Question 1092053: What amount of pure acid must be added to 300 mL of a 20% acid solution to produce a 75% acid solution? Found 2 solutions by ikleyn, greenestamps:Answer by ikleyn(52781) (Show Source):
Your equation is this "concentration" equation
= 0.75. (1)
where x is the unknown now volume of the pure acid to mix, in mL.
The numerator in the left side is the amount of the pure acid in the mixture;
the denominator is the volume of the 75% mixture,
so the ratio really is the new concentration.
To solve the equation (1), first multiply both sides by (x+300) to get
x + 0.2*300 = 0.75*(x+300).
Now simplify and complete the solution for x.
You can put this solution on YOUR website! I use a different method to solve mixture problems than what you almost always see in textbooks or on the internet.
The differences between the percentages of the ingredients and of the desired mixture exactly determine the ratio in which the two ingredients must be mixed.
In your problem, the percentages of the ingredients are 100 and 20, and the desired percentage is 75.
and
The two ingredients must be mixed in the ratio 25:55, or 5:11; the larger portion must be the 100% ingredient, since 75% is closer to 100% than to 20%. So set up a proportion knowing that you have 300 mL of the 20% acid solution:
You need to add 660 mL of pure acid to the 300 mL of 20% acid to get a mixture that is 75% acid.
