How many liters of pure acid must be mixed with 20 liters of a 15% acid solution to get a 20% mixture?

Question

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Answer 1

First, calculate the amount of acid in the 15% solution.

 

20 Liters * 0.15  yields 3 liters

 

 

Now, set up a ratio with acid as the numerator:

 

3/20 is the original ratio

 

Now, how many liters of acid to add ? I do not know, call it x

 

Then new ratio must be (3 + x)/ (20 + x)  since you are adding x liters of acid if must be added to the numerator (acid) and the solution (total liquid).

 

The point is that this new ratio must also equal a 20% soulution

 

So, (3 + x) / (20 + x) = 20/100 = 1/5 and you can cross-multiply to obtain 5(3+x) = 1(20+x)

 

Distribute the "5" and the "1" to get 15 + 5x = 20 + x

 

Now, subtract (15 + x) from both sides to obtain 4x = 5 which means x = 5/4 or 1.25

 

Therefore add 1.25 liters of pure acid to get 4.25 liters of acid in a 21.25 liter solution.

 

Check by multiplying (21.25) (.20).

 

You should get 4.25

 

 

Answer 2

Let x be the amount of pure acid needed.

 

There are a couple of things you know:

 

- You're starting with 20 liters of solution, and when you add the pure acid, you'll have 20 + x liters of solution

- The amount of acid in the 15% solution is 15% of 20 liters, or 3 liters, and when you add the pure acid, the amount of acid will be 3 + x

 

Since the concentration of acid in the new solution will be 20%, you can say that (3 + x) / (20 + x) will equal 20% or 0.2.

 

So that gives you an equation you can solve:

(3 + x) / (20 + x) = .2

(3 + x) = .2 (20 + x)

3 + x = 4 + .2x

x = 1 + .2x

.8x = 1

x = 1.25

 

So 1.25 liters of pure acid are needed