pH of Rainwater: Why Clean Rain Is Slightly Acidic and When It Becomes Acid Rain
Is perfectly clean rainwater neutral, with pH 7? No. Even in an unpolluted place, rain is slightly acidic, because carbon dioxide from the air dissolves in it and forms a weak acid. “Acid rain” is a different matter: it is rain made much more acidic by strong acids such as sulphuric and nitric acid formed from pollutant gases. Understanding the difference needs only two ideas from your ionic equilibrium chapter: Henry’s law and the weak-acid approximation.
![Two bars compare clean rain at pH 5.61 and [H⁺] 2.48 × 10⁻⁶ M with polluted rain at pH 3.70 and [H⁺] 2.0 × 10⁻⁴ M, showing about 81 times more hydrogen ions.](https://4ab.in/chemistry-calculator/blog/img/ph-of-rainwater-and-acid-rain.webp)
Step 1: CO2 dissolves
By Henry’s law, the concentration of dissolved gas is proportional to its partial pressure:
Here KH is Henry’s constant (mol L−1 atm−1) and P is the partial pressure in atm. The dissolved CO2 is written as H2CO3 in most school problems.
Step 2: the weak acid ionises
Since [H+] = [HCO3−] = x, we get x² = Ka1 × [H2CO3], so [H+] = √(Ka1 × [H2CO3]) and pH = −log[H+]. The second ionisation of carbonic acid is so weak that it is ignored here.
The constants below are typical textbook-style values for practice (25 °C): KH = 3.4 × 10−2 mol L−1 atm−1, Ka1 = 4.3 × 10−7, and CO2 partial pressure 4.2 × 10−4 atm. In an exam, always use the numbers printed in the question.
Worked examples
Example 1: pH of clean rainwater.
[H2CO3] = 3.4 × 10−2 × 4.2 × 10−4 = 1.428 × 10−5 M.
[H+] = √(4.3 × 10−7 × 1.428 × 10−5) = √(6.14 × 10−12) = 2.48 × 10−6 M.
pH = −log(2.48 × 10−6) = 6 − 0.394 = 5.61.
So unpolluted rain has pH about 5.6, which is why 5.6 is often quoted as the “natural” value, and why rain is called acid rain only when it is well below this. Check the approximation: 2.48 × 10−6 M is far larger than the 10−7 M from water itself (about 25 times), so ignoring water’s ionisation is reasonable here.
Example 2: rain polluted with a strong acid. Rainwater contains 1.0 × 10−4 M of H2SO4 (treat it as fully ionised in both steps for a simple estimate). Find the pH.
[H+] = 2 × 1.0 × 10−4 = 2.0 × 10−4 M.
pH = −log(2.0 × 10−4) = 4 − 0.301 = 3.70.
The CO2 contribution (about 2.5 × 10−6 M) is only about 1% of this, so it can be neglected.
Example 3: how much more acidic? Compare Example 2 with clean rain.
pH difference = 5.61 − 3.70 = 1.91 units. Hydrogen-ion ratio = 101.91 = about 81 times. Check using concentrations: 2.0 × 10−4 ÷ 2.48 × 10−6 = 80.6. ✓
A difference of about 2 pH units means roughly a hundred-fold change in [H+], because pH is a logarithmic scale.
Example 4: how much alkali neutralises it? A lake receives 1.00 × 106 L of the polluted rain from Example 2. How many moles of H+ does it bring, and what mass of CaCO3 neutralises it?
Moles of H+ = 2.0 × 10−4 mol L−1 × 1.00 × 106 L = 200 mol.
CaCO3 + 2H+ → Ca2+ + H2O + CO2, so moles of CaCO3 = 200 / 2 = 100 mol.
Molar mass of CaCO3 = 40.078 + 12.011 + 3(15.999) = 100.09 g mol−1. Mass = 100 × 100.09 = 10 009 g ≈ 10.0 kg.
Common mistakes
- Writing pH = 7 for clean rain. Dissolved CO2 makes it about 5.6.
- Calling any rain below pH 7 “acid rain”. The term is used for rain that is more acidic than the natural level.
- Forgetting the square root: [H+] = √(KaC), not KaC.
- Using the total CO2 pressure of 1 atm. The partial pressure of CO2 in air is a very small fraction of an atmosphere.
- Treating a pH change of 2 as “twice as acidic”. It is a hundred-fold change in [H+].
- Forgetting that H2SO4 gives two H+ per molecule in a simple strong-acid estimate.
Exam-relevance table
| Question | Method |
|---|---|
| pH of clean rain | Henry’s law → weak acid [H+] = √(KaC) |
| pH with a strong acid present | [H+] = n × C(acid); ignore CO2 if far smaller |
| Compare two pH values | Ratio = 10ΔpH |
| Neutralisation of a water body | Moles H+ → stoichiometry → mass |
Environmental chemistry chapters differ between boards; check your own syllabus for how much of this is expected.
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