These review questions are stumping me.
How much water must be added to 50 ML of a 35% acid solution to produce a mixture that is 20% acid.
Could it be as simple as 15% water?
how much acid initially?
.35*50ml= 17.5ml
How much water?
.65*60=32.5ml
.20(32.5+X)=17.5
solve for X
I got the answer to be 55 ml, but can you please explain how you knew to do that?
To solve this problem, we need to use the concept of the concentration of a solution. The concentration of a solution is usually expressed as a percentage or as a ratio of solute to solvent.
Let's break down the information given in the problem:
- You have 50 mL of a 35% acid solution, which means that in this 50 mL, there is 35% acid and 65% solvent (water).
- You want to create a mixture that is 20% acid. This means that the resulting mixture should have 20% acid and 80% solvent (water).
Now, let's find out how much water needs to be added:
1. Calculate the amount of acid in the original 50 mL solution.
In 50 mL of a 35% acid solution, the amount of acid is:
35/100 * 50 mL = 17.5 mL
2. Determine the amount of acid in the final mixture.
In the final mixture with 20% acid, the acid amount is unknown, but we know that the amount of water is 80%. Therefore, the amount of acid is 20% of the total mixture.
So, if we let "x" represent the amount of the final mixture, then we can write the equation:
0.20x = 17.5 mL
3. Solve the equation for "x".
Divide both sides of the equation by 0.20 to isolate "x":
x = 17.5 mL / 0.20 = 87.5 mL
4. Determine the amount of water needed.
To find the amount of water needed, subtract the amount of acid already present in the 50 mL solution from the total mixture amount:
Total amount of mixture - Amount of acid in the original solution = Amount of water added
87.5 mL - 17.5 mL = 70 mL
Therefore, you need to add 70 mL of water to the 50 mL 35% acid solution to produce a mixture that is 20% acid.