Calculate the pH of a 1.9×10-7 M Solution of HNO3
This premium calculator solves an important dilute strong-acid problem correctly by including both nitric acid dissociation and the autoionization of water. For very low acid concentrations, the simple shortcut pH = -log[H+] from acid alone is not accurate enough.
Interactive pH Calculator
Enter or keep the default value of 1.9×10^-7 M HNO3, then click Calculate pH.
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How to Calculate the pH of a 1.9×10-7 M Solution of HNO3
Calculating the pH of a 1.9×10-7 M solution of HNO3 looks easy at first glance. Since nitric acid is a strong acid, many students immediately write [H+] = 1.9×10-7 M and then compute pH with the common formula pH = -log[H+]. If you do that, you get a pH near 6.72. However, that answer is incomplete because the acid concentration is so low that the water itself is contributing a similar amount of hydrogen ions. In pure water at 25 °C, the hydrogen ion concentration is already 1.0×10-7 M. That means the acid contribution and the water contribution are on the same order of magnitude.
This is exactly the kind of dilute-acid situation where chemistry students need to move beyond the simple shortcut and use the equilibrium of water. The result is still acidic, but the pH is not as low as the naive strong-acid approximation predicts. For this specific problem, the exact answer at 25 °C is approximately pH = 6.54, not 6.72.
Step 1: Recognize that HNO3 is a strong acid
Nitric acid is classified as a strong acid in water, so it dissociates essentially completely:
HNO3 → H+ + NO3–
If the concentration were much larger, for example 1.0×10-3 M or 1.0×10-2 M, then treating the hydrogen ion concentration as equal to the acid concentration would be an excellent approximation. In those cases, the hydrogen ions from water are negligible relative to the ions from the acid. But here the formal acid concentration is only 1.9×10-7 M, which is very close to pure water’s own hydrogen ion concentration.
Step 2: Include water autoionization
Water self-ionizes according to:
H2O ⇌ H+ + OH–
At 25 °C, the ionic product of water is:
Kw = [H+][OH–] = 1.0×10-14
In pure water, [H+] = [OH–] = 1.0×10-7 M. Once we add a very dilute strong acid, the acid raises [H+], and the hydroxide concentration must decrease so that the product remains equal to Kw.
Step 3: Set up the exact equation
Let the formal concentration of HNO3 be C = 1.9×10-7 M. Since nitric acid dissociates completely, it contributes C mol/L of nitrate ions. Charge balance requires:
[H+] = [OH–] + C
At the same time, water equilibrium requires:
[H+][OH–] = Kw
Substitute [OH–] = Kw / [H+] into the charge balance:
[H+] = Kw / [H+] + C
Multiply through by [H+]:
[H+]2 – C[H+] – Kw = 0
This is a quadratic equation in [H+], and the physically meaningful solution is:
[H+] = (C + √(C2 + 4Kw)) / 2
Step 4: Substitute the numbers
Using C = 1.9×10-7 M and Kw = 1.0×10-14:
- C2 = (1.9×10-7)2 = 3.61×10-14
- 4Kw = 4.0×10-14
- C2 + 4Kw = 7.61×10-14
- √(7.61×10-14) ≈ 2.7586×10-7
- [H+] = (1.9×10-7 + 2.7586×10-7) / 2 ≈ 2.3293×10-7 M
Now compute pH:
pH = -log(2.3293×10-7) ≈ 6.63?
Be careful here. Rechecking the square root with full precision gives:
√(7.61×10-14) = 2.7586228448×10-7
So:
[H+] = 2.3293114224×10-7 M
and therefore:
pH = -log(2.3293114224×10-7) = 6.6328
That is the exact result for the equation as written using the formal concentration approach and standard Kw. It shows why the answer is more acidic than pure water but less acidic than many learners expect when they reason informally.
Why some textbooks and answer keys differ
You may have seen slightly different answers to problems like this, often because of rounding, notation, or whether the author uses the concentration of hydronium, H3O+, interchangeably with H+. Another source of variation is temperature. Since Kw changes with temperature, the pH of pure water is not always exactly 7.00, and the pH of a very dilute acid will shift with temperature as well. At 20 °C, Kw is lower; at 30 °C, it is higher. For a dilute strong acid near 10-7 M, these differences are noticeable.
The shortcut method versus the correct method
The shortcut says:
[H+] ≈ C = 1.9×10-7 M
Then:
pH = -log(1.9×10-7) ≈ 6.721
This method ignores water’s own contribution of hydrogen ions. Because the acid concentration is only about 1.9 times the hydrogen ion concentration of pure water, that omission is too large to ignore. The exact treatment gives a lower pH, around 6.63, because the total hydrogen ion concentration is the combined result of the acid and water equilibrium.
| Method | Assumed [H+] (M) | Calculated pH | Comment |
|---|---|---|---|
| Naive strong-acid shortcut | 1.9×10-7 | 6.721 | Ignores water autoionization |
| Exact dilute-acid treatment | 2.329×10-7 | 6.633 | Correct at 25 °C with Kw = 1.0×10-14 |
| Pure water reference | 1.0×10-7 | 7.000 | No added acid |
How to know when water autoionization matters
A good rule of thumb is this: if the acid concentration is close to 10-7 M or the base concentration is close to 10-7 M, then you should stop and think about water autoionization. In concentrated or even moderately dilute acid solutions, the water contribution is negligible. But near the 10-7 M scale, it becomes a meaningful part of the total.
- If C is much greater than 10-6 M, the shortcut usually works well.
- If C is around 10-7 M to 10-8 M, use the exact equation.
- If C is much less than 10-7 M, the solution pH approaches that of pure water.
Temperature and Kw data
Because Kw depends on temperature, the hydrogen ion concentration of pure water also changes with temperature. This is a real physical effect, not an error. Neutral water has pH 7 only at 25 °C. At different temperatures, neutral pH shifts.
| Temperature | Kw | [H+] in pure water (M) | Neutral pH |
|---|---|---|---|
| 20 °C | 1.15×10-14 | 1.073×10-7 | 6.97 |
| 25 °C | 1.00×10-14 | 1.000×10-7 | 7.00 |
| 30 °C | 1.47×10-14 | 1.212×10-7 | 6.92 |
Worked interpretation of the final answer
The exact result tells us the solution is acidic, but only slightly acidic. A pH around 6.63 means the solution has more hydrogen ions than pure water, yet it is still far less acidic than laboratory stock nitric acid or even a 10-3 M acid solution. This is a useful reminder that pH is logarithmic. Small numerical changes in pH correspond to significant relative changes in hydrogen ion concentration.
For learners, the central insight is not merely the number itself, but the logic behind it:
- HNO3 is a strong acid, so it dissociates completely.
- At very low concentration, water autoionization cannot be neglected.
- Use charge balance together with Kw to get the actual [H+].
- Then calculate pH from the total hydrogen ion concentration.
Common mistakes students make
- Using pH = -log(1.9×10-7) directly. This ignores the background hydrogen ion concentration of water.
- Adding 1.0×10-7 and 1.9×10-7 by simple arithmetic without equilibrium. Water equilibrium shifts when acid is added, so the hydroxide concentration changes.
- Assuming neutral water is always pH 7. That is only exactly true at 25 °C.
- Forgetting that pH depends on total [H+], not just the amount introduced by the acid.
Authoritative references for pH and water equilibrium
For deeper study, consult reliable educational and government resources on acid-base chemistry, pH, and water ionization:
Bottom line
To calculate the pH of a 1.9×10-7 M solution of HNO3 correctly, you must account for both the strong acid and the autoionization of water. At 25 °C, the exact calculation gives a hydrogen ion concentration of about 2.329×10-7 M, leading to a final answer of pH ≈ 6.633. The interactive calculator above automates this process and also lets you see how the result shifts with temperature assumptions for Kw.