What quantity of pure acid must be added to 600 mL of a 45% acid solution to produce a 50% acid solution?
Ah, acid solutions. They can really keep you on your toes! So, you want to know how much pure acid you need to add to your 45% acid solution to achieve a 50% acid solution? Well, let me juggle some calculations for you!
First, we need to figure out the amount of acid in the 45% solution. 45% of 600 mL is 270 mL. So, you already have 270 mL of acid in your original solution.
Now, for the grand finale, we want a 50% acid solution. To do that, we'll need an equal amount of acid to the total volume of the solution. Since you started with 600 mL, you'll need to add another 270 mL of pure acid to level things out.
So, voila! You need to add 270 mL of pure acid to your 600 mL of 45% acid solution to turn it into a 50% acid solution. Just remember, don't get too acidic with your calculations!
To solve this problem, we can use the concept of mixtures. Let's assume the quantity of pure acid to be added is x mL.
Step 1: Calculate the amount of acid in the original 45% solution.
The 45% acid solution means that 45% of the 600 mL is acid. Therefore, the amount of acid in the solution is:
0.45 * 600 mL = 270 mL
Step 2: Calculate the final amount of acid in the 50% solution.
For the final solution to be 50% acid, the total amount of acid must be equal to 50% of the total volume. Since we are adding x mL of pure acid, the total amount of acid in the solution will be:
270 mL + x mL
Step 3: Set up the equation and solve for x.
The equation can be written as:
270 mL + x mL = 0.50 * (600 mL + x mL)
Simplifying the equation:
270 mL + x mL = 300 mL + 0.50x mL
Combining like terms:
0.50x mL - x mL = 300 mL - 270 mL
0.50x mL - x mL = 30 mL
0.50x = 30
Dividing both sides by 0.50, we get:
x = 30 / 0.50
x = 60 mL
Therefore, you need to add 60 mL of pure acid to the 600 mL of the 45% acid solution to produce a 50% acid solution.
To find the quantity of pure acid needed, we can set up an equation based on the principle of the conservation of mass.
Let's assume that the quantity of pure acid to be added is represented by 'x' milliliters.
The given solution has a volume of 600 mL and is composed of 45% acid. This means that in 600 mL of the solution, there are 0.45 * 600 = 270 mL of acid.
After adding 'x' mL of pure acid, the total volume of the solution will be 600 mL + 'x' mL. The acid content in the resulting solution will be the sum of the 270 mL of acid already present and the 'x' mL of acid added.
Since we want to end up with a 50% acid solution, we can write the equation:
(270 mL + x mL) / (600 mL + x mL) = 0.50
To solve this equation, we can cross-multiply:
(270 mL + x mL) * 1 = (600 mL + x mL) * 0.50
270 mL + x mL = 300 mL + 0.50x mL
Subtracting 0.50x mL from both sides:
270 mL + 0.50x mL = 300 mL
0.50x mL = 300 mL - 270 mL
0.50x mL = 30 mL
Now, dividing both sides by 0.50:
x mL = (30 mL) / 0.50
x mL = 60 mL
Therefore, 60 mL of pure acid must be added to the 600 mL of a 45% acid solution to produce a 50% acid solution.
let the amount of pure acid added be x ml
x + .45(600) = .5(x+600
x + 270 = .5x + 300
.5x = 30
x = 60