First, we need to calculate the amount of PO43– in the 4.90×10^6 L of drinking water:
Amount of PO43– = 4.90×10^6 L * 25.0 mg/L = 1.225×10^8 mg
Since we want to remove 90.0% of the PO43–, we need to calculate the amount of PO43– remaining after treatment:
Amount of PO43– remaining = 0.1 * 1.225×10^8 mg = 1.225×10^7 mg
Next, we need to calculate the amount of Ca(OH)2 needed to precipitate this amount of PO43–:
1 mole of Ca(OH)2 can remove 3 moles of PO43–.
The molar mass of PO43– is 94.97 g/mol, so 1 mole of PO43– weighs 94.97 g.
Amount of Ca(OH)2 needed = (1.225×10^7 mg) * (1 g/1000 mg) * (1 mol PO43–/94.97 g) * (1 mol Ca(OH)2/3 mol PO43–) * (74.09 g/mol Ca(OH)2) = 31,328 g = 31.33 kg
Therefore, 31.33 kg of Ca(OH)2 is required to remove 90.0% of the PO43– from 4.90×10^6 L of drinking water containing 25.0 mg/L of PO43–.
Phosphate can be removed from drinking water supplies by treating the water with Ca(OH)2:
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How much Ca(OH)2 is required to remove 90.0% of the PO43– from 4.90×106 L of drinking water containing 25.0 mg/L of PO43–?
1 answer
1 answer