A chemist has one solu6on that is 40% sulfuric acid and one that is 10% sulfuric acid. How much of each should she use to make 20 liters of a solu6on that is 28% sulfuric acid?

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.

Explanation:

For first solution of sulfuric acid:

C₁ = 40%, V₁ = ?

For second solution of sulfuric acid:

C₂ = 10%, V₂ = ?

For the resultant solution of sulfuric acid:

C₃ = 28%, V₃ = 20L

Also,

V₁ + V₂ = V₃ = 20L ... (1)

Using

C₁V₁ + C₂V₂ = C₃V₃

40*V₁ + 10*V₂ = 28*20

So,

40V₁ + 10V₂ = 560 ... (2)

Solving 1 and 2 as:

V₂ = 20 - V₁

Applying in 2

40V₁ + 10 (20 - V₁) = 560

40V₁ + 200 - 10V₁ = 560

30V₁ = 360

V₁ = 12 L

So,

V₂ = 20 - V₁ = 8L

12 L of 40% sulfuric acid solution and 8 L of 10% sulfuric acid solution are needed to make 20 L of sulfuric acid solution.