Let's use the method of mixtures to solve this problem.
You have a 12-liter solution of 60% acid, which means it contains 60% acid and 40% water. You want to mix it with pure water to produce a mixture that is 70% water.
Let x represent the amount of pure water (in liters) you need to add to the 12-liter solution.
1. Calculate the amount of water in the 12-liter solution:
Water in 12-liter solution = 12 liters * 40% = 4.8 liters of water
2. Set up an equation for the total amount of water in the mixture:
Total water in mixture = Water in 12-liter solution + Water added
Total water in mixture = 4.8 liters + x liters
3. Set up an equation for the desired concentration (70% water) of the mixture:
Desired concentration = (Total water in mixture) / (Total volume of mixture) * 100
70% = (4.8 + x) / (12 + x) * 100
4. Solve for x:
70% = (4.8 + x) / (12 + x) * 100
First, divide both sides by 100 to remove the percentage:
0.70 = (4.8 + x) / (12 + x)
Now, cross-multiply:
0.70 * (12 + x) = 4.8 + x
Distribute the 0.70 on the left side:
8.4 + 0.70x = 4.8 + x
Subtract 0.70x and 4.8 from both sides:
8.4 - 4.8 = x - 0.70x
Simplify:
3.6 = 0.30x
Finally, divide by 0.30 to solve for x:
x = 3.6 / 0.30
x = 12
So, you should mix 12 liters of pure water with the 12-liter solution of 60% acid to produce a mixture that is 70% water.