Question

At a given temperature, \(0.45\,\text{g}\) of acetic acid in \(50\,\text{mL}\) of water is shaken with \(1.0\,\text{g}\) of charcoal and the pH of the resulting solution is \(3.0\). Assume, the adsorption of acetic acid from the aqueous solution by charcoal follows Freundlich isotherm,

\[ \frac{x}{m} = kC^{1/n} \]
If the plot of \(\log_{10}(x/m)\) against \(\log_{10} C\) gives a straight line with slope \(1\), the value of \(k\) in \(\text{L mol}^{-1}\) is _______.

Given: The molar mass of acetic acid is \(60\,\text{g mol}^{-1}\).
The acid dissociation constant of acetic acid is \(1.0 \times 10^{-5}\) at the given temperature.
\(x\) is the mass (in grams) of acetic acid adsorbed.
\(m\) is the mass (in grams) of charcoal.
\(C\) is the equilibrium concentration of acetic acid in the solution after the adsorption is complete.
\(k\) and \(n\) are constants for acetic acid–charcoal system at the given temperature.

Hint:

For adsorption problems, carefully distinguish between initial and equilibrium concentrations. Use pH to find $[H^+]$ and then $K_a$ to find the equilibrium concentration of the undissociated acid. Account for the amount adsorbed ($x$) by subtracting the equilibrium amount in solution from the initial amount.

Answer 1

Step 1: Understanding the Question:
The problem involves the adsorption of acetic acid by charcoal, which follows the Freundlich isotherm. We are given initial conditions, final pH, and information about the plot of the Freundlich isotherm in logarithmic form. We need to find the value of the constant \(k\).

Step 2: Key Formula or Approach:
1. Freundlich isotherm (logarithmic form): Given as $\log_{10}(x/m) = \log_{10}k + \frac{1}{n} \log_{10}C$.
2. From the plot information: Slope of $\log_{10}(x/m)$ vs $\log_{10}C$ is 1. This means $\frac{1}{n} = 1 \Rightarrow n=1$.
3. Acid dissociation of acetic acid: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$.
The acid dissociation constant $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$.
4. pH relation: $pH = -\log_{10}[H^+]$.

Step 3: Detailed Explanation:
Given:
- Initial mass of acetic acid = 0.45 g.
- Volume of water = 50 mL = 0.050 L.
- Mass of charcoal ($m$) = 1.0 g.
- Equilibrium pH of the solution = 3.0.
- Molar mass of acetic acid = 60 g mol\(^{-1}\).
- $K_a$ for acetic acid = $1.0 \times 10^{-5}$.
- From the plot, slope = $\frac{1}{n} = 1 \Rightarrow n = 1$.
Part 1: Calculate equilibrium concentration of $H^+$ and $CH_3COOH$ (C).
- From pH = 3.0, we get $[H^+] = 10^{-3} \text{ mol L}^{-1}$.
- For a weak acid $CH_3COOH$, in equilibrium:
\[ CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \]
Let the initial concentration of undissociated $CH_3COOH$ at equilibrium be $C$.
Then at equilibrium: $[CH_3COOH] = C - [H^+]$.
$[CH_3COO^-] \approx [H^+]$ (assuming initial dissociation is mainly from $CH_3COOH$).
Using $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$:
$1.0 \times 10^{-5} = \frac{(10^{-3})(10^{-3})}{C - 10^{-3}}$
$1.0 \times 10^{-5} = \frac{10^{-6}}{C - 10^{-3}}$
$C - 10^{-3} = \frac{10^{-6}}{1.0 \times 10^{-5}} = 10^{-1} = 0.1$
$C = 0.1 + 10^{-3} = 0.1 + 0.001 = 0.101 \text{ mol L}^{-1}$.
This is the equilibrium concentration of acetic acid in the solution. So $C = 0.101 \text{ M}$.
Part 2: Calculate mass of acetic acid in solution and adsorbed (\(x\)).
- Total initial moles of acetic acid = $\frac{0.45 \text{ g}}{60 \text{ g mol}^{-1}} = 0.0075 \text{ mol}$.
- Initial concentration of acetic acid = $\frac{0.0075 \text{ mol}}{0.050 \text{ L}} = 0.15 \text{ mol L}^{-1}$. (This is not used if we directly calculate from adsorbed amount).
- Moles of acetic acid remaining in solution at equilibrium = $C \times \text{Volume} = 0.101 \text{ mol L}^{-1} \times 0.050 \text{ L} = 0.00505 \text{ mol}$.
- Mass of acetic acid remaining in solution = $0.00505 \text{ mol} \times 60 \text{ g mol}^{-1} = 0.303 \text{ g}$.
- Mass of acetic acid adsorbed (\(x\)) = Initial mass - Mass remaining in solution
$x = 0.45 \text{ g} - 0.303 \text{ g} = 0.147 \text{ g}$.
Part 3: Calculate \(k\).
- We have \( x = 0.147 \text{ g} \).
- Mass of charcoal ($m$) = 1.0 g.
- Equilibrium concentration ($C$) = 0.101 M.
- $\frac{1}{n} = 1$, so $n=1$.
- Freundlich isotherm: $\frac{x}{m} = kC^{1/n}$ becomes $\frac{x}{m} = kC$.
\[ k = \frac{x/m}{C} = \frac{0.147 \text{ g} / 1.0 \text{ g}}{0.101 \text{ mol L}^{-1}} \]
\[ k = \frac{0.147}{0.101} \]
\[ k \approx 1.455 \]

Step 4: Final Answer:
The calculated value of \(k\) is 1.5 L mol\(^{-1}\).