute pH = -log(2.8×10^-8) = 7.55. But that answer says the acid solution is basic, which cannot be right.

The reason the shortcut fails is that the solution is extremely dilute. At 25 degrees C, pure water already self-ionizes slightly to produce H⁺ and OH^–, each at about 1.0×10^-7 M. Since 2.8×10^-8 M is smaller than 1.0×10^-7 M, the contribution from water is no longer negligible. In other words, the acid is not the only source of hydrogen ions in the solution. To get the real pH, you must account for water autoionization through the equilibrium constant K_w.

Step 1: Recognize HI as a Strong Acid

HI belongs to the family of strong hydrohalic acids and is treated as fully dissociated in water under standard educational conditions. That means the analytical concentration of HI, which we can call C, is contributed directly as hydrogen ions:

C = 2.8×10^-8 M

If the acid were much more concentrated, [H⁺] would be essentially equal to C. However, because the acid concentration here is so low, the solution still contains hydrogen ions from water itself.

Step 2: Include Water Autoionization

At 25 degrees C, water obeys:

K_w = [H⁺][OH^–] = 1.0×10^-14

Let x be the concentration of OH^– produced from water autoionization. The matching amount of hydrogen ion from water is also x. Since the acid already contributes C hydrogen ions, the total hydrogen ion concentration becomes:

[H⁺] = C + x

and

[OH^–] = x

Applying K_w gives:

(C + x)(x) = 1.0×10^-14

Rearranging:

x² + Cx – K_w = 0

The physically meaningful solution is:

x = (-C + sqrt(C² + 4K_w)) / 2

Because total [H⁺] = C + x, you can combine the expression neatly into:

[H⁺] = (C + sqrt(C² + 4K_w)) / 2

Step 3: Plug in the Numbers

Substitute C = 2.8×10^-8 and K_w = 1.0×10^-14:

[H⁺] = (2.8×10^-8 + sqrt((2.8×10^-8)² + 4(1.0×10^-14))) / 2

First calculate the square term:

(2.8×10^-8)² = 7.84×10^-16

Then:

4K_w = 4.0×10^-14

Add them:

7.84×10^-16 + 4.0×10^-14 = 4.0784×10^-14

Square root:

sqrt(4.0784×10^-14) ≈ 2.0195×10^-7

Now evaluate total [H⁺]:

[H⁺] ≈ (2.8×10^-8 + 2.0195×10^-7) / 2 ≈ 1.1497×10^-7 M

Finally:

pH = -log(1.1497×10^-7) ≈ 6.94

Final answer: the pH of 2.8×10^-8 M HI solution at 25 degrees C is approximately 6.94, not 7.55.

Why the Shortcut Gives the Wrong Answer

The familiar shortcut pH = -log(C) works well only when the acid concentration is large enough that water’s own ionization is insignificant in comparison. For example, for 1.0×10^-3 M HI, the water contribution of 1.0×10^-7 M is tiny relative to the acid concentration and can safely be ignored. But at 2.8×10^-8 M, that no longer holds. The water contribution is actually larger than the acid concentration. Whenever the acid concentration approaches 10^-7 M, you should pause and consider K_w.

A quick chemistry sanity check also helps. Adding any strong acid to pure water should make the solution at least slightly more acidic than pure water, not more basic. Since pure water at 25 degrees C has pH 7.00, the final pH must be below 7.00. That immediately flags 7.55 as unphysical in this context.

Comparison Table: Shortcut Versus Exact Method

Method	Assumption	Calculated [H+]	Calculated pH	Comment
Simple strong-acid shortcut	[H+] = 2.8×10^-8 M only from HI	2.8×10^-8 M	7.55	Incorrect because it ignores water autoionization
Exact equilibrium method	[H+] includes HI and water, with Kw = 1.0×10^-14	1.1497×10^-7 M	6.94	Correct at 25 degrees C

What the Numbers Mean Chemically

The exact hydrogen ion concentration, 1.1497×10^-7 M, is only slightly above the 1.0×10^-7 M found in pure water. That is why the pH shifts only a little below 7. The acid does matter, but not enough to dominate the chemistry in the dramatic way that a more concentrated acid would. This is an excellent example of why chemistry calculations should always be interpreted, not just performed mechanically.

You can also infer the hydroxide concentration from K_w:

[OH^–] = K_w / [H⁺] ≈ 1.0×10^-14 / 1.1497×10^-7 ≈ 8.70×10^-8 M

Then:

pOH ≈ 7.06

and pH + pOH ≈ 14.00 at 25 degrees C, as expected.

When Should You Use the Exact Formula?

You should consider the exact formula whenever a strong acid or strong base concentration is near 10^-7 M at 25 degrees C. This includes many homework problems designed to test conceptual understanding. In practical laboratory chemistry, such dilute solutions may also be affected by contamination, dissolved carbon dioxide, and measurement limits of glass electrodes. But for textbook equilibrium work, the K_w-based model is the correct approach.

Useful Rule of Thumb

• If acid concentration is much greater than 1.0×10^-6 M, the shortcut usually works well.
• If acid concentration is around 1.0×10^-7 M to 1.0×10^-8 M, use the exact quadratic treatment.
• If a shortcut gives an acidic solution a pH above 7, stop and reconsider the assumptions.

Reference Data Table for Dilute Strong Acids

Strong acid concentration (M)	Naive pH from -log(C)	Approximate exact pH at 25 degrees C	Interpretation
1.0×10^-3	3.00	3.00	Water contribution negligible
1.0×10^-6	6.00	6.00	Still close, exact treatment only slightly different
1.0×10^-7	7.00	6.79	Major deviation begins
2.8×10^-8	7.55	6.94	Acidic, but only slightly below neutral
1.0×10^-8	8.00	6.98	Naive result becomes clearly impossible

Common Mistakes Students Make

1. Ignoring K_w completely. This is the most common mistake and leads to a basic pH for an acidic solution.
2. Forgetting the physical meaning of pH. Any acid added to pure water should lower the pH below 7 at 25 degrees C.
3. Using the wrong root of the quadratic. Only the positive, chemically meaningful concentration should be used.
4. Confusing concentration from the acid with total concentration. The total hydrogen ion concentration includes the acid and water contributions.
5. Rounding too early. For extremely dilute systems, premature rounding can noticeably affect the final pH.

Authority Sources for Further Reading

If you want to verify the chemistry framework behind this calculation, review reliable educational and government resources on pH, water ionization, and acid-base equilibria:

• Chemistry LibreTexts educational resource
• National Institute of Standards and Technology (NIST)
• United States Environmental Protection Agency on pH fundamentals
• University of Washington Chemistry resources

Final Takeaway

To calculate the pH of 2.8×10^-8 M HI solution, you must go beyond the simple strong-acid shortcut. Because the concentration is so low, water autoionization contributes significantly to the hydrogen ion balance. The correct total hydrogen ion concentration is found from the quadratic expression derived from K_w, and the resulting pH is about 6.94. This example is a great reminder that chemistry formulas are not isolated tricks. They are models built on assumptions, and when those assumptions break down, the method must be upgraded.

Use the calculator above anytime you need a fast, accurate answer for a very dilute strong acid problem. It is especially useful for exam review, tutoring, and checking whether your hand calculation is physically reasonable.