ICE Table & Equilibrium Calculator
Equilibrium amounts from any K and starting mix, Q vs K, Kp ↔ Kc, and solubility from Ksp.
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- Free preview: the extent x, the converted K, the solubility or Ksp in full, with the first rows of each table and the first step; the other figures are hidden.
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ICE table
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About the ICE Table & Equilibrium Calculator
Three equilibrium calculations. ICE table: type a balanced equation, K and the starting concentrations or partial pressures; the calculator writes the equilibrium expression (leaving out pure solids, liquids and the solvent), compares Q with K to find the direction, and solves for x exactly — no small-x approximation, no quadratic to rearrange, any coefficients and any K from 10⁻⁵⁰ to 10⁵⁰ — then fills in the Initial, Change and Equilibrium rows and checks that Q equals K. Kp ↔ Kc: Kp = Kc(RT)^Δn with Δn counted from the gases of the equation, in bar or atm. Solubility: the molar solubility of a salt from its Ksp (or Ksp from the solubility), with a common ion already in the solution, and Ksp estimated from the Gibbs energies of the data table when the salt is there.
Without a pass, the free preview shows the extent x, the converted K, the solubility or Ksp in full, with the first rows of each table and the first step; the other figures are hidden; with a Pro pass, or after unlocking this result, you can copy the full answer or download it as CSV.
How to use it
- Choose ICE table, Kp ↔ Kc or Solubility (Ksp).
- For an ICE table, type the equation with states, for example
N2O4(g) <=> 2NO2(g)orCH3COOH(aq) <=> H^+(aq) + CH3COO^-(aq). - Pick concentrations (mol/L, for Kc) or partial pressures (bar or atm, for Kp), type K, then the starting amount of each substance — leave products at 0 if there are none yet.
- Read x in the dark box, then the Equilibrium row of the table; the notes say whether the small-x approximation would have held.
- For Ksp, type the salt (PbI2, Ag2CrO4, Ca3(PO4)2) and Ksp; choose a common ion and its concentration to see how much less dissolves.
- Copy the result or download it as CSV — with a Pro pass, or after unlocking this result.
Examples
N2O4(g) <=> 2NO2(g), Kp = 0.1484, 1 bar of N₂O₄
x = 0.175 bar: P(N₂O₄) = 0.825 bar, P(NO₂) = 0.350 bar
Kp worked out from the ΔfG° values of the data table.
Acetic acid at 0.1 mol/L, Ka = 1.75 × 10⁻⁵
[H⁺] = 1.314 × 10⁻³ mol/L (pH 2.88); 1.3% ionised, so the 5% approximation holds
H2(g) + I2(g) <=> 2HI(g), K = 50.5, 1 mol/L of HI only
x = −0.1098: [H₂] = [I₂] = 0.110 mol/L, [HI] = 0.780 mol/L
H2O(l) <=> H^+(aq) + OH^-(aq), Kw = 1.0 × 10⁻¹⁴, no ions at the start
x = 1.0 × 10⁻⁷ mol/L: [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol/L
Water, a pure liquid, is left out of K and is present throughout.
N2(g) + 3H2(g) <=> 2NH3(g), Kc = 0.5 at 500 K
Δn = −2, Kp = 2.893 × 10⁻⁴ (bar)
PbI₂, Ksp = 9.8 × 10⁻⁹
s = 1.348 × 10⁻³ mol/L (0.621 g/L); in 0.10 mol/L of I⁻ only 9.80 × 10⁻⁷ mol/L
AgCl
ΔG° of dissolution = 55.64 kJ/mol, so Ksp ≈ 1.79 × 10⁻¹⁰ and s = 1.34 × 10⁻⁵ mol/L
Common uses
- Equilibrium homework: ICE tables for gases and solutions, weak acids and bases, with the working shown.
- Checking whether the small-x approximation is allowed before using it in an exam.
- Converting between Kp and Kc, in bar or atm.
- Working out how much of a sparingly soluble salt dissolves, and how a common ion changes it.
How the ICE table is solved
- Each amount at equilibrium is its starting amount plus ν·x, with ν the coefficient (negative for reactants), so K = Π(amount₀ + νx)^ν.
- No amount can go below zero, which bounds x on both sides. Across that range ln Q rises steadily with x (its slope is Σν²/amount, always positive), so there is exactly one x that makes Q = K.
- The calculator finds it by bisection to full double precision, carrying the amount that nearly runs out separately so that a K of 10⁵⁰ still leaves the right 10⁻⁵⁰ of reactant. That is the same answer the quadratic formula gives when there is one, without the cancellation errors.
- Pure solids and liquids (and water as the solvent) have activity 1: they are left out of K and of the table, and are assumed to be present throughout.
Q, K and the direction of change
- Q has the form of K, with the present amounts. Q < K: the reaction shifts right, towards products, and x > 0. Q > K: it shifts left and x < 0. Q = K: it is already at equilibrium.
- With no product present Q = 0, so it can only go forward; with a reactant missing Q is infinite, so it can only go back.
- When one side holds only solids, liquids or the solvent, the reaction can start from nothing on the other side: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) or CaCO₃(s) ⇌ CaO(s) + CO₂(g) with no products at the start.
- The small-x approximation. Textbooks often take x ≪ [A]₀ for weak acids and bases. The calculator says how big x really is against the starting amount: under 5% the shortcut is fine.
Kp, Kc and solubility
- Kp = Kc(RT)^Δn, where Δn is the moles of gas of the products minus those of the reactants, R = 0.083145 L·bar/(mol·K) for Kp in bar (or 0.082057 L·atm/(mol·K) for atm) and T in kelvins. Kp in bar is the thermodynamic K of a gas reaction, the one ΔG° = −RT ln K gives.
- Ksp = [M]^a[X]^b for MₐXᵦ(s) ⇌ aM + bX. In pure water [M] = a·s and [X] = b·s, so s = (Ksp / aᵃbᵇ)^(1/(a+b)). With a common ion already present the equation is solved exactly; the solubility always drops (Le Chatelier’s principle).
- Ksp from the table: for a salt whose solid and ions are in the data table, ΔG° of dissolution = ΣΔfG°(ions) − ΔfG°(salt) and Ksp = e^(−ΔG°/RT) at 298.15 K. An error of 2 kJ/mol in the tables changes Ksp by a factor of about 2, so a measured Ksp is better when you have one.
Sources
- IUPAC, Quantities, Units and Symbols in Physical Chemistry (the Green Book), 3rd edition: Q, K°, Kp, Kc and the solubility product
- NIST, molar gas constant R (CODATA recommended values)
- CODATA, Key Values for Thermodynamics; NIST, NIST-JANAF Thermochemical Tables; Wagman et al., The NBS Tables of Chemical Thermodynamic Properties: the ΔfG° values behind the example constants and the Ksp estimates
Limitations
- Concentrations and pressures stand in for activities: fine for dilute solutions and low pressures, less so for concentrated ones.
- One equilibrium at a time: for coupled equilibria (polyprotic acids, buffers with several species) solve the main one, or use a pH calculator.
- K belongs to one temperature and to the equation as written: doubling the coefficients squares K.
- Ksp from the table is an estimate from Gibbs energies of formation; measured values can differ by a factor of 2 or more.
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Frequently asked questions
What do I get without a pass?
Without a pass, ICE Table & Equilibrium Calculator shows the extent x, the converted K, the solubility or Ksp in full, with the first rows of each table and the first step; the other figures are hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
What is an ICE table?
ICE stands for Initial, Change, Equilibrium: one row each, one column per substance in the equilibrium expression. The Change row is ν·x for each substance, and the Equilibrium row is Initial + Change; putting that row into K gives one equation in x.
When can I ignore x in the denominator?
When x is under about 5% of the starting amount — usually when K is small and the starting concentration large (roughly [A]₀/K above 400). The calculator solves the full equation anyway and tells you how big x was against the starting amount.
How do I know which way the reaction shifts?
Work out Q with the starting amounts. If Q < K the reaction goes forward (x positive); if Q > K it goes in reverse (x negative); if Q = K nothing changes.
Why are solids and liquids left out?
Their activity is 1: a pure solid or liquid has the same concentration however much of it there is, so it does not appear in K. The same goes for water as the solvent in dilute solutions.
How do I convert Kc to Kp?
Kp = Kc(RT)^Δn, with R = 0.08314 L·bar/(mol·K), T in kelvins and Δn the change in moles of gas. For N₂ + 3H₂ ⇌ 2NH₃, Δn = −2, so Kp = Kc/(RT)².
How does a common ion change solubility?
An ion that is already in the solution pushes the dissolving equilibrium back, so less of the salt dissolves. PbI₂ dissolves to 1.35 × 10⁻³ mol/L in pure water but only 9.8 × 10⁻⁷ mol/L in 0.10 mol/L NaI.