Answer:
There should be added 24.33g of EDTA
Explanation:
EDTA + Ca2+ → Ca(EDTA)
MM of EDTA = 292.24g/mole
MM of Ca2+ =40g/mole
An aqueos waste contains 20 mg/L Calcium, and collects the waste in 44 gallon drums.
The concentration of Ca2+ ions = 20mg/L = 0.02g/L
We know 1 gallon = 3.785L
The mass of calcium ions is in 44 gallons
⇒44 gallons = 3.785 * 44 = 166.54 L
Since the aqueous waste contains 0.02g/L calcium
in 44 gallons or 166.54 L ⇒ 166.54 * 0.02g = 3.308g
MM of Ca2+ = 40g/mole
⇒40g of Ca2+ reacts with 292.24g of EDTA
⇒ (292.24/40 ) * 3.308g = 24.33 g
There should be added 24.33g of EDTA