ICE Tables

Every quantitative equilibrium problem on the MCAT follows the same framework. You set up a table, plug into the K expression, and solve for an unknown. That framework is the ICE table - and once you master it, every equilibrium calculation becomes a fill-in-the-blank exercise.

What ICE Stands For

Setting Up the Table

Step 1: Write the balanced equation and the K expression.

Step 2: Create the ICE table with one column per species.

Step 3: Fill in the Initial row from the problem.

Step 4: Fill in the Change row using stoichiometric ratios. The species that increases gets +x (scaled by its coefficient), and the species that decreases gets -x (scaled by its coefficient). To determine which direction the reaction proceeds, compare Q (the reaction quotient calculated from initial concentrations) to K: if Q < K, the reaction goes forward (products increase); if Q > K, it goes backward (reactants increase).

Step 5: Add I + C to get the Equilibrium row.

Step 6: Substitute the Equilibrium expressions into the K equation and solve for x.

Worked Example 1: Starting From Reactants Only

Consider: H₂(g) + I₂(g) ⇌ 2HI(g), K_c = 50.0 at 448°C.

A flask initially contains 0.100 M H₂ and 0.100 M I₂ with no HI. Find the equilibrium concentrations.

Step 1: K_c = [HI]² / ([H₂][I₂]) = 50.0

Step 2-5: Build the ICE table. Since Q = 0 and K = 50, Q < K, so the reaction goes forward (products increase, reactants decrease).

Step 6: Plug into K:

50.0 = (2x)² / ((0.100 - x)(0.100 - x))

50.0 = (2x)² / (0.100 - x)²

Take the square root of both sides (a shortcut when both sides are perfect squares):

√50.0 = 2x / (0.100 − x)

7.07 = 2x / (0.100 - x)

7.07(0.100 - x) = 2x

0.707 = 2x + 7.07x = 9.07x

x = 0.0780

Equilibrium concentrations:

• [H₂] = 0.100 - 0.078 = 0.022 M
• [I₂] = 0.100 - 0.078 = 0.022 M
• [HI] = 2(0.078) = 0.156 M

Check: K = (0.156)² / (0.022)(0.022) = 0.0243 / 0.000484 = 50.2 (close to 50.0 - rounding accounts for the difference).

Worked Example 2: The Small-x Approximation

Consider: N₂O₄(g) ⇌ 2NO₂(g), $K_c = 4.6 \times 10^{-3}$

A flask initially contains 0.500 M N₂O₄. Find equilibrium concentrations.

$K_c = (2x)^2 / (0.500 - x) = 4x^2 / (0.500 - x)$

This leads to a quadratic, but K is very small ($10^{-3}$) compared to the initial concentration (0.500). This tells us that very little N₂O₄ will decompose, so x will be very small compared to 0.500.

The approximation: Assume x << 0.500, so 0.500 - x ≈ 0.500.

$4x^2 / 0.500 = 4.6 \times 10^{-3}$

$4x^2 = 2.3 \times 10^{-3}$

$x^2 = 5.75 \times 10^{-4}$

x = 0.024

Check the approximation: x / 0.500 = 0.024 / 0.500 = 4.8%. Since this is less than 5%, the approximation is valid.

When Can You Use the Small-x Approximation?

The approximation works when K is much smaller than the initial concentration:

ICE Tables for Ksp

ICE tables work for solubility equilibria too. The only difference: the solid does not get a column (its activity is 1).

Molar solubility is the number of moles of a salt that dissolve per liter of solution. We label it “s” in ICE tables. If s moles of PbI₂ dissolve, then s moles of Pb²⁺ appear in solution - and 2s moles of I⁻ appear (because each formula unit releases 2 iodide ions).

Example: Find the molar solubility of PbI₂ ($K_{sp} = 9.8 \times 10^{-9}$).

PbI₂(s) ⇌ Pb²+(aq) + 2I-(aq)

$K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 = (s)(2s)^2 = 4s^3$

$9.8 \times 10^{-9} = 4s^3$

$s^3 = 2.45 \times 10^{-9}$

$s = \sqrt[3]{2.45 \times 10^{-9}} \approx 1.35 \times 10^{-3}$ M

The molar solubility is $1.35 \times 10^{-3}$ M. This means $[\text{Pb}^{2+}] = 1.35 \times 10^{-3}$ M and $[\text{I}^-] = 2.70 \times 10^{-3}$ M at equilibrium.

ICE Tables With a Common Ion

When solving for solubility in a solution that already contains a common ion, the initial concentration of that ion is NOT zero.

Example: Find the molar solubility of PbI₂ in 0.10 M NaI solution.

$K_{sp} = (s)(0.10 + 2s)^2 = 9.8 \times 10^{-9}$

Since $K_{sp}$ is tiny and the common ion concentration is 0.10, s will be extremely small. Approximate: 0.10 + 2s ≈ 0.10.

$(s)(0.10)^2 = 9.8 \times 10^{-9}$

$s(0.01) = 9.8 \times 10^{-9}$

$s = 9.8 \times 10^{-7}$ M

Compare this to $1.35 \times 10^{-3}$ M in pure water - the common ion reduced solubility by over 1,000-fold.

Common ICE Table Mistakes

1. Forgetting stoichiometric ratios. If the coefficient is 2, the change is 2x, not x.
2. Wrong sign on the Change row. If the reaction goes forward, reactants decrease (-x) and products increase (+x).
3. Approximating when K is not small enough. Always check the 5% rule after solving.
4. Including a solid in the table. Solids do not appear in ICE tables or K expressions.
5. Forgetting the common ion. If a common ion is already present, its initial concentration is NOT zero.