A solution of 10% sulfuric acid and one of 25% sulfuric acid are to be used to make 500 ml of 20% sulfuric acid. How many ml of each solution needs to be used in the mixture? (round answers to the nearest ml)
amount of the 25% stuff --- x ml
amount of the 10% stuff --- 500-x ml
.25x + .1(500-x) = .2(500)
solve for x
25x + 1(500-x)=2(500)
25x + 500-x = 1000
25x-x = 1000-500
24x = 500
x 500/24 = 20.83
To solve this problem, we can use the concept of the concentration of solutions. The concentration of a solution is usually expressed as a percentage, which represents the amount of solute (in this case, sulfuric acid) dissolved in a given amount of solvent (in this case, water).
Let's solve this step by step:
Step 1: Assign variables to the unknown quantities.
Let's assume that x ml of the 10% sulfuric acid solution is needed, and y ml of the 25% sulfuric acid solution is needed to make the final mixture.
Step 2: Write down the equation based on the concentration of sulfuric acid in the final mixture.
The concentration of sulfuric acid in the 10% solution is 0.1 (10% expressed as a decimal), and in the 25% solution, it is 0.25 (25% expressed as a decimal). The resulting concentration of the 20% sulfuric acid mixture is 0.2.
The equation can be written as:
0.1x + 0.25y = 0.2(500)
Step 3: Simplify the equation.
0.1x + 0.25y = 100
Step 4: Solve the equation.
To get the values of x and y, we can use either substitution or elimination method. Let's use substitution:
From the equation 0.1x + 0.25y = 100, we can rearrange it to solve for x:
x = (100 - 0.25y) / 0.1
Step 5: Substitute the value of x into the equation.
0.1((100 - 0.25y) / 0.1) + 0.25y = 100
Simplifying further:
10 - 0.25y + 0.25y = 100
-0.25y + 0.25y = 100 - 10
0 = 90
There is no solution to this equation. It means there is no mixture of the 10% and 25% sulfuric acid solutions that can give a resulting concentration of 20%.
Therefore, it is not possible to make a 500 ml mixture of 20% sulfuric acid using 10% and 25% sulfuric acid solutions.