The Gibbs adsorption isotherm: surface excess, surfactants and the cmc

The edge of a liquid is a blurry zone, not a sharp line

Intuition

Look at a beach from the air. Where exactly does the sea end and the sand begin? Waves keep moving the edge. A liquid surface is like that at the molecular scale. The change from liquid to vapour happens over a thin, fuzzy region, and it keeps fluctuating.

Take two fluid phases, each made of one component, such as water and steam, separated by an interface. On an atomic scale the interface is ill-defined because of fluctuations. A snapshot shows concentration changing from one bulk value to the other. Averaged over time, this region is called the interphase, described by its width and by how fast composition changes inside it. So how do you measure the amount adsorbed on a moving, deformable interface? The Gibbs approach in this lecture answers that with one number, the surface excess.

Some molecules choose the surface. A second component may be more soluble in one phase, less soluble in the other, and so collects at the interface. Amphiphiles do this on purpose. They have a polar head and a non-polar tail. At the water surface the dielectric constant drops from 78 (water) to about 1 (air). At a water/oil interface it drops to 1.89 (n-hexane), 1.99 (n-decane) or 2.38 (toluene). So the tail sits in the non-polar side (air or oil) and the head stays in the water.

Surface excess counts the extra solute at the surface per square metre

If a solute increases in concentration near the surface, it is surface active, and there is a surface excess:

$\Gamma_i = \frac{n_i^{\sigma}}{A}$

$n_i^{\sigma}$ is the number of extra moles of $i$ at the interface and $A$ is the interface area. $\Gamma$ has units of mol m⁻². It can be positive (the component builds up, as surfactants do), negative (the component is depleted) or zero. The slide then asks whether $\Gamma$ and the surface tension $\gamma$ are related. The Gibbs isotherm answers yes.

Gibbs: the slope of surface tension against ln c is the surface excess

The slide puts it this way: “the slope of surface tension versus $\ln c_i$ is proportional to the surface excess of component $i$.” The constant is negative. So a positive surface excess lowers $\gamma$ and a negative one raises it. Think of it as a pay-off: the more a solute likes the surface, the more it relieves the surface’s energy, so the surface tension drops. For a single adsorbing species in an ideal solution (for example a non-ionic surfactant):

$\Gamma_i = \frac{-1}{RT}\,\frac{d\gamma}{d\ln c_i} = \frac{-c_i}{RT}\,\frac{d\gamma}{dc_i}$

Two real examples fit the pattern. Adding ethanol to water lowers $\gamma$, so ethanol has a positive surface excess. Adding water to ethanol raises $\gamma$, so water is excluded from the ethanol/air interface. Adding NaCl to water raises $\gamma$ (from about 73 to about 81 mN/m at 5 mol/L, read from the plot). So $\Gamma_{NaCl}$ is negative, and the ions stay in the bulk. The structure of ionic interfaces is still an active research area.

Try it: worked example

The surface tension of a butanol solution at 298 K is fitted by Y = 42.5 + 30.6 exp(-X/139), with Y in mN/m and X in mol/m^3. How do you get the surface excess from this fit?

Show answer and steps

AnswerDifferentiate the fit, then use Gamma = -(c/RT) d gamma/dc. The lecture gives the fit only. The number at c = 100 mol/m^3 below is a practice step, not from the lecture.

1. The Gibbs isotherm for one species is $\Gamma = -\dfrac{c}{RT}\dfrac{d\gamma}{dc}$, and here $\gamma = Y$ and $c = X$.
2. Differentiate the fit. The constant $42.5$ drops out, and the exponential gives $\dfrac{d\gamma}{dc} = -\dfrac{30.6}{139}\exp\!\left(\dfrac{-c}{139}\right)$. The slope is negative, as it must be for a positive surface excess.
3. Put it in: $\Gamma = \dfrac{c}{RT}\cdot\dfrac{30.6}{139}\exp\!\left(\dfrac{-c}{139}\right)$ (with $\gamma$ converted from mN/m to N/m).
4. Practice step (not from the lecture): at $c = 100\ \text{mol m}^{-3}$, $\exp(-100/139) = 0.487$, so $d\gamma/dc = -0.2201 \times 0.487 = -0.107\ \text{mN m}^{-1}$ per $\text{mol m}^{-3}$, which is $-1.07\times10^{-4}\ \text{N m}^{-1}$ per $\text{mol m}^{-3}$.
5. With $RT = 8.314 \times 298 = 2478\ \text{J mol}^{-1}$: $\Gamma = \dfrac{100 \times 1.07\times10^{-4}}{2478} = 4.3\times10^{-6}\ \text{mol m}^{-2}$.

The lecture says this exercise is continued in the week 13 tutorial, so only the fit is given on the slide.

Charged surfactants adsorb in pairs, which is where the factor of 2 comes from

Intuition

A charged surfactant cannot sit at the surface alone. Its charge has to be paired with an opposite charge, or the surface would build up a big electric charge. So the surfactant ion and its partner ion go to the surface together. Counting both of them in the book-keeping gives the extra 2.

For salts and ionic surfactants such as CTAB (two ions, CTA⁺ and Br⁻), the bulk chemical potentials of both ions change, equally, as you add surfactant. The Gibbs isotherm becomes

$\Gamma_i = \frac{-c_i^{+}}{2RT}\,\frac{d\gamma}{dc_i^{+}}$

The interface must stay electrically neutral, as in $\Gamma_{Na^+} = \Gamma_{Cl^-}$.

The lecture’s case study is dodecyldimethylammonium (DDA⁺) chloride in water, with and without NaCl. Surface tension falls as concentration rises, so the surfactant has a positive surface excess. With no salt the slide gives

$\Gamma_{DDA^+} = \frac{-1}{2RT}\,\frac{d\gamma}{d\ln[DDA^+]}$

Now add a lot of salt, $[NaCl] = 0.20\ \text{M}$. The full Gibbs isotherm has a term for each species:

$-d\gamma = \Gamma_{DDA^+}RT\,d\ln[DDA^+] + \Gamma_{Na^+}RT\,d\ln[Na^+] + \Gamma_{Cl^-}RT\,d\ln[Cl^-]$

When the added Na⁺ and Cl⁻ are present at much higher concentration, their concentrations hardly change while you add surfactant. Only the DDA⁺ concentration is really changing. So the last two terms vanish and

$\Gamma_{DDA^+} = \frac{-1}{RT}\,\frac{d\gamma}{d\ln[DDA^+]}$

Both forms are printed exactly like this on the slides. The factor 2 is there when two ions adsorb together and change together. It disappears when one ion is the only one changing.

Salt also raises the surface excess of the surfactant. The slide reports $\Gamma_{max} = 8.1 \times 10^{-6}\ \text{mol m}^{-2}$ for $0.20\ \text{M}$ NaCl and $3.8 \times 10^{-6}\ \text{mol m}^{-2}$ with no salt. The lecture’s own question is why the excess is higher with NaCl. The head-group size slide gives the reason: the added salt screens the electrostatic repulsion between the charged heads, so more molecules fit on the surface.

For unit conversions, the slide notes $1\ \text{dyne cm}^{-1} = 1\ \text{mN m}^{-1} = 1\ \text{mJ m}^{-2}$ and $\log_{10} x = \ln x / 2.303$. The plots are against $\log C$, so you convert before using the $\ln c$ form.

Pitfall

Do not “fix” the factor 2 between slides 15 and 16. The no-salt case has $-1/(2RT)$ and the high-salt case has $-1/(RT)$. They are both right, for the reasons above.

Surface excess gives the area each molecule takes, and salt packs heads closer

If the adsorbed layer is a monolayer, the surface excess gives the average area occupied by a molecule:

$A = \frac{1}{N_A\,\Gamma}$

The unit is m² per molecule, and $\Gamma$ is the surface excess of the surfactant ion only. The slide says the monolayer, and so the interface, is about 20 Å thick. For a monolayer of DDA⁺ in pure water, $A = 45\ \text{Å}^2$ per molecule. In $0.2\ \text{M}$ NaCl, $A = 20\ \text{Å}^2$, because the repulsion between the heads is screened.

Quick check, not from the lecture: $\Gamma = 3.8\times10^{-6}\ \text{mol m}^{-2}$ gives $A = 1/(6.022\times10^{23}\times3.8\times10^{-6}) = 4.4\times10^{-19}\ \text{m}^2$, about $44\ \text{Å}^2$. $\Gamma = 8.1\times10^{-6}$ gives about $20.5\ \text{Å}^2$. Both land close to the slide’s 45 and 20.

Exam detail

Know the two forms of the isotherm for DDA⁺ and why they differ. Know both $\Gamma_{max}$ values, both areas (45 and 20 Ų) and the reason for the change. Lecture 36’s outcomes also ask you to calculate molecular area from surface excess and vice versa.

The cmc is where the surface is full and extra surfactant makes micelles

A surfactant adsorbs at the air/water interface until it forms a monolayer. Then the surface tension shows a break at a particular concentration, the critical micelle concentration (cmc). Above it, $\gamma$ is constant as surfactant is added. The curve has two nearly straight regions: one when the monolayer forms ($\Gamma_{max}$), and one beyond the cmc. The slide’s example is C₁₆TAB, with a break near 0.9 mmol kg⁻¹, read from the plot.

Why is $\gamma$ flat above the cmc? Once micelles are present, the dissolved (monomer) concentration is fixed, like a solubility limit. Adding more surfactant makes more micelles but no extra free monomer. Micelles are not surface active, so $\Gamma = 0$. Fixing the monomer concentration fixes the surface tension, so

$\Gamma = 0, \qquad d\gamma = 0$

Recall

Why does adding salt to a charged surfactant solution let more surfactant crowd onto the surface?

The charged heads repel each other. Salt screens that repulsion, so the heads can sit closer together. The surface excess rises and the area per molecule falls, from 45 Ų to 20 Ų for DDA⁺.