mass of CaO = 24.8 kg CaO
Explanation
volume of lake = 2.97 x 104 m3
volume of lake = 2.97 x 104 m3 * (1000 L / 1 m3)
volume of lake = 2.97 x 107 L
initial pH = 4.525
initial concentration of H+ = 10-pH
initial concentration of H+ = 10-4.525
initial concentration of H+ = 2.9854 x 10-5 M
initial moles of H+ = (initial concentration of H+) * (volume of lake)
initial moles of H+ = (2.9854 x 10-5 M) * (2.97 x 107 L)
initial moles of H+ = 886.66 mol
final pH = 7.00
final concentration of H+ = 10-pH
final concentration of H+ = 10-7.0
final concentration of H+ = 1.0 x 10-7 M
final moles of H+ = (final concentration of H+) * (volume of lake)
final moles of H+ = (1.0 x 10-7 M) * (2.97 x 107 L)
final moles of H+ = 2.97 mol
moles of H+ consumed = (initial moles of H+) - (final moles of H+)
moles of H+ consumed = (886.66 mol) - (2.97 mol)
moles of H+ consumed = 883.69 mol
Balanced reaction : Ca(OH)2 + 2 H+ Ca2+ + 2 H2O
moles of Ca(OH)2 required = (moles of H+ consumed) / 2
moles of Ca(OH)2 required = (883.69 mol) / 2
moles of Ca(OH)2 required = 441.84 mol
moles of CaO required = moles of Ca(OH)2 required
moles of CaO required = 441.84 mol
mass of CaO required = (moles of CaO required) * (molar mass CaO)
mass of CaO required = (441.84 mol) * (56.0774 g/mol)
mass of CaO required = 24777.5 g
mass of CaO required = 24.8 kg
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