Let's solve the question step-by-step using the **Langmuir isotherm** in its **linear form**. --- ## **Step 1: Langmuir Isotherm Linear Form** The Langmuir isotherm equation is: \[ q_e = \frac{A_{max} K C_e}{1 + K C_e} \] Linear form: \[ \frac{C_e}{q_e} = \frac{1}{A_{max}K} + \frac{C_e}{A_{max}} \] where: - \( C_e \): equilibrium concentration of adsorbate (mg/L) - \( q_e \): amount adsorbed per g biomass (mg/g) - \( K \): Langmuir constant (L/mg) - \( A_{max} \): maximum adsorption capacity (mg/g) --- ## **Step 2: Organize Data** | \( C_e \) (mg/L) | \( q_e \) (mg/g) | \( \frac{C_e}{q_e} \) (L/g) | |------------------|------------------|-----------------------------| | 1 | .8 | 1/.8 = 1.25 | | 2 | 1.5 | 2/1.5 = 1.33 | | 5 | 3. | 5/3. = 1.67 | | 10 | 4.2 | 10/4.2 ≈ 2.38 | --- ## **Step 3: Plot \( \frac{C_e}{q_e} \) vs \( C_e \) and Linear Regression** The equation \( \frac{C_e}{q_e} = \frac{1}{A_{max}K} + \frac{C_e}{A_{max}} \) is in the form \( y = mx + b \), where: - \( y = \frac{C_e}{q_e} \) - \( x = C_e \) - Slope \( m = \frac{1}{A_{max}} \) - Intercept \( b = \frac{1}{A_{max}K} \) Let's perform linear regression on the data points: | \( x \) | \( y \) | |---------|----------| | 1 | 1.25 | | 2 | 1.33 | | 5 | 1.67 | | 10 | 2.38 | Find the slope (\( m \)) and intercept (\( b \)) using the least squares method: #### Calculate Averages \[ \bar{x} = \frac{1+2+5+10}{4} = 4.5 \] \[ \bar{y} = \frac{1.25+1.33+1.67+2.38}{4} = 1.6575 \] #### Calculate Slope \( m \): \[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \] \[ = \frac{(1-4.5)(1.25-1.6575) + (2-4.5)(1.33-1.6575) + (5-4.5)(1.67-1.6575) + (10-4.5)(2.38-1.6575)} {(1-4.5)^2 + (2-4.5)^2 + (5-4.5)^2 + (10-4.5)^2} \] Calculate: - \((1-4.5) = -3.5, (1.25-1.6575) = -.4075, \rightarrow (-3.5)*(-.4075) = 1.426\) - \((2-4.5) = -2.5, (1.33-1.6575) = -.3275, \rightarrow (-2.5)*(-.3275) = .819\) - \((5-4.5) = .5, (1.67-1.6575) = .0125, \rightarrow .5*.0125 = .00625\) - \((10-4.5) = 5.5, (2.38-1.6575) = .7225, \rightarrow 5.5*.7225 = 3.974\) Sum numerator: \(1.426 + .819 + .00625 + 3.974 = 6.225\) - \((1-4.5)^2 = 12.25\) - \((2-4.5)^2 = 6.25\) - \((5-4.5)^2 = .25\) - \((10-4.5)^2 = 30.25\) Sum denominator: \(12.25 + 6.25 + .25 + 30.25 = 49\) So, \[ m = \frac{6.225}{49} = .127 \] #### Intercept \( b \): \[ b = \bar{y} - m \bar{x} = 1.6575 - (.127 \times 4.5) = 1.6575 - .5715 = 1.086 \] --- ## **Step 4: Solve for \( A_{max} \) and \( K \)** \[ m = \frac{1}{A_{max}} \implies A_{max} = \frac{1}{m} = \frac{1}{.127} \approx 7.87 \text{ mg/g} \] \[ b = \frac{1}{A_{max} K} \implies K = \frac{1}{A_{max} b} = \frac{1}{7.87 \times 1.086} \approx \frac{1}{8.545} \approx .117 \text{ L/mg} \] --- ## **Step 5: Final Answers** ### **Summary Table** | Parameter | Value | |-------------------|-----------------| | \( K \) | .12 L/mg | | \( A_{max} \) | 7.9 mg/g | --- ### **Final Answers (rounded to 2 significant figures):** \[ \boxed{ K = .12\ \text{L/mg} } \] \[ \boxed{ A_{max} = 7.9\ \text{mg/g} } \] --- ### **Explanation** - Plotted \( \frac{C_e}{q_e} \) vs. \( C_e \), used the linear form of Langmuir. - Used linear regression to find slope and intercept. - Calculated \( K \) and \( A_{max} \) from slope and intercept. If you need the plot, let me know!