Let's break down the question step by step: ### **Given:** - **Toluene (C₇H₈)** is contaminating water at 4.25 ppm. - **Starting pH = 3.0** (unbuffered system). - **Fe(OH)₃** serves as electron acceptor, reduced to Fe²⁺. - **Toluene is oxidized:** \(\text{C}_7\text{H}_8\) → 3 mol acetate (\(\text{C}_2\text{H}_3\text{O}_2^-\)) + 1 mol CO₂. - **Question:** How many mol protons (H⁺) are liberated/consumed in this reaction? Will pH increase or decrease? --- ## **Step 1: Write the Balanced Reaction** The oxidation reaction is: \[ \text{C}_7\text{H}_8 + \text{H}_2\text{O} \rightarrow 3\ \text{C}_2\text{H}_3\text{O}_2^- + \text{CO}_2 + x\ \text{H}^+ \] Let's balance the atoms: **Toluene:** C₇H₈ **Acetate:** C₂H₃O₂⁻ - 3 acetate: 3 × C₂H₃O₂⁻ = C₆H₉O₆⁻ - 1 CO₂: C₁O₂ **Total:** - C: 6 (from acetate) + 1 (from CO₂) = 7 - H: 9 (from acetate) - O: 6 (from acetate) + 2 (from CO₂) = 8 **Reactant side:** - C₇H₈ + H₂O - C: 7 - H: 8 + 2 = 10 - O: 1 **Product side:** - Acetate + CO₂: C₇, H₉, O₈ Balance H and O by adding H⁺ as needed. --- ## **Step 2: Compare H and O Atoms** #### **Hydrogen:** - Reactants: 8 (from C₇H₈) + 2 (from H₂O) = 10 - Products: 9 (from 3 acetate) - Difference: 10 - 9 = **1 H left over** (must be released as H⁺) #### **Oxygen:** - Reactants: 1 (from H₂O) - Products: 6 (from acetate) + 2 (from CO₂) = 8 - Need 7 more O (comes from Fe(OH)₃ reduction, not from H₂O alone, but that is a redox balancing detail; in terms of protons, just focus on H) --- ## **Step 3: Proton Accounting** From the difference above, **1 proton (H⁺) is liberated per mole of toluene oxidized**. --- ## **Step 4: Effect on pH** If 1 H⁺ is released per reaction, and no buffer is present, **the solution becomes more acidic**. **pH will decrease**. --- ### **Final Answers:** **1. Number of protons (H⁺) liberated/consumed:** - **1 mole of H⁺ is liberated (released) per mole of toluene oxidized.** **2. Effect on pH:** - **The pH will decrease** (the water will become more acidic). --- **Summary Table:** | Reaction | H⁺ change | pH effect | |-----------------------------|---------------|---------------| | Toluene → 3 acetate + CO₂ | 1 H⁺ released | pH decreases |

Let's carefully analyze the problem step by step. --- ### **Given Information:** - Contaminant: Toluene (\(\mathrm{C_7H_8}\)) - Toluene amount: 4.25 ppm in water - Initial pH: 3.0 - Reaction: \(\mathrm{C_7H_8} \rightarrow 3\ \mathrm{C_2H_3O_2^-} + \mathrm{CO_2} + x\ \mathrm{H^+}\) - Toluene oxidation produces acetate and CO\(_2\). - The system involves Fe(OH)\(_3\) as an electron acceptor (reduction to Fe\(^{2+}\)), but the main focus here is on proton transfer and pH change. --- ### **Step 1: Convert ppm to mols of toluene** - 1 ppm = 1 mg/L - Toluene molar mass: \(\mathrm{C_7H_8}\): \(7 \times 12.01 + 8 \times 1.008 = 84.07 + 8.064 = 92.13\, \mathrm{g/mol}\) - Concentration: 4.25 ppm = 4.25 mg/L - Moles of toluene per liter: \[ \frac{4.25\, \mathrm{mg}}{92,130\, \mathrm{mg/mol}} \approx 4.25 \times 10^{-3} \text{ mg} / 92,130\, \text{mg/mol} \] \[ \approx \frac{4.25 \times 10^{-3}}{92,130} \approx 4.62 \times 10^{-8}\, \text{mol} \] --- ### **Step 2: Write the oxidation reaction** From the problem, the oxidation of toluene yields: \[ \mathrm{C_7H_8} \rightarrow 3\, \mathrm{C_2H_3O_2^-} + \mathrm{CO_2} + x\, \mathrm{H^+} \] --- ### **Step 3: Balance the atoms and charges** Let's balance the atoms: - **C:** Left: 7 (from toluene) Right: 3 × 2 (acetate) + 1 (CO\(_2\)) = 6 + 1 = 7 (balanced) - **H:** Left: 8 (toluene) Right: 3 × 3 (from acetate) = 9 (hydrogens in acetate) Plus possible H\(^+\) produced or consumed. - **O:** Right: 3 × 2 (acetate) + 2 (CO\(_2\)) = 6 + 2 = 8 Left: 1 (from toluene) + water molecules involved in oxidation; but since the problem states the net reaction, focus on hydrogen and proton transfer. --- ### **Step 4: Determine the number of protons** Compare H: - Left: 8 H in toluene - Right: 3 acetate molecules contain 3 × 3 = 9 H - To balance H, the net H produced or consumed: \[ 8\, \text{(left)} \quad \text{vs} \quad 9\, \text{(right)} \] Difference: \[ 9 - 8 = 1 \] This indicates **1 H\(^+\)** is released during the oxidation, because the right side has more H than the left. **Thus, for each mole of toluene oxidized, 1 mole of H\(^+\) is produced.** --- ### **Step 5: Total protons released** Number of moles of toluene in the sample: \[ 4.62 \times 10^{-8}\ \text{mol} \] Number of protons released: \[ = 1 \times 4.62 \times 10^{-8} = 4.62 \times 10^{-8}\ \text{mol} \] --- ### **Step 6: Effect on pH** - Initial pH = 3.0, which corresponds to: \[ [\mathrm{H^+}] = 10^{-3} = 1 \times 10^{-3}\ \mathrm{M} \] - The amount of protons released is very small (\(4.62 \times 10^{-8}\) mol in 1 L), which is: \[ \Delta [\mathrm{H^+}] = 4.62 \times 10^{-8}\ \mathrm{M} \] - Since the initial \([\mathrm{H^+}]\) is \(10^{-3}\) M, the relative change is negligible: \[ \frac{\Delta [\mathrm{H^+}]}{[\mathrm{H^+}]} \approx 4.62 \times 10^{-8} / 10^{-3} = 4.62 \times 10^{-5} \] which is negligible, meaning the pH change is very small. - **However**, in the real system, the oxidation releases protons, increasing acidity: \[ \boxed{\text{pH will decrease slightly}} \] --- ### **### Final answers:** | **Quantity** | **Value** | |----------------|------------| | **Protons released per mole of toluene** | **1 mol H\(^+\)** | | **Total mol of protons released in the contaminated water** | **\(4.62 \times 10^{-8}\) mol** | | **Effect on pH** | **Slight decrease (more acidic)** | --- ### **Summary:** - **Number of mol protons released:** approximately **\(4.62 \times 10^{-8}\) mol**. - **pH change:** the system's pH will **decrease slightly** due to proton release. --- **Note:** The actual pH shift is minuscule at such low contaminant levels, but theoretically, proton release causes acidification. --- **Hope this detailed step-by-step explanation helps!**

In the oxidation of toluene, each mole releases one proton (H⁺), increasing acidity. The small amount of toluene (4.25 ppm) corresponds to approximately 4.62 × 10⁻⁸ mol, thus releasing an equivalent amount of protons. This proton release causes a slight decrease in pH, making the solution more acidic. Overall, proton generation during oxidation influences the water's pH marginally, tending toward acidity.