Adsorption of acetic acid on activated charcoal follows the Freundlich isotherm for solutions. When — Surface Chemistry Chemistry Question
Question
Adsorption of acetic acid on activated charcoal follows the Freundlich isotherm for solutions. When $2\text{ g}$ of charcoal is added to $100\text{ mL}$ of $0.5\text{ M}$ acetic acid, the equilibrium concentration becomes $0.4\text{ M}$. Assuming the exponent $1/n = 1$, calculate the mass of identical charcoal (in grams) that must be added to $100\text{ mL}$ of a fresh $0.5\text{ M}$ acetic acid solution so that its final equilibrium concentration becomes $0.2\text{ M}$.
💡 Solution & Explanation
Case 1: Initial moles of acetic acid $= 0.5 \times 0.1 = 0.05\text{ mol}$. Final moles $= 0.4 \times 0.1 = 0.04\text{ mol}$. Moles adsorbed $= 0.01\text{ mol}$. Let's define $x$ in moles. $\frac{x_1}{m_1} = \frac{0.01}{2} = 0.005\text{ mol g}^{-1}$. Using Freundlich isotherm $\frac{x}{m} = kC^{1/n}$, $0.005 = k(0.4)^1 \implies k = 0.0125$. Case 2: For the final concentration $C_2 = 0.2\text{ M}$, final moles $= 0.2 \times 0.1 = 0.02\text{ mol}$. Moles adsorbed $x_2 = 0.05 - 0.02 = 0.03\text{ mol}$. Applying the isotherm: $\frac{x_2}{m_2} = kC_2 \implies \frac{0.03}{m_2} = 0.0125 \times 0.2 = 0.0025$. Solving for mass: $m_2 = \frac{0.03}{0.0025} = 12\text{ g}$.