The acidic water of a lake (V = 4 × 10⁸ L)must be neutralized with lime,CaO.It will require a minimum of ________ grams of lime to adjust the pH from 5.6 to 6.5.

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To neutralize the acidic water and adjust the pH from 5.6 to 6.5, we need to calculate the amount of lime (CaO) required. The pH scale is logarithmic, meaning a pH change from 5.6 to 6.5 represents a 10-fold decrease in the hydrogen ion concentration ([H+]). First, calculate the initial and final [H+] concentrations using the pH formula:$$ \text{pH} = -\log[H^+] $$For pH 5.6:$$ [H^+]_{\text{initial}} = 10^{-5.6} \, \text{M} $$For pH 6.5:$$ [H^+]_{\text{final}} = 10^{-6.5} \, \text{M} $$The change in [H+] needed is the difference between these two concentrations:$$ \Delta[H^+] = 10^{-5.6} - 10^{-6.5} \, \text{M} $$Given the volume of the lake is $4 \times 10^{18}$ L, convert this volume to cubic meters to match the concentration units (assuming 1 L = 1 dm$^3$). However, for simplicity and direct calculation, we can stay in liters since the concentration change will be applied directly to the volume in liters.To neutralize the [H+], lime (CaO) reacts with water to form Ca$^{2+}$ and OH$^{-}$, which then neutralizes H$^+$ ions. The molar mass of CaO is approximately 56 g/mol.The stoichiometry of the reaction is such that 1 mole of CaO neutralizes 2 moles of H$^+$ ions. Therefore, the amount of CaO needed can be calculated by first finding the moles of H$^+$ to be neutralized and then converting that to moles of CaO.$$ \text{Moles of } H^+ = \Delta[H^+] \times \text{Volume of lake in L} $$$$ \text{Moles of CaO} = \frac{1}{2} \times \text{Moles of } H^+ $$$$ \text{Mass of CaO} = \text{Moles of CaO} \times \text{Molar mass of CaO} $$Plugging in the numbers:$$ \Delta[H^+] \approx 10^{-5.6} - 10^{-6.5} \approx 2.51 \times 10^{-6} \, \text{M} $$$$ \text{Moles of } H^+ \approx 2.51 \times 10^{-6} \times 4 \times 10^{18} $$$$ \text{Moles of CaO} \approx \frac{1}{2} \times 2.51 \times 10^{12} $$$$ \text{Mass of CaO} \approx 1.255 \times 10^{12} \times 56 $$This calculation simplifies to an order of magnitude estimate, given the choices provided. The exact arithmetic might involve more precise numbers, but the key is understanding the process of converting pH change to a requirement for lime, considering the volume of the lake and the stoichiometry of the neutralization reaction. The closest match to the calculation process provided is in the order of $10^{14}$ grams, which corresponds to choice A, acknowledging that the detailed calculation steps are simplified for conceptual explanation.

The Acidic Water of a Lake (V = 4 ×