Let's solve the problem **step by step**. --- ### **Given:** 1. **Isotherm equation:** \( q = 39.3 x^{0.55} \) - \( q \) = mg gentamicin adsorbed per cm³ of carbon - \( x \) = mg gentamicin per liter solution (equilibrium concentration) 2. **Volume of activated carbon (\( V_{carbon} \)):** \( 10 \) cm³ 3. **Volume of solution (\( V_{solution} \)):** \( 6 \) liters 4. **Initial gentamicin concentration (\( C_0 \)):** \( 50 \) mg/L 5. **Find:** Percent recovery (i.e., percent of gentamicin removed from solution and adsorbed onto the carbon) --- ## **Step 1: Set up a mass balance** Let: - \( x \) = equilibrium gentamicin concentration in solution (mg/L) - \( q \) = equilibrium gentamicin adsorbed per cm³ of carbon (mg/cm³) **Initial amount of gentamicin:** \[ \text{Initial amount} = C_0 \times V_{solution} = 50 \text{ mg/L} \times 6 \text{ L} = 300 \text{ mg} \] **At equilibrium:** - Amount in solution: \( x \times V_{solution} \) - Amount adsorbed: \( q \times V_{carbon} \) \[ \text{Total gentamicin} = (\text{in solution}) + (\text{adsorbed}) \] \[ 300 = x \times 6 + q \times 10 \] --- ## **Step 2: Substitute the isotherm** Given: \( q = 39.3 x^{0.55} \) Substitute into the balance: \[ 300 = 6x + 10 \times [39.3 x^{0.55}] \] \[ 300 = 6x + 393 x^{0.55} \] --- ## **Step 3: Solve for \( x \)** \[ 6x + 393 x^{0.55} = 300 \] Let’s define \( y = x^{0.55} \Rightarrow x = y^{1/0.55} \). But it’s easier to just solve numerically for \( x \). Let's try values: ### Try \( x = 0 \): \[ 6(0) + 393(0) = 0 \] Too low. ### Try \( x = 1 \): \[ 6(1) + 393(1) = 6 + 393 = 399 \] Too high. ### Try \( x = 0.5 \): \( 0.5^{0.55} = e^{0.55 \ln 0.5} = e^{0.55 \times -0.6931} = e^{-0.3812} = 0.6827 \) \[ 6(0.5) + 393(0.6827) = 3 + 268.48 = 271.48 \] Still low. ### Try \( x = 0.6 \): \( 0.6^{0.55} = e^{0.55 \times \ln 0.6} = e^{0.55 \times -0.5108} = e^{-0.281} = 0.7549 \) \[ 6(0.6) + 393(0.7549) = 3.6 + 296.18 = 299.78 \] Very close to 300! ### Try \( x = 0.61 \): \( \ln 0.61 = -0.4930 \) \( 0.55 \times -0.4930 = -0.2712 \) \( e^{-0.2712} = 0.7627 \) \[ 6(0.61) + 393(0.7627) = 3.66 + 299.73 = 303.39 \] Now it's slightly over 300. ### Try \( x = 0.595 \): \( \ln 0.595 = -0.5194 \) \( 0.55 \times -0.5194 = -0.2857 \) \( e^{-0.2857} = 0.7515 \) \[ 6(0.595) + 393(0.7515) = 3.57 + 295.09 = 298.66 \] Still a bit low. ### Interpolate between x=0.595 (298.66) and x=0.6 (299.78): The difference between 298.66 and 299.78 is 1.12 for a 0.005 x change. So, for \( 300 - 299.78 = 0.22 \), the fraction is \( 0.22 / 1.12 = 0.196 \). So, the solution is about \( x = 0.6 + 0.196 \times 0.005 \approx 0.6 + 0.001 = 0.601 \). But let's just use \( x \approx 0.6 \) mg/L as the **equilibrium concentration**. --- ## **Step 4: Calculate amount adsorbed** \[ \text{Amount adsorbed} = q \times V_{carbon} \] First, find \( q \): \[ q = 39.3 x^{0.55} \] \[ x = 0.6 \implies x^{0.55} = 0.7549 \] \[ q = 39.3 \times 0.7549 = 29.67 \text{ mg/cm}^3 \] So, \[ \text{Total adsorbed} = 29.67 \times 10 = 296.7 \text{ mg} \] --- ## **Step 5: Percent recovery** \[ \text{Percent recovery} = \frac{\text{adsorbed gentamicin}}{\text{initial gentamicin}} \times 100\% \] \[ = \frac{296.7}{300} \times 100\% = 98.9\% \] --- ### **Final Answers** **a) The percent recovery you can expect is approximately:** \[ \boxed{98.9\%} \] **b) The equilibrium concentration in solution is approximately:** \[ \boxed{0.6\,\, \text{mg/L}} \] --- **Summary of steps:** 1. Set up the mass balance: \( 300 = 6x + 10q \) 2. Substitute the isotherm: \( q = 39.3 x^{0.55} \) 3. Solve for \( x \) numerically: \( x \approx 0.6 \) mg/L 4. Calculate \( q \): \( q \approx 29.7 \) mg/cm³ 5. Find the amount adsorbed: \( 296.7 \) mg 6. Compute percent recovery: \( 98.9\% \) --- **Final boxed answer:** \[ \boxed{98.9\%} \] recovery of gentamicin can be expected.