Question

At $T(K)$ adsorption of acetic acid on 1g of charcoal gave a Freundlich adsorption isotherm with slope of 0.5 and intercept of 1. What is the value of x, when the concentration of acetic acid is 0.1 mol $L^{-1}$? (Given: $\text{antilog}(0.5) [cite_start]= 3.162$; $\text{antilog}(0.301) = 2.0$)

• A. 0.3162
• B. 3.162
• C. 0.5
• D. 0.2

Hint:

Don't let log forms overcomplicate things. Work with the base equation $\frac{x}{m} = k \cdot C^{1/n}$ once you extract $k$ and $\frac{1}{n}$.

Answer 1

The Correct Option is B

Step 1: Recall the Freundlich isotherm.
The Freundlich equation for adsorption is \[ \frac{x}{m} = k \, C^{1/n} \] where $x$ is mass adsorbed, $m$ is mass of charcoal and $C$ is concentration.

Step 2: Take logarithms.
Taking log turns it into a straight line. \[ \log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log C \] The slope is $1/n$ and the intercept is $\log k$.

Step 3: Read the given values.
The slope is $1/n = 0.5$ and the intercept is $\log k = 1$, so $k = 10^1 = 10$. The mass $m = 1$ g and $C = 0.1$.

Step 4: Put values into the main equation.
Since $m = 1$, we get \[ x = 10 \times (0.1)^{0.5} \]

Step 5: Simplify the power.
Now $(0.1)^{0.5} = \sqrt{0.1} = \dfrac{1}{\sqrt{10}}$, so \[ x = \frac{10}{\sqrt{10}} = \sqrt{10} \]

Step 6: Use the antilog value.
Since $\sqrt{10} = 10^{0.5}$ and antilog of 0.5 is 3.162, we get \[ \boxed{x = 3.162} \]