ICE Table Solver

The simulator models a single dissociation, $\text{A}(aq) \rightleftharpoons \text{B}(aq) + \text{C}(aq)$, with $K = \frac{[\text{B}][\text{C}]}{[\text{A}]}$. This is the same shape as a weak acid ionizing, $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$, so you can read A as the acid and B and C as $\text{H}^+$ and $\text{A}^-$.

A good routine: set both sliders to the values in your problem, do steps 1 to 3 on paper, then compare your answer with the simulator. Next, raise K until step 3 flips to invalid and notice how far the approximate and exact values of $x$ drift apart.

The sliders also show a pattern worth remembering. Keep K at 1.78e-05 and lower [A]₀ from 0.50 M to 0.10 M to 0.01 M. The percentage in step 3 climbs from about 0.6% to 1.3% to about 4%. A more dilute weak acid ionizes to a larger fraction, even though [H⁺] itself gets smaller. With [A]₀ = 0.10 M, the approximation stops being valid once K passes roughly $2.5 \times 10^{-4}$, which is where $\sqrt{K[\text{A}]_0}$ reaches 5% of 0.10.

Substituting the E row into the equilibrium expression gives $$K = \frac{x \cdot x}{[\text{A}]_0 - x} = \frac{x^2}{[\text{A}]_0 - x}$$

If K is small compared with [A]₀, very little A dissociates, so $[\text{A}]_0 - x \approx [\text{A}]_0$ and $$x \approx \sqrt{K[\text{A}]_0}$$

Always check the assumption afterward: $\frac{x}{[\text{A}]_0} \times 100\% < 5\%$. If the percentage is 5% or more, rearrange to the quadratic $x^2 + Kx - K[\text{A}]_0 = 0$ and keep the positive root, $$x = \frac{-K + \sqrt{K^2 + 4K[\text{A}]_0}}{2}$$

For a weak acid, $x = [\text{H}^+]$, so $\text{pH} = -\log[\text{H}^+]$. Only aqueous species and gases appear in K; pure solids and pure liquids, including water as the solvent, are left out.

Before filling in the C row, decide which direction the reaction shifts. Here only A is present at the start, so the reaction quotient Q is 0, which is less than K, and the reaction must proceed forward: A decreases and the products increase. If a problem starts with some of every species, compare Q with K first to choose the signs.

Find the equilibrium concentrations and pH of 0.10 M acetic acid, $\text{HC}_2\text{H}_3\text{O}_2$, with $K_a = 1.8 \times 10^{-5}$.

ICE table. I: [HA] = 0.10, [H⁺] = 0, [A^-] = 0. C: $-x$, $+x$, $+x$. E: $0.10 - x$, $x$, $x$.

Set up K. $1.8 \times 10^{-5} = \frac{x^2}{0.10 - x}$. Since $K_a$ is very small next to 0.10, assume $0.10 - x \approx 0.10$: $$x = \sqrt{(1.8 \times 10^{-5})(0.10)} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}\ \text{M}$$

Check the 5% rule. $\frac{1.34 \times 10^{-3}}{0.10} \times 100\% = 1.3\%$, which is under 5%, so the approximation is valid. (The exact quadratic gives $x = 1.33 \times 10^{-3}$ M, essentially the same.)

Results. $[\text{H}^+] = [\text{C}_2\text{H}_3\text{O}_2^-] = 1.3 \times 10^{-3}$ M, $[\text{HC}_2\text{H}_3\text{O}_2] = 0.10 - 0.0013 \approx 0.099$ M, and $\text{pH} = -\log(1.34 \times 10^{-3}) = 2.87$.

To reproduce this, set [A]₀ to 0.10 M and drag K to 1.78e-05, the closest slider setting to $1.8 \times 10^{-5}$. Step 3 should read about 1.3% and valid. Now set K to 1.00e-03: the approximation gives $x = 0.010$, which is 10% of 0.10, so step 3 is marked invalid. Step 3 reports that approximate value as 10.00% of [A]₀, and the quadratic in step 4 gives the value to use instead, $x = 0.0095$ M.

ICE tables show up whenever you must find an equilibrium concentration from K: weak acid and weak base pH (Unit 8), solubility from $K_{sp}$, and gas-phase equilibria in Unit 7. Free-response questions often ask you to justify an approximation, so practice writing the 5% check in one sentence. The same table also works in reverse: if a question gives an equilibrium concentration or a measured pH, fill in the E row from that data, back out $x$, and calculate K from the equilibrium expression.