At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 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,$\frac{x}{m}=k{C}^{1/\mathrm{n }}$

${}^{}$If the plot of log₁₀(x/m) against log₁₀C gives a straight line with slope 1, the value of k in L mol⁻¹ is _______.

Given: The molar mass of acetic acid is 60 g mol⁻¹.

The acid dissociation constant of acetic acid is 1.0 × 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.

The correct answer is 1.5.

Step 1: Understand the given data
Initial mass of acetic acid, $W_{\text{initial}}=0.45\text{g}$
Volume of water, $V=50\text{mL}=0.05\text{L}$
Mass of charcoal (adsorbent), $m=1.0\text{g}$
Molar mass of acetic acid, $M=60{\text{g mol}}^{\mathrm{-1}}$
pH of the solution at equilibrium, $\text{pH}=3.0$
Acid dissociation constant of acetic acid, $K_a=1.0\times {10}^{\mathrm{-5}}$

Step 2: Calculate the equilibrium concentration of acetic acid (C)
Acetic acid (${\text{CH}}_3\text{COOH}$) is a weak monobasic acid. Its dissociation in water is given by:
${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{H}}^{+}+{\text{CH}}_3{\text{COO}}^{-}$

Given $\text{pH}=3.0$, the concentration of hydrogen ions at equilibrium is:
$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}={10}^{\mathrm{-3}}\text{M}={10}^{\mathrm{-3}}{\text{mol L}}^{\mathrm{-1}}$

For a weak acid, the dissociation constant expression is:
$K_a=\frac{{\left[{\text{H}}^{+}\right]}^2}{C}$

Rearranging to find the equilibrium concentration $C$ of acetic acid:
$C=\frac{{\left[{\text{H}}^{+}\right]}^2}{K_a}=\frac{{\left({10}^{\mathrm{-3}}\right)}^2}{1.0\times {10}^{\mathrm{-5}}}=\frac{{10}^{\mathrm{-6}}}{{10}^{\mathrm{-5}}}=0.1{\text{mol L}}^{\mathrm{-1}}$

Step 3: Calculate the mass of acetic acid remaining and adsorbed (x)
The number of moles of acetic acid remaining in 50 mL (0.05 L) of solution at equilibrium is:
$n_{\text{eq}}=C\times V=0.1{\text{mol L}}^{\mathrm{-1}}\times 0.05\text{L}=0.005\text{mol}$

The mass of acetic acid remaining in the solution is:
$W_{\text{remaining}}=n_{\text{eq}}\times M=0.005\text{mol}\times 60{\text{g mol}}^{\mathrm{-1}}=0.30\text{g}$

Therefore, the mass of acetic acid adsorbed by charcoal ($x$) is:
$x=W_{\text{initial}}-W_{\text{remaining}}=0.45\text{g}-0.30\text{g}=0.15\text{g}$

Step 4: Use Freundlich adsorption isotherm to determine k
The Freundlich adsorption isotherm equation is:
$\frac{x}{m}=k{C}^{1/n}$

Taking logarithm (base 10) on both sides:
${\log }_{10}\left(\frac{x}{m}\right)={\log }_{10}k+\frac{1}{n}$

Comparing this with the equation of a straight line, $y=mx+c$, the slope is given as 1, which means:
$\frac{1}{n}$

Substituting $\frac{1}{n}$ back into the isotherm equation:
$\frac{x}{m}=kC$

Substitute the known values ($x=0.15\text{g}$, $m=1.0\text{g}$, $C=0.1{\text{mol L}}^{\mathrm{-1}}$):
$\frac{0.15}{1.0}=k\times 0.1$

$k=\frac{0.15}{0.1}=1.5{\text{L mol}}^{\mathrm{-1}}$

Thus, the value of $k$ is 1.5.