Pure water is being added to a 25% solution of 120 milliliters of hydrochloric acid. How much water should be added to reduce it to a 15% mixture ?
here's one way. You want the concentration to be reduced by a factor of 15/25 = 3/5
So, the volume must be increased by a factor of 5/3.
5/3 * 120 = 200 ml
So, if you add 80ml of water, the concentration is reduced from 25% to 15%
Or, you can solve the equation
.25*120 + 0x = .15(120+x)
and arrive at the same answer.
To find out how much water should be added to the 25% solution of hydrochloric acid in order to achieve a 15% mixture, we need to consider the initial volume of hydrochloric acid and the target percentage.
Let's break the problem down step by step:
1. Determine the amount of hydrochloric acid in the initial solution:
The initial solution is a 25% solution, meaning it contains 25% hydrochloric acid.
So, the amount of hydrochloric acid in the solution can be calculated as follows:
25/100 * 120 ml = 30 ml
2. Calculate the volume of hydrochloric acid in the desired mixture:
In a 15% mixture, the remaining 85% will be water. Therefore, the volume of hydrochloric acid should be 15% of the total mixture volume.
Let's represent the total mixture volume by "x" ml.
So, the volume of hydrochloric acid in the desired mixture will be (15/100) * x ml.
3. Set up an equation to solve for the total volume of the mixture:
The total volume of the mixture will be the sum of the hydrochloric acid volume and the water volume added. The water volume will be the difference between the total mixture volume and the initial hydrochloric acid volume.
Therefore, we can write the equation as:
(15/100) * x = (x - 30) ml
4. Solve for the total volume of the mixture:
Let's solve the equation:
(15/100) * x = x - 30
Simplifying, we get:
0.15x = x - 30
Rearranging the equation, we get:
x - 0.15x = 30
Solving further, we find:
0.85x = 30
x = 30 / 0.85 ≈ 35.294 ml
5. Calculate the amount of water to add:
The amount of water to add should be the difference between the total volume of the mixture and the initial hydrochloric acid volume.
Therefore, the amount of water to add will be:
35.294 ml - 120 ml ≈ -84.706 ml
Since the calculated amount of water needed to achieve a 15% mixture is negative, it means that it is not possible to achieve a 15% mixture by adding pure water to the initial 25% solution of hydrochloric acid.