What mass of limestone do I need to neutralize a lake with pH 5.6 and a volume of #4.2 × 10# L?

Step 1: Determine the H⁺ concentration.

Step 2: Determine the H⁺ moles.

Step 3: Determine the CaCO3 mass.

2H + CaCO₃ → Ca²⁺ + H₂O + CO₂

To neutralize the lake, we need to calculate the amount of limestone needed to raise its pH to a neutral level, which is around pH 7. The neutralization reaction involves the reaction of calcium carbonate (limestone) with the acidic components in the lake water, typically hydrogen ions (H+).

First, we need to determine the amount of acid present in the lake water. Since the lake water has a pH of 5.6, we can calculate the concentration of hydrogen ions ((H^+)) using the formula:

[ \text{pH} = -\log[H^+] ]

[ [H^+] = 10^{-\text{pH}} ]

[ [H^+] = 10^{-5.6} ]

Next, we need to consider the balanced chemical equation for the neutralization reaction between calcium carbonate (CaCO3) and hydrogen ions:

[ \text{CaCO}_3 + 2H^+ \rightarrow \text{Ca}^{2+} + \text{CO}_2 + \text{H}_2\text{O} ]

From the balanced equation, we see that 1 mole of calcium carbonate reacts with 2 moles of hydrogen ions.

Now, we can calculate the moles of hydrogen ions in the lake water, and then determine the moles of calcium carbonate needed to neutralize them. Finally, we'll use the molar mass of calcium carbonate to find the mass needed.

[ \text{Volume of lake water} = 4.2 \times 10^6 , \text{L} ]

[ [H^+] = 10^{-5.6} ]

[ \text{Moles of } H^+ = [H^+] \times \text{Volume of lake water} ]

[ \text{Moles of } \text{CaCO}_3 = \frac{1}{2} \times \text{Moles of } H^+ ]

[ \text{Mass of } \text{CaCO}_3 = \text{Moles of } \text{CaCO}_3 \times \text{Molar mass of } \text{CaCO}_3 ]

By performing these calculations, you can find the mass of limestone needed to neutralize the lake.