Question

The surface tension ($\gamma$) of a solution, prepared by mixing 0.02 mol of an organic acid in 1 L of pure water, is represented as \[ \gamma^* - \gamma = A \log(1 + Bc) \] where $\gamma^*$ is the surface tension of pure water, $A = 0.03 \, \mathrm{N\,m^{-1}}$, $B = 50 \, \mathrm{mol^{-1}\,L}$, and $c$ is concentration in mol L$^{-1}$. The excess concentration of the organic acid at the surface of the liquid, determined by Gibbs adsorption equation at 300 K, is $n \times 10^{-6} \, \mathrm{mol\,m^{-2}}$. The value of $n$ (rounded off to two decimal places) is ________.

Hint:

Use the Gibbs adsorption isotherm $\Gamma = -\frac{1}{RT}\frac{d\gamma}{d(\ln c)}$ for surface-active solutes — a decrease in surface tension indicates positive adsorption.

Answer 1

Correct Answer: 2.6

Step 1: Recall Gibbs adsorption equation. \[ \Gamma = -\frac{1}{RT} \frac{d\gamma}{d(\ln c)} \] Given: $\gamma^* - \gamma = A \log(1 + Bc)$ \[ \Rightarrow \frac{d\gamma}{d(\ln c)} = \frac{d\gamma}{dc} \times c = -\frac{ABc}{(1 + Bc)} \times c \] Step 2: Substitute into Gibbs equation. \[ \Gamma = \frac{A B c^2}{RT(1 + Bc)} \] Step 3: Substitute given values. \[ A = 0.03, \, B = 50, \, c = 0.02, \, R = 8.314, \, T = 300 \] \[ \Gamma = \frac{0.03 \times 50 \times (0.02)^2}{8.314 \times 300 \times (1 + 1)} = 1.94 \times 10^{-6} \, \mathrm{mol\,m^{-2}} \] Step 4: Conclusion. The excess concentration n = 1.94.