How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?
amount of pure acid to be added -- x L
1(x) + .3(60-x) = .5(60)
10x + 3(60-x) = 5(60)
10x + 180 - 3x = 300
7x = 120
x = 120/7 or appr 17.14 L
To solve this problem, we can break it down into two parts:
1. Finding the amount of 100% pure acid needed.
2. Finding the amount of 30% acid solution needed.
Let's begin with the first part:
1. Finding the amount of 100% pure acid needed:
Let's assume that x liters of pure acid need to be added.
Now, the concentration of the pure acid is 100%, so the amount of acid in x liters would be x liters * 100% = x liters.
Now, let's move on to the second part:
2. Finding the amount of 30% acid solution needed:
We know that the total volume of the final solution is 60L and it needs to be 50% acid. Since the other part consists of pure acid, the concentration of the 30% acid solution would be (50% - 100%) = -50%.
Let's assume that y liters of 30% acid solution need to be added.
The amount of acid contributed by the 30% acid solution will be y liters * 30% = y * 0.3 liters.
Now, let's set up an equation based on the amount of acid in the final solution:
Amount of acid from pure acid + Amount of acid from 30% acid solution = Total amount of acid in the final solution
x liters + y * 0.3 liters = 60L * 50%
Now, let's convert the percentages into decimals:
x liters + y * 0.3 liters = 60L * 0.5
Simplifying further:
x + 0.3y = 30
Now we have two equations:
x + 0.3y = 30 (1)
x + y = 60 (2)
We can solve this system of equations using substitution or elimination.
Let's solve it using elimination:
Multiply equation (2) by 0.3:
0.3x + 0.3y = 18 (3)
Now subtract equation (1) from equation (3):
0.3x + 0.3y - x - 0.3y = 18 - 30
-0.7x = -12
Dividing both sides by -0.7:
x = -12 / -0.7
x = 17.14
Therefore, approximately 17.14 liters of pure acid must be added to the 30% acid solution to make a 60L solution that is 50% acid.