How much pure acid must be added to 6 milliliters of a 5% acid solution to produce a 40% acid solution? Please Show All Work!
To solve this problem, you need to find out how much pure acid needs to be added to the given 6 milliliters of a 5% acid solution.
Here's the step-by-step process:
Step 1: Determine the amount of acid in the initial 5% acid solution.
The initial solution is 6 milliliters of a 5% acid solution. Since the concentration is given as a percentage, it means that 5% of the solution's volume is pure acid.
Therefore, the amount of acid in the initial solution is (5/100) * 6 milliliters.
Step 2: Calculate the total volume of the final 40% acid solution.
The final solution is a 40% acid solution. Let's assume the volume of the final solution is "V" milliliters.
Step 3: Calculate the amount of acid in the final solution using the given concentration.
The desired concentration is 40%. It means that 40% of the final solution's volume is pure acid.
So, the amount of acid in the final solution is (40/100) * V milliliters.
Step 4: Set up an equation equating the amount of acid in the initial solution and the final solution.
The amount of acid in the initial solution should be equal to the amount of acid in the final solution.
(5/100) * 6 = (40/100) * V
Step 5: Solve the equation for V.
To find V, we need to solve the equation obtained in step 4:
(5/100) * 6 = (40/100) * V
Multiplying both sides of the equation by 100 to remove the fractions:
5 * 6 = 40V
30 = 40V
Dividing both sides by 40:
V = 30/40
V = 0.75 milliliters
Therefore, you need to add 0.75 milliliters of pure acid to the initial 6 milliliters of the 5% acid solution to obtain a 40% acid solution.
c1 v1 + c2 v2 = c3 (v1+v2)
5 x 6 + 100(Pure acid) x v2 =
40(6+v2)
30 + 100v2 = 240 + 40v2
60v2 = 210
v2 = 21/6 = 7/2 =
3.5ml is the answer