Question

A lake has been acidified to a pH of 4.525. If the lake has a volume of 2.97x104 cubic meters, what mass, in kg, of CaO must be added to raise the pH to a value of 7.00? Hint: The reaction of CaO with water is: CaO + H2O– Ca(OH)2 kg CaO

Answer 1

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+ $\rightarrow$ 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