How much pure acid must be added to a 30% acid solution to make a 60L solution which is 50% acid?

Did that one last night

http://www.jiskha.com/display.cgi?id=1391650592

Oh okay. Will look at the link. Thanks.

Hi Reiny, I already saw the link. But how did you get the 10x on the second part? Did you multiply everything by 10? Why so? Why not 100?

To determine the amount of pure acid that needs to be added, we can follow these steps:

Step 1: We need to calculate the amount of acid in the current 30% solution. The 30% acid solution has 30 parts acid in every 100 parts of the solution. Therefore, in a 60L solution, the amount of acid is: 30/100 * 60 = 18L.

Step 2: We need to determine the amount of pure acid needed in the final 60L solution with a concentration of 50%. Let's assume we need x liters of pure acid.

Step 3: Now, we can set up an equation to solve for x. Since we know the amount of acid in the 30% solution (18L) and the desired concentration in the final solution (50%), we can write the equation as:

(18L + x) / 60L = 50/100

Step 4: Solve the equation for x. Multiply both sides of the equation by 60 to eliminate the denominators:

18L + x = 30L

Step 5: Subtract 18L from both sides of the equation:

x = 30L - 18L

Step 6: Simplify the equation:

x = 12L

Therefore, you need to add 12 liters of pure acid to the 30% acid solution in order to obtain a 60L solution with a concentration of 50% acid.