At a given temperature, 0.45 g of acetic acid in 50 mL of water is shaken with 1.0 g of charcoal and

Q: Why do we assume $[CH_3COO^-]$ is equal to $[H^+]$?

In a pure aqueous solution of a weak acid, every molecule that dissociates produces one acetate ion and one hydrogen ion. Since the contribution of $H^+$ from water autoionization is negligible at pH 3, we can safely assume they are equal.

Q: Why did we approximate $C - 10^{-3}$ to just $C$?

The value of $10^{-3}$ (0.001) is very small compared to 0.1. Subtracting it gives 0.099, which is practically 0.1. In competitive exams, such minor approximations are standard and expected to simplify calculations without losing significant accuracy.

Q: What does the slope of 1 physically mean in the Freundlich isotherm?

A slope of 1 means that $1/n = 1$, or $n = 1$. This indicates that the extent of adsorption ($x/m$) is directly proportional to the concentration ($C$). This linear relationship typically occurs at low concentrations.

Q: Why did we use the total volume of 50 mL for the final concentration calculation? Doesn't the charcoal take up space?

The volume of 1.0 g of solid charcoal is negligible compared to 50 mL of water. Therefore, we assume the volume of the solution remains constant at 50 mL throughout the process.

Q: Can the Freundlich isotherm be used for all concentration ranges?

No, the Freundlich isotherm is an empirical model that works well at low to moderate concentrations. At very high concentrations, the adsorption sites on the charcoal become saturated, and the amount adsorbed plateaus, which this model fails to predict accurately.

Question

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,

$\frac{x}{m} = kC^{1/n}$

If the plot of $\log_{10}(x/m)$ against $\log_{10}C$ gives a straight line with slope 1, the value of k in $	ext{L mol}^{-1}$ is ____.

Given: The molar mass of acetic acid is $60 	ext{ g mol}^{-1}$.
The acid dissociation constant of acetic acid is $1.0 	imes 10^{-5}$ 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.

Sigma Percentile · Accepted Answer

### Analyzing the Setup
We start with a classic physical chemistry scenario: a solute (acetic acid) dissolved in a solvent (water), interacting with a solid adsorbent (charcoal). The charcoal acts like a molecular sponge, pulling acetic acid molecules out of the bulk solution and onto its surface.

### The Master Equation: Ionic Equilibrium
The problem gives us a crucial clue: the final pH of the solution is 3.0. This is our gateway to finding the equilibrium concentration of acetic acid.

Since $	ext{pH} = 3.0$, we know that the concentration of hydrogen ions, $[H^+]$, is $10^{-3} 	ext{ M}$.

Acetic acid is a weak acid, and it dissociates partially:
$$CH_3COOH ightleftharpoons CH_3COO^- + H^+$$

Using the acid dissociation constant ($K_a$), we can set up the equilibrium expression:
$$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$$

Given $K_a = 1.0 	imes 10^{-5}$ and assuming $[CH_3COO^-] \approx [H^+] = 10^{-3} 	ext{ M}$, we can solve for the concentration of undissociated acetic acid. 
$$1.0 	imes 10^{-5} = \frac{10^{-3} 	imes 10^{-3}}{C - 10^{-3}}$$

Solving this gives $C - 10^{-3} = 0.1 	ext{ M}$. For all practical purposes, the total equilibrium concentration $C$ is approximately **$0.1 	ext{ M}$**.

### Calculating the Adsorbed Mass
Now, let's figure out how much acetic acid was actually "sponged up" by the charcoal.

Initially, we had $0.45 	ext{ g}$ of acetic acid. Converting this to moles (using the molar mass of $60 	ext{ g/mol}$):
$$n_{	ext{initial}} = \frac{0.45}{60} = 0.0075 	ext{ moles}$$

The initial molarity in $50 	ext{ mL}$ ($0.05 	ext{ L}$) of water was:
$$M_{	ext{initial}} = \frac{0.0075}{0.05} = 0.15 	ext{ M}$$

After adsorption, the concentration dropped to $0.1 	ext{ M}$. The change in concentration is $0.15 - 0.1 = 0.05 	ext{ M}$.

The moles of acetic acid adsorbed are:
$$n_{	ext{adsorbed}} = 0.05 	ext{ M} 	imes 0.05 	ext{ L} = 0.0025 	ext{ moles}$$

Converting this back to mass gives us our $x$:
$$x = 0.0025 	ext{ moles} 	imes 60 	ext{ g/mol} = \mathbf{0.15 	ext{ g}}$$

### Final Calculation: The Freundlich Isotherm
We are now ready to use the Freundlich adsorption isotherm:
$$\frac{x}{m} = k C^{1/n}$$

Taking the base-10 logarithm of both sides yields a linear equation:
$$\log_{10}\left(\frac{x}{m}ight) = \log_{10} k + \frac{1}{n} \log_{10} C$$

The problem states that the plot of $\log_{10}(x/m)$ versus $\log_{10} C$ has a slope of 1. Therefore, $\frac{1}{n} = 1$.

Substituting our known values ($x = 0.15 	ext{ g}$, $m = 1.0 	ext{ g}$, $C = 0.1 	ext{ M}$):
$$\log_{10}\left(\frac{0.15}{1.0}ight) = \log_{10} k + 1 	imes \log_{10}(0.1)$$

$$\log_{10}(0.15) = \log_{10} k - 1$$

$$\log_{10}(0.15) = \log_{10}(k 	imes 0.1)$$

$$0.15 = 0.1 k$$

$$k = \mathbf{1.5}$$

This elegant problem beautifully weaves together concepts from equilibrium and surface chemistry!

Enter Numerical Value:

Visualized Solution

$S e t u p & O r i e n t a t i o n$

$I ni t ia l C o n ce n t r a t i o n$

$E q u i l ib r i u m C o n ce n t r a t i o n (C)$

$A d sor b e d M a ss (x)$

$F r e u n d l i c h I so t h er m$

$F ina l C a l c u l a t i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Adsorption

Analyzing the Setup We start with a classic physical chemistry scenario: a solute (acetic acid) dissolved in a solvent (water), interacting with a solid adsorbent (charcoal)

The Master Equation

Calculating the Adsorbed Mass

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup & Orientation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Concentration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equilibrium Concentration (C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Adsorbed Mass (x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Freundlich Isotherm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward