E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaWeak Acid Calculations: pH, Ka and Percentage Ionization
E2.4 E3.7 E3.4 A1.13About 75 minutes
Learning goals- I can use an ICE table and Ka to calculate [H₃O⁺], pH and percentage ionization of a weak acid.
- I can decide when the "x is small" approximation is valid, and solve the quadratic when it is not.
- I can calculate Ka from the initial concentration and measured pH of a weak acid.
- I can explain, with calculations, why dilution increases the percentage ionization of a weak acid.
Words to know
ICE table
A table of Initial, Change and Equilibrium concentrations used to solve equilibrium problems.
small-x approximation
Treating C − x as C when x (the amount that ionizes) is tiny compared with C. Valid when the percentage ionization is 5% or less.
5% rule
The test for the small-x approximation: if x ÷ C × 100% ≤ 5%, the approximation is acceptable.
quadratic formula
x = [−b ± √(b² − 4ac)] ÷ 2a, used to solve x² + Ka·x − Ka·C = 0 when the approximation fails (take the positive root).
How acidic is the vinegar in your cupboard?
The label on white vinegar in Canada usually says 5% acetic acid. Acetic acid is weak, so you can't just take −log of its concentration. In this lesson you will predict the pH of vinegar from its label using Ka, and then work backwards: measure a pH and find Ka.
The tool is the ICE table you used in the unit Chemical Equilibrium (ICE Tables II: Finding Equilibrium Concentrations from K and ICE Tables III: The Small-K Approximation). Weak-acid problems are the most common kind of equilibrium calculation in chemistry, and you will meet them again in first-year university courses in chemistry, biology and environmental science.
The setup
For a weak acid HA with initial concentration C:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
Let x be the amount of HA that ionizes (in mol/L). Each HA that ionizes forms one H₃O⁺ and one A⁻. (The H₃O⁺ from water's own ionization, 10⁻⁷ mol/L or less, is ignored.)
| [HA] | [H₃O⁺] | [A⁻] |
|---|
| Initial | C | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | C − x | x | x |
Substitute into Ka:
Ka = x² ÷ (C − x)
The small-x approximation
This is the small-K approximation and 5 % rule from ICE Tables III: The Small-K Approximation, now applied to Ka. Weak acids usually ionize only a few percent, so x is tiny compared with C, and C − x ≈ C. Then:
Ka ≈ x² ÷ C, so x = √(Ka × C)
After solving, always check that the percentage ionization (x ÷ C × 100%) is 5% or less. If it is, the error from the approximation is small enough to ignore at the precision of the Ka values. If not, go back and solve the full quadratic:
x² + Ka·x − Ka·C = 0, so x = [−Ka + √(Ka² + 4·Ka·C)] ÷ 2 (only the positive root makes sense).
A quick pre-check: if C ÷ Ka ≥ 500, the approximation will always pass the 5% test.
Worked exampleExample 1: pH of 0.10 mol/L acetic acid
Find [H₃O⁺], pH and percentage ionization of 0.10 mol/L CH₃COOH(aq). Ka = 1.8 × 10⁻⁵.
- Equation: CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻. ICE: equilibrium concentrations are 0.10 − x, x, x.
- Ka = x² ÷ (0.10 − x) = 1.8 × 10⁻⁵.
- Pre-check: C ÷ Ka = 0.10 ÷ 1.8 × 10⁻⁵ ≈ 5600, which is ≥ 500, so assume 0.10 − x ≈ 0.10.
- x² = 1.8 × 10⁻⁵ × 0.10 = 1.8 × 10⁻⁶, so x = 1.34 × 10⁻³ mol/L.
- Check: 1.34 × 10⁻³ ÷ 0.10 × 100% = 1.3% ≤ 5% ✓
- pH = −log(1.34 × 10⁻³) = 2.87.
[H₃O⁺] = 1.3 × 10⁻³ mol/L, pH = 2.87, 1.3% ionized.
Worked exampleExample 2: When the approximation fails (HF)
Find the pH of 0.10 mol/L HF(aq). Ka = 6.6 × 10⁻⁴.
- Ka = x² ÷ (0.10 − x) = 6.6 × 10⁻⁴.
- Pre-check: C ÷ Ka = 0.10 ÷ 6.6 × 10⁻⁴ ≈ 150, which is less than 500. Try the approximation anyway: x = √(6.6 × 10⁻⁴ × 0.10) = 8.1 × 10⁻³, which is 8.1% of C. That fails the 5% test.
- Solve the quadratic: x² + (6.6 × 10⁻⁴)x − 6.6 × 10⁻⁵ = 0.
- x = [−6.6 × 10⁻⁴ + √((6.6 × 10⁻⁴)² + 4(6.6 × 10⁻⁵))] ÷ 2 = [−6.6 × 10⁻⁴ + √(2.644 × 10⁻⁴)] ÷ 2 = [−6.6 × 10⁻⁴ + 1.626 × 10⁻²] ÷ 2 = 7.80 × 10⁻³ mol/L.
- pH = −log(7.80 × 10⁻³) = 2.11. Percentage ionization = 7.8%.
pH = 2.11 (the approximation would have given 2.09, a small but real error).
Worked exampleExample 3: Finding Ka from a measured pH
A 0.20 mol/L solution of an unknown monoprotic weak acid has a pH of 2.62. Find Ka and the percentage ionization.
- [H₃O⁺] = 10⁻²·⁶² = 2.40 × 10⁻³ mol/L. This is x.
- At equilibrium: [H₃O⁺] = [A⁻] = 2.40 × 10⁻³ mol/L and [HA] = 0.20 − 0.0024 = 0.1976 mol/L.
- Ka = (2.40 × 10⁻³)² ÷ 0.1976 = 2.9 × 10⁻⁵.
- Percentage ionization = 2.40 × 10⁻³ ÷ 0.20 × 100% = 1.2%.
Ka = 2.9 × 10⁻⁵, 1.2% ionized. (Here you do not need an approximation: all three equilibrium concentrations are known.)
Worked exampleExample 4: The pH of vinegar from its label
White vinegar is 5.0% acetic acid by mass-volume (5.0 g per 100 mL). Estimate its pH. Ka = 1.8 × 10⁻⁵; M(CH₃COOH) = 60.05 g/mol.
- Concentration: 5.0 g per 100 mL is 50 g per litre. 50 g ÷ 60.05 g/mol = 0.83 mol, so C = 0.83 mol/L.
- C ÷ Ka ≈ 46 000, so the approximation is safe: x = √(1.8 × 10⁻⁵ × 0.83) = 3.9 × 10⁻³ mol/L.
- Check: 3.9 × 10⁻³ ÷ 0.83 × 100% = 0.47% ✓
- pH = −log(3.9 × 10⁻³) = 2.41.
Vinegar's pH is about 2.4. Measured values for 5% vinegar are usually close to this.
Dilution and percentage ionization
The table below uses the full quadratic for acetic acid at four concentrations. As the acid is diluted tenfold, [H₃O⁺] falls by only about a factor of 3 (√10 ≈ 3.2), so the pH rises by only about 0.5 per tenfold dilution, not 1. The percentage ionization climbs, just as Le Châtelier's principle predicts.
| C (mol/L) | [H₃O⁺] (mol/L) | pH | % ionization |
|---|
| 1.0 | 4.2 × 10⁻³ | 2.37 | 0.42% |
| 0.10 | 1.3 × 10⁻³ | 2.88 | 1.3% |
| 0.010 | 4.2 × 10⁻⁴ | 3.38 | 4.2% |
| 0.0010 | 1.3 × 10⁻⁴ | 3.90 | 12.5% |
Just for funThe "p" in pH was introduced by the Danish chemist Søren Sørensen in 1909, and historians still argue about what he meant by it. "Power" and the German "Potenz" are popular guesses. Whatever it stands for, it saves chemists from writing a lot of negative exponents.
Hands-on activityPredict, then test, the pH of kitchen acids
You need: white vinegar (check the % on the label), lemon juice, a calculator, red cabbage indicator from the first lesson, and pH test strips if you have them (sold at pharmacies, pool supply and garden stores).
Grown-up note: help with the cabbage indicator (knife and hot water). Do not drink the samples.
- Use Example 4 to predict the pH of your vinegar from its label percentage.
- Lemon juice contains about 5 to 6% citric acid (M = 192.12 g/mol). Treat citric acid as a monoprotic acid with Ka1 = 7.4 × 10⁻⁴ and a concentration of 0.29 mol/L. Is the small-x approximation valid? Calculate the pH.
- Test each liquid with a pH strip (or with cabbage indicator, which will only show "strongly acidic"). Record the results.
- Make a tenfold dilution of the vinegar (1 tablespoon in 9 tablespoons of water). Predict the new pH with the approximation, then test it.
What to look for: vinegar ≈ 2.4 (prediction) and about 2.4 to 3 on a strip. For lemon juice, C ÷ Ka ≈ 390, below 500, and the approximation gives x = 0.0146 (5.0%, right at the limit). The quadratic gives x = 0.0143 mol/L and pH ≈ 1.8, while measured lemon juice is usually about 2 to 2.6, because other substances in the juice affect the result. The diluted vinegar should be about 0.5 pH unit higher (about 2.9), not a full unit. Pause here and think: why is a real food sample harder to predict than a lab solution?
Your turn
1Find the pH of 0.050 mol/L lactic acid (Ka = 1.4 × 10⁻⁴). Check the approximation.
Approximation: x = √(1.4 × 10⁻⁴ × 0.050) = 2.65 × 10⁻³, which is 5.3%: fails. Quadratic: x² + 1.4 × 10⁻⁴x − 7.0 × 10⁻⁶ = 0 gives x = 2.58 × 10⁻³ mol/L, pH = 2.59.
2Find the pH of 0.050 mol/L HClO (Ka = 2.9 × 10⁻⁸).
C ÷ Ka is huge, so x = √(2.9 × 10⁻⁸ × 0.050) = 3.8 × 10⁻⁵ mol/L. pH = 4.42.
3Find the pH and percentage ionization of 0.25 mol/L formic acid, HCOOH (Ka = 1.8 × 10⁻⁴).
C ÷ Ka ≈ 1400. x = √(1.8 × 10⁻⁴ × 0.25) = 6.7 × 10⁻³ mol/L. pH = 2.17; ionization = 6.7 × 10⁻³ ÷ 0.25 = 2.7% ✓.
4A 0.150 mol/L solution of a weak acid has pH 2.85. Find Ka.
[H₃O⁺] = 10⁻²·⁸⁵ = 1.41 × 10⁻³. Ka = (1.41 × 10⁻³)² ÷ (0.150 − 0.00141) = 1.3 × 10⁻⁵.
5Find the pH of 0.50 mol/L HCN (Ka = 6.2 × 10⁻¹⁰).
x = √(6.2 × 10⁻¹⁰ × 0.50) = 1.76 × 10⁻⁵ mol/L. pH = 4.75.
6What concentration of acetic acid has a pH of 3.00?
x = 1.0 × 10⁻³. Ka = x² ÷ (C − x), so C − x = (1.0 × 10⁻³)² ÷ 1.8 × 10⁻⁵ = 0.056, and C = 0.056 + 0.001 = 0.057 mol/L.
More practice: warm-up
1In an ICE table for a weak acid, what does x represent?
The concentration of acid that ionizes, which equals [H₃O⁺] (and [A⁻]) at equilibrium.
2Write Ka in terms of x and C.
3What is the maximum percentage ionization for which the small-x approximation is accepted?
4If [H₃O⁺] = 1.0 × 10⁻³ mol/L in a 0.010 mol/L acid, what is the percentage ionization?
5Find x = √(Ka × C) for Ka = 1.0 × 10⁻⁶ and C = 0.010 mol/L.
x = √(1.0 × 10⁻⁸) = 1.0 × 10⁻⁴ mol/L.
More practice: on your own
1Find the pH of 0.020 mol/L benzoic acid (Ka = 6.3 × 10⁻⁵). Is the approximation valid?
Approximation: x = √(1.26 × 10⁻⁶) = 1.12 × 10⁻³ = 5.6% of C: not valid. Quadratic: x = [−6.3 × 10⁻⁵ + √((6.3 × 10⁻⁵)² + 4(6.3 × 10⁻⁵)(0.020))] ÷ 2 = 1.09 × 10⁻³ mol/L, pH = 2.96.
2Find the pH of 0.0050 mol/L HF (Ka = 6.6 × 10⁻⁴).
C ÷ Ka ≈ 7.6, far below 500. Quadratic: x² + 6.6 × 10⁻⁴x − 3.3 × 10⁻⁶ = 0, x = 1.52 × 10⁻³ mol/L, pH = 2.82 (30% ionized; the approximation would give 2.74).
3Aspirin (acetylsalicylic acid) is a weak monoprotic acid. A 0.010 mol/L solution has pH 2.76. Find Ka.
[H₃O⁺] = 10⁻²·⁷⁶ = 1.74 × 10⁻³. Ka = (1.74 × 10⁻³)² ÷ (0.010 − 0.00174) = 3.03 × 10⁻⁶ ÷ 0.00826 = 3.7 × 10⁻⁴.
4Find the pH of 0.10 mol/L H₃PO₄ using only Ka1 = 7.1 × 10⁻³.
C ÷ Ka ≈ 14, so use the quadratic: x² + 7.1 × 10⁻³x − 7.1 × 10⁻⁴ = 0, x = 2.33 × 10⁻² mol/L, pH = 1.63.
5A student dilutes 0.10 mol/L acetic acid to 0.0010 mol/L and says the pH rises from 2.87 to 4.87. What is the correct new pH? Explain the error.
Using the quadratic for 0.0010 mol/L: x = 1.25 × 10⁻⁴ mol/L, pH = 3.90. The student treated acetic acid like a strong acid. For a weak acid, a 100-fold dilution raises pH by only about 1 unit, because more of the acid ionizes as it is diluted.
Challenge
1Show that for a weak acid where the approximation holds, pH = ½(pKa − log C). Use it to find the pH of 0.10 mol/L acetic acid (pKa = 4.74).
x² = Ka·C, so −log x = ½(−log Ka − log C), which is pH = ½(pKa − log C). For acetic acid: ½(4.74 − (−1)) = ½(5.74) = 2.87 ✓.
2Two acids have the same concentration, 0.10 mol/L. Acid A has pH 2.00 and acid B has pH 4.00. Find Ka for each and the ratio of their Ka values.
A: x = 0.010, Ka = (0.010)² ÷ (0.10 − 0.010) = 1.1 × 10⁻³. B: x = 1.0 × 10⁻⁴, Ka = (1.0 × 10⁻⁴)² ÷ (0.10 − 0.0001) = 1.0 × 10⁻⁷. Ratio ≈ 1.1 × 10⁴: a 2-unit pH difference at the same concentration corresponds to about a 10 000-fold difference in Ka (when the approximation holds).
3What volume of 5.0% vinegar (0.83 mol/L acetic acid) must be diluted to 1.00 L to make a solution with pH 3.00?
From the Your turn question, pH 3.00 needs C = 0.057 mol/L. V₁ = c₂V₂ ÷ c₁ = 0.057 × 1.00 ÷ 0.83 = 0.068 L, about 68 mL.
Quiz
1. For a weak acid with Ka = 4.0 × 10⁻⁶ at C = 0.10 mol/L, what is [H₃O⁺] using the approximation?
2. When must you solve the quadratic instead of using the small-x approximation?
3. A 0.10 mol/L weak acid has pH 3.00. What is Ka?
4. Diluting a weak acid tenfold raises its pH by about
5. In the ICE table for 0.20 mol/L HA, what is the equilibrium [HA]?
6. Which acid is most likely to need the quadratic formula at 0.10 mol/L?
Big idea (E2.4, E3.7): the pH of a weak acid comes from an ICE table and Ka. When less than 5% ionizes, x = √(Ka × C); otherwise solve the quadratic. Working backwards from a measured pH gives Ka.
How to help: Ask your teen to talk you through the flowchart for one problem: "What's x? Can we approximate? How do you know?" Insist on the 5% check every time; it is the step students skip most often. The kitchen activity compares calculated and measured pH values; discuss why real foods differ from lab solutions.
Common difficulties:
- Treating a weak acid as strong (pH = −log C).
- Skipping the 5% check, or calculating it with the wrong numbers (it is x ÷ C, not Ka ÷ C).
- Choosing the negative root of the quadratic.
- Rounding [H₃O⁺] early and then reporting a pH that doesn't follow; keep extra digits until the end.