Equilibrium Constant Calculator

Compute Kc from stoichiometric coefficients and equilibrium concentrations, or solve any of them in reverse.

Covers nine molar concentration scales from M to yM and solves for K, any concentration, or any coefficient.

Updated October 1, 2026

CreatorFrank Zhao

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What this calculator works out

At equilibrium a reversible reaction has stopped changing, and the ratio of products to reactants has settled at a fixed value for that temperature. This calculator evaluates that ratio for a general reaction of the form

a[\mathrm{A}] + b[\mathrm{B}] \rightleftharpoons c[\mathrm{C}] + d[\mathrm{D}]

Nine quantities describe that one reaction: four stoichiometric coefficients, four equilibrium molar concentrations, and the equilibrium constant itself. Only eight of them are independent. Fill in eight and the ninth follows, which is why this tool can work in either direction — it will happily hand you $K$ from measured concentrations, or hand you the missing concentration back from a known $K$ , or even recover a coefficient you left blank.

Two of the nine quantities are deliberately left at zero by default. That is not a placeholder — a coefficient of zero is the correct way to say a species does not appear on that side of the equation at all, and it has real consequences for your answer.

You will find this useful whenever an equilibrium question comes out of laboratory work rather than out of a textbook: checking a reported constant against your own titration data, recovering the concentration you forgot to record, or sanity-checking a literature value before you build a buffer or set a process operating point on it.

How to use it

Worked through with a simple association reaction in which all four stoichiometric coefficients are 1. Values shown are the ones the calculator produces.

Balance the equation before anything else

The coefficients are exponents, not weights, so they have to be the correct small integers. If you are not sure of them, run the equation through the chemical equation balancer first.

Enter four coefficients and four concentrations

Any eight of the nine boxes are enough. Set the unit beside each concentration to whatever scale your data is in, and leave the one box you want to solve for empty.

Read the result in the box that fills itself in

The solved field turns blue and shows the full precision it was computed at. Click into it and the thousands separators disappear so you can edit it cleanly; click away and they come back.

Worked example: finding K

For $\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$ with all coefficients set to 1 and equilibrium concentrations of 0.200 M, 0.150 M, 0.300 M and 0.450 M:

K = \frac{[\mathrm{C}]^{c} \times [\mathrm{D}]^{d}}{[\mathrm{A}]^{a} \times [\mathrm{B}]^{b}}

= \frac{[\mathrm{C}] \times [\mathrm{D}]}{[\mathrm{A}] \times [\mathrm{B}]}

= \frac{(0.300)(0.450)}{(0.200)(0.150)}

= \frac{0.135}{0.030}

= 4.5

Now blank $[\mathrm{D}]$ , type $K = 4.5$ instead, and the calculator returns $[\mathrm{D}] = 0.450\ \mathrm{M}$ — the same number, recovered from the other side of the equation.

The equations behind the answer

The calculator starts from the mass action expression and rearranges it. Concentrations are normalised to moles per litre first, so the unit you enter each species in cancels out of the ratio and leaves $K$ properly dimensionless.

K = \frac{[\mathrm{C}]^{\,c} \times [\mathrm{D}]^{\,d}}{[\mathrm{A}]^{\,a} \times [\mathrm{B}]^{\,b}}

The four reverse directions

A product concentration comes out as an $c$ -th root, and a reactant concentration as an $a$ -th root. Recovering a coefficient is the same algebra run through logarithms instead, because the coefficient sits in the exponent:

[\mathrm{A}] = \sqrt[a]{\frac{[\mathrm{C}]^{\,c} \times [\mathrm{D}]^{\,d}}{K \times [\mathrm{B}]^{\,b}}}

a = \frac{\ln\!\left(\frac{[\mathrm{C}]^{\,c} \times [\mathrm{D}]^{\,d}}{K \times [\mathrm{B}]^{\,b}}\right)}{\ln[\mathrm{A}]}

The two cases where no answer exists

These are the two situations that look like a bug but are not. In both, the mathematics genuinely has no solution, so the calculator leaves the field blank and explains why instead of showing you a number.

A coefficient of zero removes its species from the equation

If $a = 0$ then $[\mathrm{A}]^{0} = 1$ no matter what $[\mathrm{A}]$ is. The concentration genuinely cannot be recovered, and the field reports that its coefficient must be greater than zero.

A concentration of exactly 1 M cannot give a coefficient

The coefficient comes from a ratio of two logarithms. $\ln(1) = 0$ , so $[\mathrm{A}] = 1\ \mathrm{M}$ leaves $a$ completely undetermined, and the field says so.

Pure phases are not part of the expression

A solid such as calcium carbonate or a solvent such as water appears in your written equation but not in $K$ , because their activity is effectively constant and cancels. If you include one, you are adding a term that should not be there.

Reading K and the traps that change it

Magnitude tells you which side is favoured

A value of 4.5 means products are favoured at equilibrium; a value of 0.001 means reactants are. But it is the distance from 1 that matters, not the raw number, and that distance is best read on a logarithmic scale: each factor of ten moves the equilibrium meaningfully further toward one side.

$K = 0.001$

Reactants strongly favoured

Three orders of magnitude below 1, so little product forms.

$K = 4.5$

Products moderately favoured

Less than one order of magnitude above 1 — a partial shift.

$K = 250$

Products strongly favoured

Two and a half orders of magnitude above 1, so conversion runs well forward.

Four input traps, and what each one does

Forgetting a coefficient

Change $a$ from 1 to 2 while keeping the same four concentrations and $K$ jumps from 4.5 to 22.5. Every coefficient you enter is being used as an exponent.

Leaving in a species that is not there

With $d = 0$ the calculator returns $K = 10$ whether $[\mathrm{D}]$ is 0.1 or 5, because that concentration has left the expression entirely.

A mass concentration instead of a molarity

Every box wants moles per litre. If your data is in grams per litre, convert it first with the molarity calculator rather than entering the number directly.

Fighting the unit scales

You do not have to make your data match. Entering every concentration in millimolar, or all in micromolar, or mixing nanomolar for one species with millimolar for another, all return the same 4.5 — each species is normalised on its own before the ratio is taken.

K belongs to the temperature, not to your starting mixture. Doubling every concentration changes nothing about the value of

K

. Adding a catalyst changes nothing either. If your two runs at the same temperature give different constants, the cause is in the concentrations — most often a species that was left out of the expression, or one that should not have been in it.

When it fits, and when it does not

Where it fits

Concentrations in, K out

You have equilibrium molar concentrations for up to four species and want $K$ . This is the case the tool is built for.

K in, a concentration out

You know $K$ and need one equilibrium concentration back to complete a mass balance or a yield calculation.

Acid-base and buffer work

Here the equilibrium constant is the acid dissociation constant and the concentrations are small. For the rearranged buffer form, the Henderson-Hasselbalch calculator goes straight from a ratio to a pH.

Where it does not

More than two species per side

Reshape the equation first, or split it into several smaller reactions and evaluate each on its own.

Gaseous equilibria at pressure

The expression you want there is built from partial pressures, not from molar concentrations.

Concentrated, high-ionic-strength solutions

Activities stop tracking concentrations, so what you compute here is a concentration quotient rather than the true thermodynamic constant.

Kinetics questions

This gives you the equilibrium position, never how fast the reaction gets there.

Frequently asked questions

Why does K have no units?

Because each concentration is measured against the same standard concentration, 1 mol/L, and the factors cancel between products and reactants. The calculator applies that normalisation to every species, which is also why you may enter them on different molar scales without changing the answer.

Why is a coefficient 0 instead of blank?

Because zero is a meaningful stoichiometric coefficient: it declares that the species is absent from that side of the equation. Leaving it blank would instead read as a value you have yet to supply, and the calculator would try to solve for it.

Can I get Ka from a pKa table?

Yes, and the pKa calculator does the conversion in both directions. Once you have $K_a$ as a number, treat it as $K$ here for a reaction of the form acid ⇌ conjugate base plus hydrogen ion.

My two runs gave different K values. What should I check first?

Confirm both runs were at the same temperature, then check the expression itself: a missing coefficient, a species that should have been excluded because it is a pure solid or the solvent, or a concentration entered in the wrong unit. These account for far more discrepancies than measurement noise does.

Limitations

This tool does algebra on the numbers you give it. It cannot tell you whether they were measured well.

Results are worth exactly as much as the concentrations behind them.

It assumes ideal behaviour, so concentrated or high-ionic-strength solutions give a concentration quotient rather than a true activity-based constant.

It assumes a single temperature and carries no dependence on pressure.

It has no notion of phase. Excluding pure solids and the solvent is your responsibility, and if you forget, the calculator cannot flag it.

For buffer preparation, process design, or any decision with safety or cost attached, confirm the result against your own validated data or with a qualified chemist before acting on it.

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