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Adsorption Isotherm Calculator (Langmuir & Freundlich)

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Understanding the Langmuir and Freundlich adsorption isotherm calculator — concept and principle

An adsorption isotherm describes how much of a substance an adsorbent such as activated carbon, zeolite or resin holds at equilibrium for a given concentration in the surrounding liquid. The calculator evaluates the two most widely used models, Langmuir and Freundlich, at your equilibrium concentration and estimates the mass of adsorbent needed to bring a batch of water from its initial concentration down to that level.

Adsorption isotherm and adsorbent doseAt 20 mg/L, K_L × C = 1 so the Langmuir surface is half full at 75 of 150 mg/g; removing 80 g from 1,000 L needs 1.07 kg of adsorbent.Half of the adsorption sites are filledfilledemptyq_max = 150 mg/gall sites fullLangmuir q = 750.05 × 20 = 1 → half fullFreundlich q = 46.812 × 20^(1/2.2)1,000 L × (100 − 20) mg/L = 80 g removed80,000 mg ÷ 75 mg/g = 1.07 kgadsorbent needed (Langmuir)
At Ce = 1/K_L the Langmuir surface is exactly half covered: 75 of 150 mg/g.

How it is calculated

q is in mg of solute per g of adsorbent, C_e and C₀ in mg/L, V in litres, q_max in mg/g and K_L in L/mg. The units of K_F depend on n and are those of the data it was fitted to.

The principle

Langmuir pictured a surface with a fixed number of identical sites, each holding one molecule. At equilibrium the rate of adsorption onto empty sites equals the rate of desorption from filled ones, which gives the saturating curve above: nearly linear at low concentration and levelling off at q_max when the surface is full (a monolayer). Freundlich's equation is empirical; it fits heterogeneous surfaces where sites differ in energy, so uptake keeps rising with concentration without a clear plateau. An n above 1 indicates favourable adsorption. The dose formula is a mass balance: what leaves the water must end up on the adsorbent.

Worked example

  1. Inputs: C_e = 20 mg/L, q_max = 150 mg/g, K_L = 0.05 L/mg, K_F = 12, n = 2.2, C₀ = 100 mg/L, V = 1,000 L.
  2. K_L × C_e = 0.05 × 20 = 1, so q = 150 × 1 ÷ 2 = 75 mg/g (half of capacity).
  3. Freundlich: q = 12 × 20^(1/2.2) = 12 × 3.903 = 46.83 mg/g.
  4. Solute removed = (100 − 20) × 1,000 = 80,000 mg; adsorbent = 80,000 ÷ 75 = 1,066.7 g = 1.067 kg.
How Langmuir q (mg/g) changes with Langmuir qmax (mg/g)How Langmuir q changes with Langmuir qmax05010015037.575112150188225300Example: 75Langmuir qmax (mg/g)
Each point is the calculator re-run with only “Langmuir qmax (mg/g)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Langmuir qmax (mg/g)Langmuir q (mg/g)
37.518.75
7537.5
11256.25
15075
18893.75
225112.5
300150

How to read the result

When K_L × C_e = 1 the Langmuir surface is exactly half covered, so 1/K_L is the concentration at half capacity. The two models give different uptakes because they were fitted separately; the closer each matches your measured data in the concentration range of interest, the more trustworthy it is. A lower target C_e means lower q and therefore more adsorbent.

Common mistakes

Keep in mind: Isotherms come from equilibrium tests on a specific adsorbent, water matrix and temperature. Competing substances, contact time and pH can lower real capacity, so full-scale dosing needs pilot tests and engineering judgement.

Frequently asked questions

Which model should I choose?

Fit both to your data and compare the error; Langmuir suits surfaces that saturate, Freundlich often fits activated carbon over a moderate range.

What does the Langmuir separation factor tell me?

R_L = 1/(1 + K_L C₀) between 0 and 1 indicates favourable adsorption.

Sources: I. Langmuir, Journal of the American Chemical Society 40 (1918) 1361 · H. Freundlich, Zeitschrift für physikalische Chemie 57 (1906) 385

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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