Worked example 1
A solution has [OH⁻] = 2.5 × 10⁻⁴ M at 25 C. Find [H₃O⁺], pH and pOH.
Try it first: Start from Kw rather than from a pH formula.
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Acids and BasesCourse level: High School / HonorsCourse level: AP Chemistry
Autoionization of Water, pH and pOH
What you'll be able to do: Convert freely among [H₃O⁺], [OH⁻], pH and pOH and classify a solution as acidic, neutral or basic.
Introduction
Water itself is very slightly ionized, and that tiny equilibrium sets the whole pH scale. Once you have Kw, every acid-base number in this unit connects to every other one.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Write the autoionization equilibrium and the expression for Kw
- Convert between [H₃O⁺], [OH⁻], pH and pOH
- Classify solutions as acidic, neutral or basic
- Apply significant-figure rules to logarithmic quantities
Water ionizes itself
In any sample of pure water a very small fraction of molecules transfer a proton to a neighbour: 2 H₂O ⇌ H₃O⁺ + OH⁻.
At 25 C this gives [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ M, which is about one ionized molecule in every 500 million.
Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 C
Kw links the two ions
Kw is an equilibrium constant, so it holds in every aqueous solution, not just pure water. Add acid and [H₃O⁺] rises, which forces [OH⁻] down so the product stays 1.0 × 10⁻¹⁴. Neither ion is ever truly zero.
The rigorous equilibrium relation is Kw = a(H₃O⁺)a(OH⁻). The concentration product [H₃O⁺][OH⁻] approximates it in dilute solutions where activity coefficients are near one. The numerical 1.0 × 10⁻¹⁴ applies at 25 °C under this approximation. It is not the value of Kw at every temperature or a promise of ideal behavior in concentrated acid.
If you know one ion concentration, divide Kw by it to get the other.
The p-function
Because these concentrations span many orders of magnitude, chemists take negative logarithms. pH = -log[H₃O⁺], pOH = -log[OH⁻], pKw = 14.00. Taking -log of the Kw expression gives pH + pOH = 14.00 at 25 C.
| Known quantity | Next calculation |
|---|---|
| [H₃O⁺] | pH = −log₁₀[H₃O⁺] |
| pH | [H₃O⁺] ≈ 10−pH M in the dilute model |
| [OH⁻] | pOH = −log₁₀[OH⁻], then pH = pKw − pOH |
| One ion concentration | Other ion ≈ Kw divided by the known concentration |
For pH 4.25, [H₃O⁺] ≈ 5.6 × 10⁻⁵ M. At 25 °C, [OH⁻] ≈ 1.8 × 10⁻¹⁰ M and pOH = 9.75. Check that the logarithmic values sum to 14.00 and the concentrations multiply to Kw. A minus sign belongs in the exponent when inverting pH.
pH + pOH = pKw = 14.00 at 25 C
Reading the scale
At 25 C, pH < 7 is acidic, pH = 7 is neutral and pH > 7 is basic. Each pH unit is a factor of ten in [H₃O⁺], so a solution at pH 3 is one hundred times more acidic than one at pH 5. pH values below 0 and above 14 are perfectly possible for concentrated solutions.
A pH difference ΔpH corresponds to a hydronium activity ratio of 10−ΔpH. Comparing pH 3.0 with 5.0 therefore gives 100 times greater hydronium activity for the lower-pH sample, not twice as much. That statement concerns the current solution's acidity, not whether its molecular acid has the larger Ka. High-concentration solutions can lie outside pH 0–14, but direct concentration-based calculations become less reliable there.
Temperature matters
Autoionization is endothermic, so heating water increases Kw. At 60 C, Kw is about 9.6 × 10⁻¹⁴ and neutral water has a pH near 6.5. The water is still neutral because the two ion concentrations are still equal.
Neutral is defined by [H₃O⁺] = [OH⁻], not by the number 7.
Significant figures in logs
In a pH the digits before the decimal point only encode the power of ten. [H₃O⁺] = 4.2 × 10⁻⁵ M has two significant figures, so pH = 4.38 is written with two decimal places.
For logarithms, concentration significant figures become pH decimal places. Thus 4.2 × 10⁻⁵ M gives pH 4.38; conversely pH 5.72 gives 1.9 × 10⁻⁶ M. Keep guard digits during Kw division and logarithms, then round at the end. With temperature-specific pKw = 13.54, neutral pH is 6.77, and pH 7.00 is basic. Neutrality means equal hydronium and hydroxide activities in the idealized model, not a temperature-independent numeral on the meter.
Compare pH with the neutral point at that temperature
Neutral means [H₃O⁺] = [OH⁻], not always pH 7. At a different temperature Kw and pKw change, so neutral pH becomes pKw/2. A solution can have pH 7 and be basic if the neutral pH at that temperature is below 7.
Use the temperature-specific Kw before applying pH+pOH=pKw.
The neutral benchmark is pKw/2.
Key ideas
Definition
Autoionization
Proton transfer between two water molecules producing H₃O⁺ and OH⁻.
Rule
Kw
Kw is the product of ion activities; in the dilute concentration approximation [H₃O⁺][OH⁻] ≈ 1.0 × 10⁻¹⁴ at 25 °C. Use the temperature-specific value.
Rule
pH
pH = -log[H₃O⁺]; [H₃O⁺] = 10-pH.
Rule
pH + pOH
Equals 14.00 at 25 C only.
Rule
Neutrality
[H₃O⁺] = [OH⁻]; the pH of neutrality changes with temperature.
Equation
Ion product of water
Kw = [H₃O⁺][OH⁻]
- Kw = 1.0 × 10⁻¹⁴ at 25 C
Equation
pH definition
pH = -log[H₃O⁺]
- [H₃O⁺] = hydronium concentration in mol/L
Equation
Inverse of pH
[H₃O⁺] = 10-pH
- pH = measured or calculated pH
Worked examples
Worked example 2
What is [H₃O⁺] in a solution of pH 5.72?
Try it first: Undo the logarithm.
0 of 3 steps revealed.
Worked example 3
At 50 C, Kw = 5.5 × 10⁻¹⁴. What is the pH of pure water, and is it acidic?
Try it first: In pure water the two ion concentrations are still equal.
0 of 3 steps revealed.
Worked example 4
At a temperature where pKw=13.54, neutral pH is 6.77. Is a solution of pH 7.00 acidic, basic, or neutral?
Try it first: Compare [H₃O⁺] with the neutral concentration, or pH with 6.77.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Assuming pH 7 is always neutral. | Kw changes with temperature, so neutral pH is 7.00 only at 25 C. | Compare [H₃O⁺] with [OH⁻]. |
| Treating pH as a linear scale. | Each unit is a factor of ten in concentration. | pH 2 is 1000 times more acidic than pH 5. |
| Reporting pH 4.3812 from a concentration with two significant figures. | Only decimal places count in a log. | Match decimal places to significant figures. |
| Saying [OH⁻] is zero in an acidic solution. | Kw forces both ions to be present at all times. | Divide Kw by [H₃O⁺]. |
| Using pH + pOH = 14 at other temperatures. | That sum equals pKw, which is 14.00 only at 25 C. | Recompute pKw from the given Kw. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
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What you should now know
Water autoionizes: 2 H₂O ⇌ H₃O⁺ + OH⁻, with Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 C. Because the product is fixed, knowing either ion concentration gives the other. pH = -log[H₃O⁺] and pOH = -log[OH⁻], so pH + pOH = 14.00 at 25 C. A solution is acidic when [H₃O⁺] > [OH⁻] (pH < 7 at 25 C), neutral when they are equal, and basic when [OH⁻] is larger. Neutral means equal ion concentrations, not pH exactly 7: autoionization is endothermic, so at higher temperature Kw rises and neutral pH falls below 7. In pH values only the digits after the decimal point are significant.
- Water autoionizes and Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 C
- Knowing one ion concentration always gives the other
- pH = -log[H₃O⁺] and pH + pOH = 14.00 at 25 C
- Neutral means equal ion concentrations, not pH 7
- Only decimal places are significant in a pH value
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