Question

A chemist has two large containers of sul- furic acid solution, with different concentrations of acid in each container. Blending 300 mL of the first solution and 600 mL of the second gives a mixture that is 15% acid, whereas blending 100 mL of the first with 500 mL of the sec- ond gives a 12 12 % acid mixture. What are the concentrations of sulfuric acid in the original containers?

Answer 1

Answer:

Acid 1 = 25%

Acid 2 = 10%

Explanation:

You can actually solve this using an equation system.

First, we know that the first mixture was made using 300 mL of the first acid (Let's call it X) and 600 mL of the second acid (Call it Y) giving a 15% acid concentration. The second mixture was made with 100 mL of X and 500 mL of Y giving a 12.5% concentration of mixed acid.

You should remember that concentration and volumen are relationed with the number of moles, so, if you multiply concentration with volume, you'll get the moles of that solution. If we apply the same principle here, we can know the original concentrations of both acids.

Writting concentrations as moles:

Moles X: 300X and 100X

Moles Y: 600Y  and 500Y

Using these expressions we can know the original concentrations:

(1) 300X + 600Y = 15*900 ----------> 300X + 600Y = 13500

(2) 100X + 500Y = 12.5*600 -------> 100x + 500Y = 7500

solving the value of X by sustitution we have:

From (1):

300X= 13500 - 600Y ----> X = 13500 - 600Y / 300 (3)

Replacing (3) in (2):

100(13500 - 600Y/300) + 500Y = 7500

4500 - 200Y + 500Y = 7500

300Y = 3000

Y = 3000/300 -----> Y = 10

Replacing this value in equation (3):

X = 13500 - 600(10) / 300

X = 25

Therefore the values of the original solutions are 25% for the first acid, and 10% the second acid.