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

The concentrations of sulfuric acid in the original containers are 57% and 150%.

Step-by-step explanation:

To solve this problem, we can set up a system of equations using the given information. Let's use x to represent the concentration of the first solution and y to represent the concentration of the second solution.

We have the following two equations:

0.15(300) + 0.12(100) = 0.01(x)

0.15(600) + 0.12(500) = 0.01(y)

Simplifying these equations, we get:

45 + 12 = 0.01x

90 + 60 = 0.01y

57 = 0.01x

150 = 0.01y

Dividing both sides by 0.01, we find that x = 5700 and y = 15000. Therefore, the concentrations of sulfuric acid in the original containers are 5700/100 = 57% and 15000/100 = 150%.

Answer 2

Acid 1 = 25%

Acid 2 = 10%

Step-by-step 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.