To solve this problem, we can use the following steps:
Step 1: Calculate the amount of H+ ions in the lake.
pH is a measure of the concentration of hydrogen ions (H+) in a solution. The lower the pH, the higher the concentration of H+ ions. The pH of the lake is 4.045, so we can calculate the concentration of H+ ions using the following equation:
pH = -log[H+]
4.045 = -log[H+]
[H+] = 7.27 × 10^-5 mol/L
Step 2: Calculate the amount of hydroxide ions (OH-) needed to neutralize the H+ ions.
To raise the pH to 7.00, we need to add enough hydroxide ions (OH-) to neutralize the H+ ions in the lake. The balanced chemical equation for this reaction is:
Ca(OH)2 + 2H+ → Ca2+ + 2H2O
For every 2 H+ ions, we need 1 Ca(OH)2 molecule to neutralize them. The amount of Ca(OH)2 needed can be calculated using the following equation:
mol Ca(OH)2 = 0.5 × [H+] × V / 1000
where V is the volume of the lake in liters (8.99 × 10^7 L), and we divide by 1000 to convert from cubic meters to liters.
mol Ca(OH)2 = 0.5 × (7.27 × 10^-5 mol/L) × (8.99 × 10^7 L) / 1000
mol Ca(OH)2 = 3275.5 mol
Step 3: Calculate the mass of CaO needed to produce the required amount of Ca(OH)2.
To produce Ca(OH)2, we need to react CaO with water:
CaO + H2O → Ca(OH)2
The molar mass of CaO is 56.08 g/mol, so the mass of CaO needed can be calculated using the following equation:
mass CaO = mol CaO × molar mass CaO
mol CaO = mol Ca(OH)2 / 1 (from the balanced equation)
mass CaO = (mol Ca(OH)2 / 1) × (molar mass CaO / 1 mol CaO)
mass CaO = 3275.5 mol × 56.08 g/mol
mass CaO = 183,812.8 g = 183.8 kg (to 3 significant figures)
Therefore, 183.8 kg of CaO must be added to raise the pH of the lake from 4.045 to 7.00.