How many milliliters of pure acid must be added to 150 milliliters of a 70% solution of acid to obtain a 80% solution?
let the amount to be added be x ml
.7(150) + x = .8(150+x)
times 10
7(150) + 10x = 8(150+x)
solve for x
To solve this problem, we need to determine how much pure acid we need to add to the existing 70% acid solution in order to achieve a desired concentration of 80%.
Let's break down the given information:
- We have a 150 milliliters of a 70% acid solution.
- We want to add some pure acid (100% concentration) to this solution.
- The desired concentration of the final solution is 80%.
To find the amount of pure acid needed, we can use the concept of concentration and apply the following formula:
(C1 * V1) + (C2 * V2) = C3 * V3
Where:
C1 refers to the concentration of the initial solution,
V1 refers to the volume of the initial solution,
C2 refers to the concentration of the pure acid we add,
V2 refers to the volume of the pure acid we add,
C3 refers to the concentration of the final solution, and
V3 refers to the volume of the final solution.
In this case, we have:
C1 = 70% = 0.7
V1 = 150 milliliters
C2 = 100% = 1.0
C3 = 80% = 0.8
We need to solve for V2, the volume of pure acid we need to add.
Substituting the given values into the formula, we have:
(0.7 * 150) + (1.0 * V2) = 0.8 * (150 + V2)
Now we can solve for V2:
(0.7 * 150) + V2 = 0.8 * (150 + V2)
105 + V2 = 120 + 0.8V2
V2 - 0.8V2 = 120 - 105
0.2V2 = 15
V2 = 15 / 0.2
V2 = 75
Therefore, you would need to add 75 milliliters of pure acid to the 150 milliliters of the 70% acid solution to obtain an 80% acid solution.