Question

Phosphate can be removed from drinking-water supplies by treating water with Ca(OH)2: Ca(OH)2(aq) + PO43-(aq) Ca5OH(PO4)3(s) + OH-(aq) How much Ca(OH)2 is required to remove 90% of the PO43- from 4.5 x 106 L of drinking water containing 25 mg/L of PO43-?

Answer 1

Balanced reaction of Ca(OH)2 and PO4-3 is

5 Ca(OH2) aq + 3PO4^3- = Ca5OH(PO4)₃ (s) +9 OH-

so 3 mole of PO4^3- requires 5 moles of Ca(OH)2

drinking water contains 25 mg/L PO4^3-, water volume= 4.5*10⁶L

mass of water= Volume* concentration of PO4^3-=4.5*10⁶L *25 mg/L=112.5*10⁶ mg

since 1000mg= 1gm, 112.5*10⁶ mg= 112.5*10⁶ mg*10^-3 gm=112.5*1000gm

molar mass of PO43- = 31+4*16= 95 g/mole, moles of PO4^3- = 112.5*1000/95 =1184.2 moles

since 90% removal is envisaged, mole of PO4-3 to be removed= 1184.2*0.9= 1066 moles

3 mole of PO4^3- requies 5 mole of Ca(OH)2'

1066 moles of PO4-3 required 5*1066/3= 1777 moles, molar mass of Ca(OH)2= 40+2*17= 74

mass of Ca(OH)2 required= 1777*74 gm =131498 gm=131.5 Kg