Conversion of % by volume?

Question

A chemistry question: How do I convert a solution of x% v/v (by volume) aqueous solution into y% v/v?

To make my question more clear, say I have a 50% v/v aqueous ammonium hydroxide solution. How would I determine how much water to add to it to change it to 2.5% v/v aqueous solution?

I do not want an answer to the ammonium hydroxide question. I want a process to follow to figure out this type of problem. Thanks.

Assuming I start off with j ml. I want to add k ml of water (solvent).

Answer 1

The actual volume of the liquid of interest is percentage*amount.

Say I have 1liter of 50% alcohol, thats .5*1L=.5L of alcohol.

The percentage itself is the volume of interest, e.g. alcohol, divided by the total volume. Thus if I have x liters of % concentration and add V extra solvent. the new percentage is:

x%/(V+x)

e.g if I have 1liter of the 50% alcohol above and add 1l of water, the new concentration is:

(1L).5/(2L) = .25 = 25% alcohol

Knowing two different concentrations is not enough, you need to know the volumes you are working with, starting with either the amount desired, amounts you have, etc.

From the above, if you have j mL of % concentration of something and add k mL of water, the resulting solution has a concentration of

%(j)/(j+k)

by volume.

If we call this percentage %y, and the percentage of the original solution %x then

%y = (%x)j/(j+k) and

j = (%y)k/((%x) - (%y))

Thus in your example, if we want to use %x = .5 and %y = .025, then

j = 0.0526k.

This gives the ratio of the two liquids, e.g. if I start with 100mL of water, I would add 5.26mL of ammonium chloride to get 105.26mL of 2.5% solution.

If we want 5mL total in the end, then j+k=5mL. Or k=5mL-j. Using the above then we need j = (5mL - j)*0.0526 or

j = 0.0526*5mL/(1.0526) = .25mL. So to make 5mL of your 2.5% solution, take .25mL of the 50% amhyd and add water until you get up to 5mL.

Answer 2

Possibly:
.50 + .50 = 2.5 + 97.5 (percentages of solutions)

97.5 - .50 = 47.5

47.5/.50 = .95 % increase

So, you would add 195% of the amount of water.