Question

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

Answer 1

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"`

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"`

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"_3`