Question

500mL of a dilute phosphoric acid solution was prepared. If the concentration dropped from 36% to 2%, how much water was needed for the dilution?

Answer 1

We have V₁S₁ =V₂S₂  or V₂ = (V₁S₁ / S₂) ................. (1)

where V₁ = Initial volume, S₁ = Initial concentration, V₂ = Final volume, S₂ = Final concentration

Here V₁ = 500 mL, S₁ = 36% = 0.36, S₂ = 2% = 0.02

Now putting the values of all the terms in above equation (1) we get

V₂ = {(500 x 0.36) / 0.02} mL = 9000 mL

The amount of water that was needed for the dilution is (9000 - 500) mL = 8500 mL = 8.5 L

Therefore the amount of water that was needed for the dilution is 8.5 L.