Question 1

What is an ICE table and how do I use it?

Accepted Answer

An ICE table lists, for every substance in the equilibrium expression, its Initial amount, the Change as the reaction proceeds, and the Equilibrium amount. The change is set by the coefficients: if the amount of one product grows by x, another product with coefficient 2 grows by 2x and a reactant with coefficient 3 falls by 3x. OpenStax (Chemistry 2e, section 13.4) gives four steps: decide the direction by comparing Q with K; build the table; put the equilibrium terms into the K expression and solve for x; check that Q equals K. Pure solids and liquids are left out of the expression. Their row shows a dash.

Question 2

Worked example: how much PCl₅ decomposes?

Accepted Answer

OpenStax Example 13.9: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has Kc = 0.0211, and the mixture starts as 1.00 M PCl₅ only. Q₀ = 0 is below K, so the reaction runs forward: PCl₅ falls by x and each product rises by x. Then x² ÷ (1.00 − x) = 0.0211, a quadratic whose positive root is x = 0.135 M, so [PCl₅] = 0.865 M (the book rounds it to 0.87) and [PCl₃] = [Cl₂] = 0.135 M. The calculator finds that root numerically inside the allowed range, so it works for any coefficients, not only for quadratics, and it shows the table with the x terms.

Question 3

When can I assume x is small, and what is the 5% rule?

Accepted Answer

When K is small compared with the starting concentration, x is tiny next to the starting amount and (C − x) can be replaced by C, which gives x = √(K·C) and avoids the quadratic. OpenStax section 14.3 accepts the shortcut when x turns out to be less than 5% of the starting concentration. It holds for formic acid, 0.534 M with Ka = 1.8 × 10⁻⁴: x = 9.8 × 10⁻³ M is 1.8% of it. It fails for the bisulfate ion, 0.50 M with Ka = 1.2 × 10⁻²: the shortcut gives 0.077 M, 15% of the start, while the exact answer is 0.072 M (pH 1.14). The calculator shows the shortcut beside the exact answer and says whether the rule passes. Try both examples in the weak acid tab.

Question 4

How do I calculate K from experimental data?

Accepted Answer

Use the “K from data” tab: give the starting amounts and one amount measured at equilibrium. The measured change fixes x, the table follows, and K = Q at the equilibrium amounts. OpenStax Example 13.7: I₂ + I⁻ ⇌ I₃⁻ starts at 1.000 × 10⁻³ M for both reactants, and 6.61 × 10⁻⁴ M of I₂ is left at equilibrium. So x = 3.39 × 10⁻⁴ M and Kc = 3.39 × 10⁻⁴ ÷ (6.61 × 10⁻⁴)² = 776. If you know all the equilibrium amounts, the equilibrium constant calculator computes K from them directly.

Question 5

How do I find the pH and percent ionization of a weak acid?

Accepted Answer

For HA ⇌ H⁺ + A⁻ with starting concentration C, Ka = x² ÷ (C − x), where x = [H⁺] = [A⁻]; then pH = −log x and the percent ionization is x ÷ C × 100 (OpenStax section 14.3). Formic acid, 0.534 M, Ka = 1.8 × 10⁻⁴: x = 9.7 × 10⁻³ M, pH 2.01, 1.8% ionized. For a weak base, B + H₂O ⇌ BH⁺ + OH⁻ gives x = [OH⁻] and pOH, and pH = 14.00 − pOH at 25°C, the same relation as in our pH calculator. Percent ionization falls as the solution gets more concentrated. The tab can also work backward: Ka from a measured pH. Water’s own 10⁻⁷ M of H⁺ is ignored, as in OpenStax, which is fine unless the acid is extremely dilute or weak. For a buffer use the buffer pH calculator.

Question 6

Why may my textbook’s answer differ in the last digit?

Accepted Answer

Textbooks round at each step and the calculator does not. OpenStax lists 0.06, 1.06 and 1.88 M for H₂ + I₂ ⇌ 2HI with K = 50.5 from 1.00 M H₂ and 2.00 M I₂; solving exactly gives 0.065, 1.065 and 1.870 M. For the 0.0325 M ammonia solution it prints 7.56 × 10⁻⁴ M and 2.33%, which are the shortcut values; the exact ones are 7.48 × 10⁻⁴ M and 2.30%. And Ka for 0.0516 M HNO₂ at pH 2.34 works out to 4.4 × 10⁻⁴ (4.5 × 10⁻⁴ from the rounded [H⁺] the book uses) although it prints the tabulated 4.6 × 10⁻⁴. Trust the unrounded value from the calculator and round it to the digits your course asks for.

ICE table calculator