A 2% solution of Hydrochloric acid is required for a procedure. A 5% solution is availabe. How much of the 5% solution will be required to make 1 L of solution?
Are these 5% and 2% mass-volume?
To dilute a 5% solution to a 2% solution, we need to dilute it by a factor of 2.5 (2/5 = 1/2.5).
So a 5% solution contains 5 g HCl/100 mL solution. So if we end up with 1000 mL of solution, how many g HCl must be there to be 2%. X/1000 = 0.02 = 2,000 g. With a solution that is 5 g/100, how much must we use to obtain 2,000 g? That will be 2000/5 = 400. So we take 400 g of the 5% solution, make to 1000 mL, and we have a 2% solution.
See if you can come up with some helpful hints for the other posts that concern concn.
To find out how much of the 5% solution is required to make 1 L of a 2% solution, we need to use a principle called the "method of alligation" or "alligation rule". This method involves mixing two solutions with different concentrations to obtain a desired concentration.
Let's represent the required 2% solution as "A" and the available 5% solution as "B". We can set up the following table to apply the alligation rule:
A B
------|-------
2% 5%
The difference between 5% and 2% is 3%, which means the 5% solution needs to be diluted with a 0% (pure water) solution in some ratio to achieve the desired 2% concentration.
Let the ratio of the 5% solution to the 0% solution be x:(1-x), where x is the fraction of the 5% solution required.
So, according to the alligation rule:
(5-2) / (2-0) = x / (1-x)
Simplifying the equation, we have:
3 / 2 = x / (1-x)
Cross-multiplying gives:
2x = 3 - 3x
Combining like terms:
2x + 3x = 3
5x = 3
Dividing by 5:
x = 3/5
Therefore, to make 1 L of a 2% solution using a 5% solution and water, you would need to mix 3/5 L (or 600 mL) of the 5% solution with 2/5 L (or 400 mL) of water.