How much pure water must be mixed with 10 liters of a 25% acid solution to reduce it to a 10% acid solution?
Original Mixture: 10 liters, 25% acid = 0.25
Added mixture : x liters, 0% acid = 0 since it is pure water.
Total mixture: Volume = (10 + x), % = 10% = 0.1
Using mixtures formula:
Volume₁ %₁ + Volume₂%₂ = Total VolumeTotal%
100.25 + x0 = (10 + x)0.10
2.5 + 0 = 100.10 + 0.1x
2.5 = 1 + 0.1x
1 + 0.1x = 0.250
0.1x = 2.5 - 1
0.10 = 1.5
x = 1.5/0.1
x = 15
Amount of pure water to be added is 15 liters.
Hope this helps.