Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given t

Q: Why do we take the logarithm of the Freundlich isotherm equation?

Taking the logarithm transforms the power-law relationship $\frac{x}{m} = K \cdot C^{1/n}$ into a linear equation $\log\left(\frac{x}{m}\right) = \log K + \frac{1}{n} \log C$. This makes it much easier to solve for the unknown constants $K$ and $n$ using standard linear algebra techniques, like solving simultaneous equations.

Q: How did we calculate $\log 4$ and $\log 16$ using only $\log 2 = 0.3$?

We use the power rule of logarithms: $\log(a^b) = b \cdot \log a$. Since $4 = 2^2$, $\log 4 = 2 \cdot \log 2 = 2 \cdot 0.3 = 0.6$. Similarly, $16 = 2^4$, so $\log 16 = 4 \cdot \log 2 = 4 \cdot 0.3 = 1.2$.

Q: Can we solve this without taking logarithms?

Yes, you can set up ratios like $\frac{4}{10} = \frac{K \cdot 10^{1/n}}{K \cdot 16^{1/n}}$, but solving for $1/n$ this way often leads to messy fractional exponents that are hard to compute without a calculator. The logarithmic method is cleaner and directly utilizes the given $\log 2$ value.

Q: What does the slope of the graph represent physically?

The slope, $1/n$, represents the intensity of adsorption. It typically ranges between 0 and 1. A value closer to 0 indicates that adsorption is independent of concentration (saturation), while a value closer to 1 indicates a linear dependence.

Q: How did we calculate the final antilog of $1.2$ so easily?

We reversed the logarithmic power rule! We know $1.2 = 4 \cdot 0.3$. Since $0.3 = \log 2$, we can write $1.2 = 4 \cdot \log 2 = \log(2^4) = \log 16$. Therefore, the antilog of $1.2$ is exactly $16$.

Question

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from $10	ext{ mg g}^{-1}$ and $16	ext{ mg g}^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be $4	ext{ mg g}^{-1}$ and $10	ext{ mg g}^{-1}$, respectively. At this temperature, the concentration (in $	ext{mg g}^{-1}$) of adsorbed phenol from $20	ext{ mg g}^{-1}$ aqueous solution of phenol will be ___.
Use : $\log_{10} 2 = 0.3$

Sigma Percentile · Accepted Answer

### The Magic of Adsorption

Imagine you have a dry sponge and you place it in a puddle of water. The sponge eagerly soaks up the liquid, holding it within its porous structure. In the realm of surface chemistry, we observe a remarkably similar phenomenon known as **adsorption**. This occurs when a solid surface (the adsorbent) accumulates molecules of a gas or liquid (the adsorbate) on its exterior.

To mathematically model how much adsorbate is captured at a given concentration, scientists rely on the **Freundlich Adsorption Isotherm**. It provides a beautiful, empirical relationship:

$$ \frac{x}{m} = K \cdot C^{\frac{1}{n}} $$

Here, $\frac{x}{m}$ represents the mass of adsorbate per unit mass of adsorbent, $C$ is the equilibrium concentration, and $K$ and $n$ are constants specific to the system at a given temperature. This equation is the master key to unlocking our problem.

### Linearizing the Curve: The Power of Logarithms

Dealing with fractional exponents like $\frac{1}{n}$ can be algebraically tedious. How do we simplify this? We bring down the power using the magic of logarithms! By taking the base-10 logarithm on both sides, we transform the exponential curve into a crisp, straight line:

$$ \log\left(\frac{x}{m}ight) = \log K + \frac{1}{n} \log C $$

Notice how this perfectly mirrors the classic equation of a straight line, $y = mx + c$. Our y-axis is $\log\left(\frac{x}{m}ight)$, our x-axis is $\log C$, our slope is $\frac{1}{n}$, and our y-intercept is $\log K$. This linear form makes it incredibly easy to solve for our unknown constants using experimental data.

### Decoding the Constants: Using Experimental Data

The problem provides us with two sets of experimental observations. Let's plug in the reality of our experiment. For our first data point, when the concentration $C$ is $10$, the adsorbed amount $\frac{x}{m}$ is $4$. Substituting these into our linear equation gives:

$$ \log 4 = \log K + \frac{1}{n} \log 10 $$

The problem generously provides a handy tool: $\log 2 = 0.3$. Since $4 = 2^2$, we can easily calculate $\log 4 = 2 	imes 0.3 = 0.6$. And we know that $\log 10 = 1$. So, our first equation simplifies beautifully to:

$$ 0.6 = \log K + \frac{1}{n} \quad \dots (1) $$

Now, let's move to the second experimental observation. When the concentration increases to $16$, the adsorbed amount goes up to $10$. Substituting this gives:

$$ \log 10 = \log K + \frac{1}{n} \log 16 $$

Again, $\log 10 = 1$. What about $\log 16$? Well, $16 = 2^4$. Therefore, $\log 16 = 4 	imes \log 2 = 4 	imes 0.3 = 1.2$. Our second equation is now:

$$ 1 = \log K + \frac{1.2}{n} \quad \dots (2) $$

Look at what we have built: a neat system of two linear equations with two variables, $\log K$ and $\frac{1}{n}$. The easiest way to solve this is to eliminate $\log K$ by subtracting equation (1) from equation (2):

$$ 1 - 0.6 = \left(\log K + \frac{1.2}{n}ight) - \left(\log K + \frac{1}{n}ight) $$

$$ 0.4 = \frac{0.2}{n} \implies \frac{1}{n} = 2 $$

We've found our slope! With the slope in hand, finding the intercept is a walk in the park. Let's substitute $\frac{1}{n} = 2$ back into our simpler first equation:

$$ 0.6 = \log K + 2 \implies \log K = -1.4 $$

We now have all the constants that define our specific adsorption system.

### The Final Prediction: Calculating the Unknown

We have successfully decoded the isotherm for this specific fly ash and phenol system. The question now asks us to predict the future: what will be the adsorbed amount when the concentration is $20$? Let's set up the final equation by substituting $C = 20$, $\log K = -1.4$, and $\frac{1}{n} = 2$ into our logarithmic isotherm equation:

$$ \log\left(\frac{x}{m}ight) = -1.4 + 2 \log 20 $$

This is the final stretch. Let's calculate $\log 20$. Since $20 = 2 	imes 10$, we have $\log 20 = \log 2 + \log 10 = 0.3 + 1 = 1.3$. Multiplying that by our slope gives $2 	imes 1.3 = 2.6$. Now, add our intercept:

$$ \log\left(\frac{x}{m}ight) = -1.4 + 2.6 = 1.2 $$

How do we find the final value of $\frac{x}{m}$? We take the antilog:

$$ \frac{x}{m} = 10^{1.2} $$

Here is a brilliant mathematical trick to solve this without a calculator: recognize that $1.2 = 4 	imes 0.3$. Since we were given that $10^{0.3} = 2$, we can rewrite the expression:

$$ 10^{1.2} = 10^{4 	imes 0.3} = (10^{0.3})^4 = 2^4 = 16 $$

And there we have it! The final adsorbed concentration is exactly **$16	ext{ mg g}^{-1}$**.

Enter Numerical Value:

Visualized Solution

$Freundlich Adsorption Isotherm$

$Linearizing the Equation$

$Applying the First Data Point$

$Applying the Second Data Point$

$Solving for the Slope (\frac{1}{n})$

$Solving for the Intercept (lo g K)$

$Setting up for the Target Concentration$

$Final Calculation$

The Sigma Insight: Adsorption

The Magic of Adsorption

Linearizing the Curve

Decoding the Constants

The Final Prediction

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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; } Freundlich Adsorption Isotherm

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the First Data Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Second Data Point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for the Slope (n1​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for the Intercept (logK)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up for the Target Concentration

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