A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid solution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?

The ratio acid to water in the mixture is 2:3

Explanation:

Let the volume of 20% acid solution used to make the mixture = x units

So, the volume of 50% acid solution used to make the mixture = 2x units

Total volume of the mixture = x + 2x = 3x units

For 20% acid solution:

C₁ = 20%, V₁ = x

For 50% acid solution:

C₂ = 50%, V₂ = 2x

For the resultant solution of sulfuric acid:

C₃ = ?, V₃ = 3x

Using

C₁V₁ + C₂V₂ = C₃V₃

20*x + 50*2x = C₃*3x

So,

20 + 50*2 = C₃*3

Solving

120 = C₃*3

C₃ = 40 %

Thus, for the resultant mixture,

Acid percentage = 40%

Water percentage = (100 - 40) % = 60%

Ratio acid to water in the mixture = 40:60 = 2:3