What mass of limestone do I need to neutralize a lake with pH 5.6 and a volume of 4.2 × 104.2×10 L?

1 Answer
Ernest Z.
Dec 13, 2014

You need 530 kg of limestone to neutralize the lake.

Explanation:

Step 1. Calculate the concentration of H⁺

["H"^+] = 10^("-pH")" mol/L" =10^-5.6" mol/L" = 2.5 × 10^-6" mol/L"[H+]=10-pH mol/L=10−5.6 mol/L=2.5×10−6 mol/L

Step 2. Calculate the moles of H⁺

Moles = 4.2 ×10^9" L" × (2.5 × 10^-6" mol")/"1 L" = 1.05 × 10^4" mol"4.2×109 L×2.5×10−6 mol1 L=1.05×104 mol

Step 3. Calculate the mass of CaCO₃

CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂

Mass of CaCO₃ = 1.05 × 10^4" mol H"^+ × ("1 mol CaCO"_3)/("2 mol H"^+) × ("100.09 g CaCO"_3)/("1 mol CaCO"_3) × ("1 kg CaCO"_3)/("1000 g CaCO"_3) = "530 kg CaCO"_31.05×104 mol H+×1 mol CaCO32 mol H+×100.09 g CaCO31 mol CaCO3×1 kg CaCO31000 g CaCO3=530 kg CaCO3

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