Calculate the pH of 6.7 × 10-8 M HCl
Use this premium calculator to solve the pH of a very dilute hydrochloric acid solution correctly, including the effect of water autoionization when needed.
Expert Guide: How to Calculate the pH of 6.7 10 8 M HCl
When students search for how to “calculate the pH of 6.7 10 8 m hcl wolfgram,” they are usually trying to evaluate the pH of a very dilute hydrochloric acid solution written more clearly as 6.7 × 10-8 M HCl. This looks simple at first glance because HCl is a strong acid, and strong acids are often taught to dissociate completely. In an introductory problem, you might immediately say that the hydrogen ion concentration equals the acid concentration, so [H+] = 6.7 × 10-8 M. If you then apply pH = -log[H+], you would get a pH above 7. That result is physically suspicious for an acid solution.
The reason is subtle but important. At extremely low acid concentrations, the contribution of hydrogen ions from water itself can no longer be ignored. Pure water at 25°C already contains about 1.0 × 10-7 M H+ and 1.0 × 10-7 M OH– because of water autoionization. Since 6.7 × 10-8 M is smaller than 1.0 × 10-7 M, the water contribution matters. That is why the simple shortcut fails and the exact equilibrium treatment is preferred.
Quick Answer
For a 6.7 × 10-8 M HCl solution at 25°C, the exact pH is approximately 6.86. This is acidic, but only slightly acidic. It is not 7.17, even though that is what the naive strong-acid-only shortcut suggests.
Why the Simple Method Gives the Wrong Intuition
For ordinary strong acid problems, we use the assumption:
For this problem, that would mean:
Then:
But a pH of 7.17 would imply a slightly basic solution, which cannot be correct after adding hydrochloric acid. The issue is that the approximation silently ignores the 1.0 × 10-7 M hydrogen ion concentration coming from water. That assumption works well for concentrated acids like 10-3 M HCl or 10-2 M HCl, but not for 10-8 M scale acid solutions.
The Correct Exact Method
Let the formal concentration of HCl be C = 6.7 × 10-8 M. Because HCl is a strong acid, it contributes essentially that amount to the hydrogen ion balance. However, water also contributes a small additional amount. Let x be the concentration of H+ and OH– created from water autoionization beyond the acid source.
Then:
[OH-] = x
K_w = [H+][OH-] = 1.0 × 10^-14
Substitute the expressions into the Kw relationship:
This gives a quadratic equation:
Using the quadratic formula:
But since total hydrogen ion concentration is C + x, we can write it directly as:
Now insert the numbers:
Evaluating this yields:
Now compute pH:
Step-by-Step Summary
- Recognize that HCl is a strong acid and dissociates essentially completely.
- Notice the concentration is only 6.7 × 10-8 M, which is close to the 10-7 M hydrogen ion level of pure water.
- Reject the shortcut [H+] = C if it produces an impossible chemical interpretation.
- Use the water equilibrium condition Kw = [H+][OH–].
- Solve the quadratic expression for the total hydrogen ion concentration.
- Take the negative logarithm to obtain pH.
Comparison: Approximate vs Exact Result
| Method | Assumption | Calculated [H+] | Calculated pH | Chemical Interpretation |
|---|---|---|---|---|
| Simple strong-acid approximation | [H+] = 6.7 × 10-8 M only | 6.7 × 10-8 M | 7.17 | Implies basic behavior, which is not physically reasonable for HCl solution |
| Exact method with water autoionization | [H+] includes both acid and water contribution | 1.3896 × 10-7 M | 6.86 | Correctly predicts a slightly acidic solution |
How Close Is This to Neutral Water?
Neutral water at 25°C has pH 7.00. A solution of 6.7 × 10-8 M HCl has pH about 6.86, so it is only mildly more acidic than pure water. This makes intuitive sense because the added strong acid concentration is very tiny.
| Solution | [H+] at 25°C | pH | Comment |
|---|---|---|---|
| Pure water | 1.0 × 10-7 M | 7.00 | Reference neutral point at 25°C |
| 6.7 × 10-8 M HCl, exact treatment | 1.3896 × 10-7 M | 6.86 | Slightly acidic because acid adds to water background |
| 1.0 × 10-6 M HCl | Approximately 1.1 × 10-6 M | About 5.96 | Now acid strongly dominates over water |
| 1.0 × 10-3 M HCl | Approximately 1.0 × 10-3 M | 3.00 | Water contribution becomes negligible |
Common Mistakes Students Make
- Ignoring water autoionization: This is the biggest mistake in very dilute acid or base calculations.
- Assuming every strong acid problem is trivial: Strong dissociation is only one part of the chemistry. Background water equilibrium can still matter.
- Thinking pH above 7 is acceptable for an acid solution: If your answer suggests that adding HCl makes water basic, revisit the setup.
- Forgetting that pH depends on temperature through Kw: The common value 1.0 × 10-14 is for 25°C.
When Can You Safely Ignore Water?
A useful practical rule is this: if the acid concentration is much larger than 1.0 × 10-7 M, then the contribution of water is usually negligible. For instance, with 10-4 M HCl, the water contribution is tiny relative to the acid contribution, so the simple strong-acid formula works extremely well. With 10-8 M HCl, however, the acid concentration is actually smaller than the hydrogen ion level in pure water, so ignoring water is not justified.
Why This Problem Appears Often in Online Solvers
This kind of problem is popular because it tests whether you understand the limitations of pH shortcuts. It often appears in textbook homework, tutoring sites, and computational tools such as Wolfram-style query engines. The keyword phrase “wolfgram” is usually a misspelling of “Wolfram,” but the chemistry challenge remains the same: calculate the true pH of an ultra-dilute strong acid solution.
Physical Interpretation of the Result
The exact pH of 6.86 tells us that the solution is just a little more acidic than pure water. That may seem strange at first because HCl is a strong acid, but strength and concentration are different ideas. HCl is strong because it dissociates completely. The solution is only weakly acidic because the amount added is extremely small. In other words, complete dissociation of a very tiny quantity still produces only a very tiny shift away from neutrality.
Useful Formula for Very Dilute Strong Acids
For a dilute strong monoprotic acid with formal concentration C at 25°C, a convenient exact expression is:
This formula is especially useful when C is around 10-8 to 10-6 M, where the transition from water-dominated to acid-dominated behavior occurs. Once C becomes much larger than 10-7 M, the result approaches [H+] ≈ C.
Authoritative Chemistry References
For deeper background on pH, acid-base chemistry, and water equilibrium, consult these reliable academic and government resources:
- LibreTexts Chemistry for detailed educational explanations from academic contributors.
- U.S. Environmental Protection Agency (.gov) pH overview for practical interpretation of pH scales.
- U.S. Geological Survey (.gov) pH and Water for accessible scientific context.
- University of California, Berkeley Chemistry (.edu) for university-level chemistry resources.
Final Takeaway
If you need to calculate the pH of 6.7 × 10-8 M HCl, the exact answer is about 6.86, not 7.17. The difference comes from the fact that pure water already contains hydrogen ions at about 1.0 × 10-7 M. Whenever the acid concentration becomes comparable to that value, water autoionization must be included. This is the correct expert-level interpretation and the result your calculator above computes automatically.