E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaThe Ionization of Water: Kw, pH and pOH
E3.5 E3.4 E2.4 A1.13About 70 minutes
Learning goals- I can write the equilibrium equation and the Kw expression for the autoionization of water.
- I can use Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C to find [H₃O⁺] or [OH⁻].
- I can convert among [H₃O⁺], [OH⁻], pH and pOH with correct significant digits.
- I can calculate the pH of strong acid and strong base solutions, including mixtures.
- I can explain why neutral water has a pH different from 7 at temperatures other than 25 °C.
Words to know
autoionization of water
The reaction of water with itself: 2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq).
ion product of water, Kw
The equilibrium constant Kw = [H₃O⁺][OH⁻]. At 25 °C, Kw = 1.0 × 10⁻¹⁴.
pH
pH = −log[H₃O⁺], the negative logarithm of the hydronium ion concentration in mol/L.
pOH
pOH = −log[OH⁻]. At 25 °C, pH + pOH = 14.00.
neutral
A solution in which [H₃O⁺] = [OH⁻]. Its pH is 7.00 only at 25 °C.
Pure water is not just H₂O
Very pure water conducts electricity, just barely. Something in it must be charged. The ions come from water itself: once in a while, one water molecule passes a proton to another.
2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
In Brønsted-Lowry terms, one water molecule is the acid and the other is the base (water is amphiprotic). This reaction is called the autoionization (or self-ionization) of water. It is a dynamic equilibrium that lies very far to the left, but it never stops, and it is present in every aqueous solution you will ever study.
Remember
From Grade 11: pH = −log[H₃O⁺], each pH unit is a factor of 10, and in pure water at 25 °C, [H₃O⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L. From the unit Chemical Equilibrium: pure liquids and solids are left out of an equilibrium constant expression because their concentrations do not change.
The ion product of water, Kw
Write the equilibrium constant expression for the autoionization. Liquid water is left out, so:
Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C
This is one of the most useful equations in the course. It applies to every aqueous solution at 25 °C, not just pure water:
- In pure water, [H₃O⁺] = [OH⁻] = √(1.0 × 10⁻¹⁴) = 1.0 × 10⁻⁷ mol/L.
- Add acid, and [H₃O⁺] rises. By Le Châtelier's principle the autoionization shifts left, and [OH⁻] falls so that the product stays 1.0 × 10⁻¹⁴.
- Add base, and [OH⁻] rises while [H₃O⁺] falls.
So if you know one of the two concentrations, Kw gives you the other: [OH⁻] = Kw ÷ [H₃O⁺] and [H₃O⁺] = Kw ÷ [OH⁻].
pOH and the 14 rule
Just as pH = −log[H₃O⁺], we define pOH = −log[OH⁻]. Taking −log of both sides of Kw = [H₃O⁺][OH⁻] gives:
pH + pOH = pKw = 14.00 (at 25 °C)
With these four quantities you can always move around the square shown below.
Strong acids and strong bases
Strong acids and strong bases ionize or dissociate completely, so their ion concentrations come straight from stoichiometry:
- Strong monoprotic acid (HCl, HBr, HI, HNO₃, HClO₄): [H₃O⁺] = c(acid).
- Group 1 hydroxide (NaOH, KOH, LiOH): [OH⁻] = c(base).
- Group 2 hydroxide (Ca(OH)₂, Sr(OH)₂, Ba(OH)₂): each formula unit gives two OH⁻, so [OH⁻] = 2 × c(base).
The tiny contribution from water's own autoionization (10⁻⁷ mol/L or less) is ignored unless the acid or base is extremely dilute (see the Challenge).
Significant digits
In a pH or pOH, only the digits after the decimal point are significant. The number of decimal places equals the number of significant digits in the concentration: [H₃O⁺] = 2.5 × 10⁻² mol/L (2 sig. digits) gives pH = 1.60 (2 decimal places).
Worked exampleExample 1: A strong acid
Find [H₃O⁺], pH, [OH⁻] and pOH for 0.025 mol/L HCl(aq) at 25 °C.
- HCl is a strong acid, so [H₃O⁺] = 0.025 mol/L = 2.5 × 10⁻² mol/L.
- pH = −log(2.5 × 10⁻²) = 1.602… → 1.60 (two decimal places for two significant digits).
- [OH⁻] = Kw ÷ [H₃O⁺] = (1.0 × 10⁻¹⁴) ÷ (2.5 × 10⁻²) = 4.0 × 10⁻¹³ mol/L.
- pOH = 14.00 − 1.60 = 12.40. Check: −log(4.0 × 10⁻¹³) = 12.40 ✓
[H₃O⁺] = 2.5 × 10⁻² mol/L, pH 1.60, [OH⁻] = 4.0 × 10⁻¹³ mol/L, pOH 12.40.
Worked exampleExample 2: A Group 2 hydroxide
Find the pH of 0.0050 mol/L Ba(OH)₂(aq) at 25 °C.
- Ba(OH)₂(s) → Ba²⁺(aq) + 2 OH⁻(aq), so [OH⁻] = 2 × 0.0050 = 0.010 mol/L.
- pOH = −log(0.010) = 2.00.
- pH = 14.00 − 2.00 = 12.00.
pH = 12.00. (Forgetting the factor of 2 would give the wrong answer, 11.70.)
Worked exampleExample 3: From pH back to concentrations
Human blood has a pH of about 7.40. Find [H₃O⁺] and [OH⁻], using Kw at 25 °C.
- [H₃O⁺] = 10⁻⁷·⁴⁰ = 3.98 × 10⁻⁸ → 4.0 × 10⁻⁸ mol/L (two decimal places in the pH gives two significant digits).
- pOH = 14.00 − 7.40 = 6.60.
- [OH⁻] = 10⁻⁶·⁶⁰ = 2.5 × 10⁻⁷ mol/L.
[H₃O⁺] = 4.0 × 10⁻⁸ mol/L, [OH⁻] = 2.5 × 10⁻⁷ mol/L. Blood is slightly basic. (Body temperature is 37 °C, where Kw is a little larger, so these values are approximate.)
Worked exampleExample 4: Mixing a strong acid and a strong base
25.0 mL of 0.100 mol/L HCl is mixed with 15.0 mL of 0.100 mol/L NaOH. Find the pH of the mixture.
- Moles of H₃O⁺: n = cV = 0.100 mol/L × 0.0250 L = 2.50 × 10⁻³ mol.
- Moles of OH⁻: 0.100 mol/L × 0.0150 L = 1.50 × 10⁻³ mol.
- They neutralize 1:1 (H₃O⁺ + OH⁻ → 2 H₂O). H₃O⁺ is in excess: 2.50 × 10⁻³ − 1.50 × 10⁻³ = 1.00 × 10⁻³ mol left.
- Total volume = 25.0 + 15.0 = 40.0 mL = 0.0400 L. [H₃O⁺] = 1.00 × 10⁻³ ÷ 0.0400 = 0.0250 mol/L.
- pH = −log(0.0250) = 1.602 → 1.602 (three significant digits, three decimal places).
pH = 1.602. Always use the total volume after mixing. You will use exactly this method for titration curves.
Kw changes with temperature
The autoionization of water is endothermic (breaking an O–H bond needs energy). By Le Châtelier's principle, heating shifts it to the right, so Kw increases with temperature, as the table shows.
At 50 °C, pure water has pH 6.63, but it is still neutral, because [H₃O⁺] still equals [OH⁻]. "Neutral" means equal concentrations; pH 7 is only the neutral point at 25 °C. In this course, assume 25 °C unless a question says otherwise.
| Temperature | Kw | pH of neutral water |
|---|
| 0 °C | 1.14 × 10⁻¹⁵ | 7.47 |
| 25 °C | 1.0 × 10⁻¹⁴ | 7.00 |
| 50 °C | 5.5 × 10⁻¹⁴ | 6.63 |
| 100 °C | 5.1 × 10⁻¹³ | 6.15 |
Just for funIn pure water at 25 °C, only about 1 water molecule in every 550 million has passed a proton to a neighbour at any moment. Water is a very reluctant acid, and an equally reluctant base.
Hands-on activityA tenfold dilution ladder
You need: red cabbage indicator (from the first lesson), lemon juice, 5 clear glasses, a tablespoon, tap water, a marker and tape for labels.
Grown-up note: an adult should help make the cabbage indicator (knife and hot water). Do not drink the mixtures.
- Label the glasses 1 to 5. Put 2 tablespoons of lemon juice in glass 1 (one tablespoon stays there and one moves on).
- Put 9 tablespoons of water in each of glasses 2 to 5.
- Move 1 tablespoon from glass 1 into glass 2 and stir. That is a tenfold dilution. Then move 1 tablespoon from glass 2 into glass 3, and so on to glass 5.
- Add 2 tablespoons of cabbage indicator to each glass and line them up on white paper.
- Predict: if lemon juice were a strong acid, each tenfold dilution would raise the pH by exactly 1 unit. Record the colours and compare them to the colour guide (red to pink in acids, purple near neutral).
What to look for: the colour moves from red toward pink and then purple as you dilute. The change is gradual, smaller than you might expect from five tenfold steps. That is because citric acid is a weak acid: when you dilute it, a larger fraction of its molecules ionize, so [H₃O⁺] falls by less than a factor of 10 each time. You will calculate exactly how much in the lesson on weak acid calculations.
Your turn
1Write the autoionization equation and the Kw expression for water.
2 H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq); Kw = [H₃O⁺][OH⁻] (water is left out because it is a pure liquid).
2Find [OH⁻] in a solution with [H₃O⁺] = 3.2 × 10⁻⁴ mol/L.
[OH⁻] = (1.0 × 10⁻¹⁴) ÷ (3.2 × 10⁻⁴) = 3.1 × 10⁻¹¹ mol/L.
3Find the pH and pOH of 0.0040 mol/L HNO₃.
[H₃O⁺] = 4.0 × 10⁻³ mol/L. pH = −log(4.0 × 10⁻³) = 2.40; pOH = 14.00 − 2.40 = 11.60.
4Find the pH of 0.15 mol/L KOH.
[OH⁻] = 0.15 mol/L. pOH = −log 0.15 = 0.82. pH = 14.00 − 0.82 = 13.18.
5Find the pH of 0.020 mol/L Ca(OH)₂ (assume it all dissolves).
[OH⁻] = 2 × 0.020 = 0.040 mol/L. pOH = 1.40. pH = 12.60.
6A solution has pOH 4.75. Find its pH, [H₃O⁺] and [OH⁻].
pH = 14.00 − 4.75 = 9.25. [H₃O⁺] = 10⁻⁹·²⁵ = 5.6 × 10⁻¹⁰ mol/L. [OH⁻] = 10⁻⁴·⁷⁵ = 1.8 × 10⁻⁵ mol/L.
More practice: warm-up
1What is the value of Kw at 25 °C?
3A solution has pH 3.00. Is it acidic, basic or neutral at 25 °C?
pOH = 14.00 − 8.30 = 5.70.
5Find the pH of 1.0 × 10⁻³ mol/L HCl.
[H₃O⁺] = 1.0 × 10⁻³ mol/L, so pH = 3.00.
More practice: on your own
1Find the pH after 2.00 g of NaOH is dissolved in enough water to make 500.0 mL of solution.
n = 2.00 g ÷ 40.00 g/mol = 0.0500 mol. [OH⁻] = 0.0500 ÷ 0.5000 L = 0.100 mol/L. pOH = 1.000, so pH = 13.000.
210.0 mL of 0.50 mol/L HCl is diluted to 250.0 mL. Find the new pH.
c₂ = c₁V₁/V₂ = 0.50 × 10.0 ÷ 250.0 = 0.020 mol/L. pH = −log 0.020 = 1.70.
330.0 mL of 0.200 mol/L NaOH is mixed with 20.0 mL of 0.200 mol/L HNO₃. Find the pH.
n(OH⁻) = 6.00 × 10⁻³ mol; n(H₃O⁺) = 4.00 × 10⁻³ mol. Excess OH⁻ = 2.00 × 10⁻³ mol in 0.0500 L = 0.0400 mol/L. pOH = 1.398, pH = 12.602.
4Rainwater has pH 5.6 and a sample of acid rain has pH 4.3. How many times greater is [H₃O⁺] in the acid rain?
The difference is 1.3 pH units, so the ratio is 10¹·³ ≈ 20 times greater.
5At 50 °C, Kw = 5.5 × 10⁻¹⁴. Find [H₃O⁺] and pH of pure water at 50 °C. Is it acidic?
[H₃O⁺] = √(5.5 × 10⁻¹⁴) = 2.3 × 10⁻⁷ mol/L; pH = 6.63. It is neutral, not acidic, because [H₃O⁺] = [OH⁻].
Challenge
1A student says the pH of 1.0 × 10⁻⁸ mol/L HCl is 8.00. Explain the error and find a better value. (Hint: let water contribute x of both H₃O⁺ and OH⁻; then (1.0 × 10⁻⁸ + x)(x) = 1.0 × 10⁻¹⁴.)
An acid cannot have a basic pH. The acid's H₃O⁺ is smaller than water's own 1.0 × 10⁻⁷, so water's contribution can't be ignored. Solve x² + (1.0 × 10⁻⁸)x − 1.0 × 10⁻¹⁴ = 0: x = 9.5 × 10⁻⁸. [H₃O⁺] = 1.0 × 10⁻⁸ + 9.5 × 10⁻⁸ = 1.05 × 10⁻⁷ mol/L, so pH ≈ 6.98, very slightly acidic.
2How many grams of Ba(OH)₂ (M = 171.35 g/mol) must dissolve in 2.00 L of water to give a pH of 12.50?
pOH = 1.50, so [OH⁻] = 10⁻¹·⁵⁰ = 0.0316 mol/L. [Ba(OH)₂] = 0.0316 ÷ 2 = 0.0158 mol/L. n = 0.0158 × 2.00 = 0.0316 mol. m = 0.0316 × 171.35 ≈ 5.42 g.
3Use Le Châtelier's principle and the Kw data in this lesson to explain whether the autoionization of water is endothermic or exothermic.
Kw grows from 1.14 × 10⁻¹⁵ at 0 °C to 5.1 × 10⁻¹³ at 100 °C, so heating shifts the equilibrium to the right. A system shifts right when heated only if the forward reaction absorbs heat, so the autoionization is endothermic (and neutralization, its reverse, is exothermic).
Quiz
1. What is [OH⁻] in a solution with [H₃O⁺] = 1.0 × 10⁻³ mol/L at 25 °C?
2. What is the pH of 0.010 mol/L Sr(OH)₂?
3. A solution has pH 5.25 at 25 °C. What is its pOH?
4. How many significant digits does the pH 3.47 represent in [H₃O⁺]?
5. Pure water at 60 °C has a pH of about 6.5. What is true about it?
6. Adding a strong acid to water causes [OH⁻] to
Big idea (E3.5, E3.4): water ionizes slightly, and in every aqueous solution at 25 °C, [H₃O⁺][OH⁻] = Kw = 1.0 × 10⁻¹⁴ and pH + pOH = 14.00. Knowing one of [H₃O⁺], [OH⁻], pH or pOH gives all the others.
How to help: Draw the four-box square from the lesson on a card and keep it beside your teen's calculator. Give a value ("pOH is 3.20") and ask for the other three. Check that they use the 10ˣ key (often SHIFT + log) to undo a logarithm.
Common difficulties:
- Forgetting the factor of 2 for Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂.
- Using the volume of only one solution after mixing instead of the total volume.
- Significant digits: in a pH, count only the digits after the decimal point.
- Thinking pH 7 always means neutral. That is only true at 25 °C.