\( 1<\begin{array}{l}\text { Consider a situation in which acid rain has heavily polluted a lake to a level of pH 5. An unscrupulous chemical } \\ \text { company dumps some acid into the lake illegally. Assume that the lake contains } 100 \text { million gallons of water and that } 1 \\ \text { the company dumps } 100,000 \text { gallons of acid with pH 3. Complete parts (a) through (d) below. }\end{array} \)

To solve the problem, we need to analyze the situation step by step, focusing on the pH levels and the volumes of water and acid involved. ### Known Conditions: 1. The lake has a volume of \( V_L = 100 \) million gallons. 2. The initial pH of the lake is \( \text{pH}_L = 5 \). 3. The company dumps \( V_A = 100,000 \) gallons of acid with a pH of \( \text{pH}_A = 3 \). ### Step 1: Convert pH to Hydrogen Ion Concentration The pH scale is logarithmic, and the concentration of hydrogen ions \([H^+]\) can be calculated using the formula: \[ [H^+] = 10^{-\text{pH}} \] #### For the lake: \[ [H^+]_L = 10^{-5} \text{ M} \] #### For the acid: \[ [H^+]_A = 10^{-3} \text{ M} \] ### Step 2: Calculate the Total Volume of the Mixture The total volume after dumping the acid is: \[ V_T = V_L + V_A = 100,000,000 + 100,000 = 100,100,000 \text{ gallons} \] ### Step 3: Calculate the Total Moles of Hydrogen Ions To find the total moles of hydrogen ions in both the lake and the acid, we need to convert the volumes from gallons to liters (1 gallon = 3.78541 liters). #### Convert volumes: \[ V_L = 100,000,000 \text{ gallons} \times 3.78541 \text{ L/gallon} \approx 378,541,000 \text{ L} \] \[ V_A = 100,000 \text{ gallons} \times 3.78541 \text{ L/gallon} \approx 378,541 \text{ L} \] #### Calculate moles of hydrogen ions: \[ \text{Moles of } H^+ \text{ in the lake} = [H^+]_L \times V_L = 10^{-5} \times 378,541,000 \] \[ \text{Moles of } H^+ \text{ in the acid} = [H^+]_A \times V_A = 10^{-3} \times 378,541 \] ### Step 4: Calculate Total Moles of Hydrogen Ions Now we can calculate the total moles of hydrogen ions in the mixture: \[ \text{Total moles of } H^+ = \text{Moles from lake} + \text{Moles from acid} \] ### Step 5: Calculate the New Concentration of Hydrogen Ions Finally, we can find the new concentration of hydrogen ions in the total volume: \[ [H^+]_{new} = \frac{\text{Total moles of } H^+}{V_T} \] ### Step 6: Calculate the New pH The new pH can be calculated using: \[ \text{pH}_{new} = -\log([H^+]_{new}) \] Now, let's perform the calculations for the moles of hydrogen ions and the new pH. I'll calculate the moles of hydrogen ions in the lake and the acid, and then find the new pH. Calculate the value by following steps: - step0: Calculate: \(10^{-3}\times 378541\) - step1: Evaluate the power: \(\frac{1}{1000}\times 378541\) - step2: Multiply: \(\frac{378541}{1000}\) Calculate or simplify the expression \( 10^{-5} * 378541000 \). Calculate the value by following steps: - step0: Calculate: \(10^{-5}\times 378541000\) - step1: Multiply the terms: \(378541000\times 10^{-5}\) - step2: Express with a positive exponent: \(378541000\times \frac{1}{10^{5}}\) - step3: Rewrite the expression: \(1000\times 378541\times \frac{1}{10^{5}}\) - step4: Rewrite the expression: \(10^{3}\times 378541\times \frac{1}{10^{5}}\) - step5: Reduce the numbers: \(378541\times \frac{1}{10^{2}}\) - step6: Multiply: \(\frac{378541}{10^{2}}\) - step7: Simplify: \(\frac{378541}{100}\) Calculate or simplify the expression \( 3785.41 + 378.541 \). Calculate the value by following steps: - step0: Calculate: \(3785.41+378.541\) - step1: Add the numbers: \(4163.951\) Calculate or simplify the expression \( 4163.951 / (100000000 + 100000) * 3.78541 \). Calculate the value by following steps: - step0: Calculate: \(\frac{4163.951}{\left(100000000+100000\right)}\times 3.78541\) - step1: Remove the parentheses: \(\frac{4163.951}{100000000+100000}\times 3.78541\) - step2: Add the numbers: \(\frac{4163.951}{100100000}\times 3.78541\) - step3: Divide the terms: \(\frac{378541}{9100000000}\times 3.78541\) - step4: Convert the expressions: \(\frac{378541}{9100000000}\times \frac{378541}{100000}\) - step5: Multiply the fractions: \(\frac{378541\times 378541}{9100000000\times 100000}\) - step6: Multiply the terms: \(\frac{143293288681}{9100000000\times 100000}\) - step7: Multiply the terms: \(\frac{143293288681}{910000000000000}\) Calculate or simplify the expression \( -\log(0.000157) \). Calculate the value by following steps: - step0: Calculate: \(-\log_{10}{\left(0.000157\right)}\) - step1: Convert the expressions: \(-\log_{10}{\left(\frac{157}{1000000}\right)}\) - step2: Rewrite the expression: \(-\log_{10}{\left(157\right)}+6\) ### Step-by-Step Solution 1. **Calculate Moles of Hydrogen Ions in the Lake:** \[ \text{Moles of } H^+ \text{ in the lake} = 10^{-5} \times 378,541,000 \approx 3785.41 \text{ moles} \] 2. **Calculate Moles of Hydrogen Ions in the Acid:** \[ \text{Moles of } H^+ \text{ in the acid} = 10^{-3} \times 378,541 \approx 378.541 \text{ moles} \] 3. **Total Moles of Hydrogen Ions:** \[ \text{Total moles of } H^+ = 3785.41 + 378.541 \approx 4163.951 \text{ moles} \] 4. **Total Volume of the Mixture:** \[ V_T = 100,000,000 + 100,000 = 100,100,000 \text{ gallons} \approx 378,541,000 \text{ L} \] 5. **New Concentration of Hydrogen Ions:** \[ [H^+]_{new} = \frac{4163.951}{378,541,000} \approx 0.000157 \text{ M} \] 6. **Calculate the New pH:** \[ \text{pH}_{new} = -\log(0.000157) \approx 3.8041 \] ### Final Answer The new pH of the lake after the acid is dumped is approximately **3.80**. If you have any further questions or need additional parts (a) through (d) completed, please let me know!