Question

100 million gallon per day of water of pH 8 and alkalinity (HCO3^-) of 200 mg/L as CaCO₃. If it is required to lower the pH from 8 to 7 by using 1 g/L of HCl stock solution, then how much HCl is required (gallon/day) while considering the above mentioned alkalinity?

Answer 1

Molar Mass of CaCO3 = 100 g/mol

=> Molar concentration of CaCO3 = 200*10^-3/100=2*10^-3 mol/litre=HCO3- concentration

Total Moles of HCO3- = Concentration*Total water qty = 2*10^-3*100*10^6*3.7=740000 moles/day

Final moles of HCO3- desired = 10^(Final pH-Initial pH)*Initial moles = 10^(7-8)*740000=74000 moles/day

Total HCO3- moles consumed = 740000-74000=666000 moles/day

CaCO3+2HCl=CaCl2+H2CO3

=>One mole CaCO3 requires 2 moles HCl

=> HCl moles required = 2*666000=1332000 moles/day

HCl molar mass = 36.5 g/mol

=>HCl g required = 1332000*36.5=48618000 g/day

Given, HCl solution density = 1g/l

3.7 litre=1gallon

=>HCl solution required = 48618000/1/3.7=13.14 million gallon/day