A mixture of 10% acid and 90% water is added to 5 liters of pure acid. The final mixture is 40% water.?

Answer 1

You start with X liters of the mixture

The total amount of water, divided by the total amount of acid/water mixture is 40%. We then solve for the amount of mixture that we had prior to adding the 5 L of pure acid.

.9x / (x + 5L) = .4

.9x = .4x + 2 L

.5x = 2L

x = 4L

You started with 4 Liters of the mixture

Answer 2

The question cannot be answered without more information. If the mixture was initial mixture were 95 liters the final mixture would be 85.5% water. The only way the final mixture would be 40% water is if you used 4 liters of the initial mixture. It would be 5.4 liters of pure acid and 3.6 liters of water giving you the 60/40 mixture.

Answer 3

This is really a chemistry question more than a food question, but:
It depends on whether you mean percent by volume or mass. Do you know the molarity? Is the volume of the initial mixture 1 liter? Is the acid a weak acid, or a strong acid? I wish I could help more, but this is an extremely vague question. This might help:
http://www.shodor.org/UNChem/basic/ab/index.html

Answer 4

This sounds like the type of question I envisioned when entering the beer, wine & spirits section