how much pure acid must be added to 15 ounces of an acid solution which is 40% acid in order to produce a solution which is 50% acid?
Thank you
x ounces + 15 ounces = 15+x ounces in jar
x ounces + .4(15) = .5 (15+x)
x + 6 = 7.5 + .5 x
.5 x = 1.5
x = 3
You are welcome.
To solve this problem, we'll use a basic formula:
(amount of pure acid) / (total amount of solution) = desired concentration
Let's break it down step-by-step:
Step 1: Define the given information
- Amount of acid solution: 15 ounces
- Concentration of the acid solution: 40% acid
- Concentration of the desired solution: 50% acid
Step 2: Convert percentages to decimal values
- 40% = 0.4 (decimal equivalent)
- 50% = 0.5 (decimal equivalent)
Step 3: Set up the equation
Let x represent the amount of pure acid to be added. Since the total solution volume is the sum of the initial solution and the pure acid added:
(0.4 * 15) + x = (0.5 * (15 + x))
Step 4: Solve the equation
Start by simplifying both sides of the equation:
6 + x = 7.5 + 0.5x
Next, isolate the variables on one side of the equation by subtracting 0.5x from both sides:
x - 0.5x + 6 = 7.5
0.5x + 6 = 7.5
Subtract 6 from both sides:
0.5x = 1.5
Finally, divide both sides by 0.5 to solve for x:
x = 1.5 / 0.5
x = 3
So, you need to add 3 ounces of pure acid to the 15 ounces of the 40% acid solution to obtain a solution that is 50% acid.