Autoionization of water and Kw
A Brønsted–Lowry acid is a proton (H⁺) donor and a base is a proton acceptor (2.12.1). Water can both give and accept a proton, so it is amphoteric两性, and protons transfer between water molecules:
At 25 °C, Kw = 1.0 × 10⁻¹⁴. Every aqueous solution contains both H₃O⁺ and OH⁻, and the product of their concentrations equals Kw (H₃O⁺ is often written simply as H⁺).
pH and pOH
At 25 °C, pH + pOH = 14.00. The pH scale is logarithmic: each decrease of 1 in pH multiplies [H₃O⁺] by 10, and a decrease of 2 multiplies it by 100. The number of decimal places in a pH equals the number of significant figures in [H₃O⁺] (2.12.3).
Drag the pH of two solutions and compare their [H⁺] and [OH⁻]:
Neutral doesn’t always mean pH 7
The autoionization of water is endothermic, so as the temperature rises, Kw increases and pKw decreases. A solution is neutral when [H₃O⁺] = [OH⁻], which gives pH = pKw ÷ 2. Only at 25 °C is that 7.00. Drag the temperature to see:
Kw data: measured values at 0, 25, 50, 75, and 100 °C (Bandura & Lvov, 2006); temperatures in between are interpolated linearly. For the acid and base solutions, the H⁺ or OH⁻ from water itself is ignored.
Example 1: pH from [OH⁻]
At 25 °C a solution has [OH⁻] = 2.5 × 10⁻⁴ M. Find the pOH, the pH and [H₃O⁺].
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Calculate the pOH. The pOH is the negative log of [OH⁻]:
Significant figures: 2.5 has two, so the pOH gets two decimal places.
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Calculate the pH. At 25 °C, pH + pOH = pKw = 14.00:
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Calculate [H₃O⁺]. Use the ion product of water, Kw = [H₃O⁺][OH⁻]:
From the pH instead: 10−10.40 = 4.0 × 10⁻¹¹ M, the same result.
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Classify. [OH⁻] = 2.5 × 10⁻⁴ > [H₃O⁺] = 4.0 × 10⁻¹¹, so the solution is basic (pH > 7 says the same thing, since this is 25 °C).
Example 2: a lower pH, yet basic
Two samples are measured: P at 25 °C has pH 6.60, and Q at 75 °C has pH 6.50 (at 75 °C, pKw = 12.70; data from Bandura & Lvov 2006). Classify each as acidic, neutral or basic.
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P (25 °C): find both ion concentrations (in M).
[H₃O⁺] > [OH⁻], so P is acidic.
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The neutral point for Q. A neutral solution has [H₃O⁺] = [OH⁻], so pH = pOH = pKw ÷ 2:
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Q (75 °C): find both ion concentrations (in M). Here pH + pOH = 12.70, not 14.00:
[OH⁻] > [H₃O⁺], so Q is basic; its pH of 6.50 is also above the neutral point of 6.35 at this temperature.
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Conclusion. Q has a lower pH and a higher [H₃O⁺] than P, yet it is basic. Away from 25 °C, compare [H₃O⁺] with [OH⁻] (or the pH with pKw ÷ 2), not the pH with 7.
- pH = 3.2
pH < 7.00.
- pOH = 4.0
pH = 10.0.
- [H₃O⁺] = 1.0 × 10⁻⁷ M
Equal to [OH⁻].
- [OH⁻] = 1.0 × 10⁻³ M
[OH⁻] > [H₃O⁺].
- [H₃O⁺] = 5.0 × 10⁻⁹ M
Smaller than 10⁻⁷; pH ≈ 8.30.
- pOH = 12.5
pH = 1.5.
- [H₃O⁺] = 100 × [OH⁻]
More H₃O⁺.
- pH = pOH
Equal amounts of the two ions.
Common mistakes
Mistake 1: “A neutral solution always has pH 7.” Neutral means [H₃O⁺] = [OH⁻]. Only at 25 °C is the neutral pH 7.00; at higher temperatures it is below 7.
Mistake 2: “A pH lower by 1 is twice as acidic.” pH is logarithmic: a pH lower by 1 means 10 times the [H₃O⁺]; lower by 3 means 1000 times.
Mistake 3: “pH can only be between 0 and 14.” Concentrated strong acids and bases can go beyond that range. For 2.0 M HCl, pH = −log 2.0 = −0.30.
Quick check
At 25 °C, what is the pH of a solution with [H₃O⁺] = 3.0 × 10⁻⁵ M?
pH = −log(3.0 × 10⁻⁵) = 4.52. C is the pOH.
Solution X has pH 2.0 and solution Y has pH 5.0. [H₃O⁺] in X is how many times that in Y?
A pH difference of 3 is a factor of 10³ = 1000 in [H₃O⁺], and X has the larger value.
At 0 °C, pKw = 14.95. A solution at this temperature has pH 7.20. It is:
At 0 °C the neutral pH is 14.95 ÷ 2 = 7.48. 7.20 is lower: [H₃O⁺] = 6.3 × 10⁻⁸ M and [OH⁻] = 1.8 × 10⁻⁸ M, so the solution is acidic even though its pH is above 7.
Which statement is correct?
The autoionization equilibrium is always present, so [H₃O⁺][OH⁻] = Kw.
At 25 °C a solution has pOH = 2.70. [H₃O⁺] is about:
pH = 14.00 − 2.70 = 11.30, so [H₃O⁺] = 10⁻¹¹·³⁰ = 5.0 × 10⁻¹² M. A is [OH⁻].
In the reaction NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, water acts as:
Water gives H⁺ to NH₃ and becomes OH⁻, so it is a Brønsted–Lowry acid. In its reaction with HCl, water is a base.
AP-style free response
At 25 °C, the pH values of three liquids are measured: rainwater 5.60, seawater 8.10 and gastric juice 1.50.
(a) Calculate [H₃O⁺] in each liquid.
(b) [H₃O⁺] in gastric juice is how many times that in rainwater?
(c) Calculate [OH⁻] in seawater.
(d) Classify each liquid as acidic or basic, and give your reasoning.
(e) Kw is 1.0 × 10⁻¹⁴ at 25 °C and 5.5 × 10⁻¹⁴ at 50 °C. Is the autoionization of water endothermic or exothermic? Explain using Le Châtelier’s principle.
Model answer and scoring
(a) Rainwater 10⁻⁵·⁶⁰ = 2.5 × 10⁻⁶ M; seawater 10⁻⁸·¹⁰ = 7.9 × 10⁻⁹ M; gastric juice 10⁻¹·⁵⁰ = 3.2 × 10⁻² M (each pH has two decimal places, so each concentration has two significant figures).
(b) The pH difference is 5.60 − 1.50 = 4.10, so the factor is 10⁴·¹⁰ ≈ 1.3 × 10⁴.
(c) pOH = 14.00 − 8.10 = 5.90, so [OH⁻] = 10⁻⁵·⁹⁰ ≈ 1.3 × 10⁻⁶ M.
(d) At 25 °C the dividing line is pH 7 (where [H₃O⁺] = [OH⁻]): rainwater and gastric juice have pH < 7 and are acidic; seawater has pH > 7 and is basic.
(e) Endothermic. Kw increases as the temperature rises, so the equilibrium 2H₂O ⇌ H₃O⁺ + OH⁻ shifts to the right. By Le Châtelier’s principle, raising the temperature shifts an equilibrium in the endothermic direction, so the forward reaction (ionization) is endothermic.
Scoring: (a) concentrations from pH; (b) factor from the pH difference; (c) pH + pOH = 14; (d) reasoning; (e) endothermic, deduced from Kw increasing with temperature and explained by the shift.
Summary
- Kw = [H₃O⁺][OH⁻], which is 1.0 × 10⁻¹⁴ at 25 °C; pH + pOH = pKw (14.00 at 25 °C).
- A difference of 1 in pH is a factor of 10 in [H₃O⁺].
- Neutral: [H₃O⁺] = [OH⁻] and pH = pKw ÷ 2. As the temperature rises, Kw increases and the neutral pH falls below 7.
AP tip: For pH, the significant figures are the decimal places. To compare two solutions, find the pH difference first, then write “[H₃O⁺] differs by a factor of 10(pH difference).” When a question isn’t at 25 °C, decide acidic or basic by comparing [H₃O⁺] with [OH⁻], not by checking whether the pH is below 7.