Surface Chemistry

Concept:The problem involves adsorption of acetic acid on charcoal following Freundlich isotherm, where the slope of $\log(x/m)$ vs $\log C$ is $1$, and we need to find the constant $k$ using equilibrium data from pH and initial concentration.Explanation:Step 1: Calculate initial moles of acetic acid.$\text{Moles} = \frac{0.45\ \text{g}}{60\ \text{g mol}^{-1}} = 0.0075\ \text{mol}$.Volume is $50\ \text{mL} = 0.050\ \text{L}$, so initial molarity: $M_{\text{initial}} = \frac{0.0075}{0.050} = 0.15\ \text{M}$.Step 2: After adsorption, pH $= 3$, so $[\text{H}^+] = 10^{-3}\ \text{M}$.Acetic acid dissociation: $\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H}^+$.At equilibrium, $[\mathrm{CH_3COO^-}] = [\mathrm{H}^+] = 10^{-3}\ \text{M}$, and let $C$ be the equilibrium concentration of undissociated acid.Using $K_a = 1.0 \times 10^{-5}$:$K_a = \frac{(10^{-3})(10^{-3})}{C - 10^{-3}}$.Solve: $1.0 \times 10^{-5} = \frac{10^{-6}}{C - 0.001}$, giving $C - 0.001 = 0.1$, so $C = 0.101\ \text{M}$.Step 3: Calculate the amount of acetic acid adsorbed.Moles initially: $0.0075\ \text{mol}$ in $50\ \text{mL}$.Moles at equilibrium: $0.101\ \text{mol L}^{-1} \times 0.050\ \text{L} = 0.00505\ \text{mol}$.Moles adsorbed: $0.0075 - 0.00505 = 0.00245\ \text{mol}$.Mass adsorbed $x$: $0.00245\ \text{mol} \times 60\ \text{g mol}^{-1} = 0.147\ \text{g}$.Mass of charcoal $m = 1.0\ \text{g}$, so $\frac{x}{m} = 0.147$.Step 4: Apply Freundlich isotherm: $\frac{x}{m} = k C^{1/n}$.Take logs: $\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log C$.Given slope $= 1$, so $\frac{1}{n} = 1$.Thus $\log(0.147) = \log k + \log(0.101)$.Rearrange: $\log k = \log(0.147) - \log(0.101) = \log\left(\frac{0.147}{0.101}\right)$.$\frac{0.147}{0.101} = 1.455$, so $\log k = \log(1.455)$, hence $k = 1.455 \ \text{L mol}^{-1} \approx 1.45$.Answer:The value of $k$ is $1.45\ \text{L mol}^{-1}$.