How to Use ICE Tables

ICE tables are bookkeeping for equilibria

An equilibrium problem asks: given starting concentrations and a K, what are the final concentrations? An ICE table is the bookkeeping system that turns that question into a single algebraic equation. Three rows — Initial, Change, Equilibrium — track every species through the reaction. Once the table is filled, you substitute into the K expression, solve for x, and read off the answers.

The table works for every flavor of equilibrium you’ll meet: weak acids, weak bases, Ksp solubility, gas-phase Kc/Kp, complex ion formation, and common-ion problems. The setup is identical; only the K expression changes.

How to set one up

1. Write the balanced equation and the K expression.

Take 0.100 M acetic acid, Ka = 1.8 × 10⁻⁵.

CH₃COOH ⇌ H⁺ + CH₃COO⁻, Ka = [H⁺][CH₃COO⁻]/[CH₃COOH].

2. Fill in the Initial row. Concentrations before any reaction. For a weak acid in water, [H⁺] and [A⁻] start at 0 (technically 10⁻⁷ from water, but that’s negligible unless your acid is very weak or very dilute).

3. Fill in the Change row. Use stoichiometry. For every x mol/L of HA that dissociates, 1x of H⁺ and 1x of A⁻ appear. Reactants get −, products get +. If a coefficient isn’t 1, the change reflects it (more on this below).

4. Add to get the Equilibrium row.

	CH₃COOH	H⁺	CH₃COO⁻
I	0.100	0	0
C	−x	+x	+x
E	0.100 − x	x	x

5. Substitute and solve. Ka = x · x / (0.100 − x) = 1.8 × 10⁻⁵.

The small-x approximation

If Ka is small relative to your initial concentration (rule of thumb: Ka/C₀ < 10⁻³, or equivalently x < 5% of C₀), you can drop x from the denominator: 0.100 − x ≈ 0.100. Then x² = (1.8 × 10⁻⁵)(0.100) = 1.8 × 10⁻⁶, so x = 1.34 × 10⁻³.

Check: 1.34 × 10⁻³ / 0.100 = 1.34%, well under 5%. Approximation valid. pH = −log(1.34 × 10⁻³) = 2.87.

When the approximation fails

For 0.0010 M HF with Ka = 6.8 × 10⁻⁴: Ka/C₀ = 0.68, way above the cutoff. You have to solve the quadratic.

x²/(0.0010 − x) = 6.8 × 10⁻⁴ → x² + 6.8 × 10⁻⁴ x − 6.8 × 10⁻⁷ = 0.

Quadratic formula: x = [−6.8 × 10⁻⁴ + √(4.62 × 10⁻⁷ + 2.72 × 10⁻⁶)] / 2 = 5.52 × 10⁻⁴ M.

pH = 3.26.

If you’d used the approximation: x = √(Ka × C₀) = 8.25 × 10⁻⁴, which is 82.5% of the initial concentration — clearly nonsense, the acid would be more than 80% dissociated.

Worked examples

Weak base. 0.250 M NH₃, Kb = 1.8 × 10⁻⁵.

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. ICE setup is identical to the weak acid case. Kb = x²/(0.250 − x) ≈ x²/0.250 = 1.8 × 10⁻⁵. x = 2.12 × 10⁻³ M = [OH⁻]. Check: 0.85% < 5%, valid. pOH = 2.67, pH = 11.33.

Gas-phase with a coefficient. N₂O₄ ⇌ 2 NO₂, Kc = 4.63 × 10⁻³ at 25 °C. Start with 0.500 mol N₂O₄ in 2.00 L → [N₂O₄]₀ = 0.250 M.

	N₂O₄	NO₂
I	0.250	0
C	−x	+2x
E	0.250 − x	2x

The +2x is the part students forget. The coefficient of NO₂ is 2, so its change is twice as large as the change in N₂O₄.

Kc = (2x)²/(0.250 − x) = 4x²/(0.250 − x) = 4.63 × 10⁻³. Approximate: x = 0.0170, but that’s 6.8% — over the threshold. Solve exactly: 4x² + 4.63 × 10⁻³ x − 1.158 × 10⁻³ = 0 → x = 0.01644. [N₂O₄] = 0.234 M, [NO₂] = 0.0329 M.

Common-ion effect. 0.100 M acetic acid + 0.200 M sodium acetate, Ka = 1.8 × 10⁻⁵.

	CH₃COOH	H⁺	CH₃COO⁻
I	0.100	0	0.200
C	−x	+x	+x
E	0.100 − x	x	0.200 + x

The acetate doesn’t start at 0 — it starts at 0.200 M from the salt. Ka ≈ x(0.200)/0.100 = 1.8 × 10⁻⁵ → x = 9.0 × 10⁻⁶. pH = 5.05.

Compare to the pure 0.100 M acid (pH 2.87): the common ion suppresses dissociation by Le Chatelier’s principle. This is exactly the buffer calculation, derived from first principles instead of jumping to Henderson-Hasselbalch.

Traps to watch for

Forgetting stoichiometric coefficients in the C row. If a product has a coefficient of 2, its change is 2x and its squared term in K is (2x)² = 4x². Skipping the 4 is a classic factor-of-4 error.

Blindly applying the approximation. Always do the 5% check at the end. If x/C₀ > 5%, the approximation lied to you and you have to solve the quadratic. Modern calculators handle quadratics in seconds — there’s no excuse to leave a 30% error in your answer.

Initial concentrations of zero when they’re not. Common-ion problems, problems where you mix two solutions, problems with a salt already present — make sure your I row reflects what’s actually there.

Solid and liquid concentrations. Pure solids and pure liquids have activity = 1 and don’t appear in K (or in the ICE table). For Ksp problems with PbCl₂(s) ⇌ Pb²⁺ + 2 Cl⁻, the solid isn’t in the table; only the dissolved ions are.

Wrong K direction. If you write the equation reversed, Kreverse = 1/Kforward. Don’t grab a Ka from a table and use it as if it described the reverse reaction.

Practice problems (use the Equilibrium Calculator to check):

1. pH of 0.500 M HNO₂ (Ka = 4.5 × 10⁻⁴)?
2. Equilibrium concentrations for H₂ + I₂ ⇌ 2 HI, Kc = 54.3, with [H₂]₀ = [I₂]₀ = 0.500 M?
3. pH of 0.150 M pyridine (Kb = 1.7 × 10⁻⁹)?
4. Molar solubility of PbCl₂ (Ksp = 1.7 × 10⁻⁵)?
5. pH of 0.300 M NH₃ + 0.200 M NH₄Cl (Kb = 1.8 × 10⁻⁵)?