Dilution calculations are easy when you're in the middle of your chemistry class in college. However, if you haven't done one in quite some time, it's easy to make mistakes. One thing that is important to remember is that there is a difference between mass and volume percentages.

Dilution calculations are easy whenyou're in the middle of your chemistry class in college. However, if youhaven't done one in quite some time, it's easy to make mistakes. Onething that is important to remember is that there is a difference between massand volume percentages. Rather than trying to remember chemicalquantities that you may have forgotten (molarity and molality and others), I'llshow you how to simply think through such a calculation (but I hope you haven'tforgotten all of your chemistry!). Let's illustrate the calculationprocedure with the following example:

A hydrogen peroxide (H_{2}O_{2})solution is commercially available as a 35% by mass solution in water (H_{2}O). You wish to prepare two solutions, one at 10% by mass hydrogen peroxide and oneat 10% by volume hydrogen peroxide. The densities you'll need are: d(commercial solution) =9.4 lb/gal, d (water) = 8.3 lb/gal, d (hydrogenperoxide) = 11.98 lb/gal.

__10% by mass hydrogenperoxide solution__

1. Choose a basis for thecalculation

BASIS: 10.0 lbs of 10% by mass hydrogen peroxide

10.0 lbs of solution = 9.0 lbs water + 1.0 lb hydrogenperoxide

2. Measuring is done by volume,calculate the volume of solution which contains 1.0 lb of H_{2}O_{2}.

lbs hydrogen peroxide/gal solution = d (commercial solution)x mass % of solution

lbs hydrogen peroxide/gal solution = (9.4 lb/gal)(0.35) =3.29 lb hydrogen peroxide/gal solution

3. Scale up or down to find thevolume of solution needed for 1.0 lb hydrogen peroxide

__1.0 lbs hydrogen peroxide__ = __3.29 lbshydrogen peroxide__ X gal solution 1.0 gallons solutions

X = 0.304 gallons solution, therefore 1.0 lbs hydrogenperoxide = 0.304 gal solutions

4. Find the mass of water in theamount of solution needed to make 1.0 lb of hydrogen peroxide

lbs solution = (volume solution) x (solution density) =(0.304 gal solution)(9.4 lbs/gal) = 2.86 lbs soln

lbs water = (lbs solution) x (mass % water) = (2.86 lbssolution)(0.65) = 1.86 lbs water

5. Find the __extra__ waterthat needs to be added

According to our basis, we need a total of 9.0 lbs of water. In theamount of solution that contains our needed weight of hydrogen peroxide, wefound that there are 1.86 lbs of water, so we still need:

9.0 lbs water - 1.86 lbs water = 7.14 lbs water

volume __extra__ water = (mass __extra__ water)/(density of water)

volume __extra__ water = (7.14 lbs water)/(8.3 lbs water/gallon) = 0.860 galwater

6. Find your "MAGIC NUMBER"

For a 10% by mass hydrogen peroxide solution being prepared from a 35% by masshydrogen peroxide solution, you need to add 0.860 gal of water for every 0.304gal of solution:__0.860 gal water__ = 2.829

0.304 gal solution

*When preparing these solutions, multiple the volume ofsolution of 2.829 to find the volume of water to be added.*

__10% by volumehydrogen peroxide solution__

1. Choose a basis for thecalculation

BASIS: 1 gallon of 10% by volume hydrogen peroxide

1.0 gallon solution = 0.9 gallons water + 0.1 gallonshydrogen peroxide

2. Find the volume of hydrogenperoxide in 1.0 gallon of solution

Since the solution concentration is on a mass basis, findthe mass of hydrogen peroxide in one gallon, then convert to volume:

lbs hydrogenperoxide/gallon solution = d (commercial solution) x mass % solution

lbs hydrogen peroxide/gallon solution = (9.4 lb/gal)(0.35) =3.29 lb hydrogen peroxide/gal solution

gal hydrogen peroxide = lb hydrogen peroxide/d (hydrogenperoxide)

gal hydrogen peroxide = (3.29 lb hydrogen peroxide)/(11.98lb H_{2}O_{2}/gal H_{2}O_{2})

gal hydrogen peroxide = 0.275 gal hydrogen peroxide/galsolution

3. Find the amount of waterneeded

From part 2, we know that there is 0.275 gal hydrogenperoxide/gal solution. Therefore:

gal water/gal solution = 1.0 gal solution - 0.275 galhydrogen peroxide

gal water/gal solution = 0.725 gal water

With a total of0.275 gal hydrogen peroxide, the total volume of solution is 0.275/0.10 =2.75. So, the total volume of water needed for a 10% by volume hydrogenperoxide solution is:

__Extra__ water needed = Total soln volume - H_{2}O_{2}volume - water volume already in solution

__Extra__ water needed = 2.75 gal - 0.275 gal - 0.725 gal= 1.75 gal water

4. Find you "MAGIC NUMBER"

For a 10% by volume hydrogen peroxide solution beingprepared from a 35% by mass hydrogen peroxide solution, you need to add 1.75gal water for every gallon of solution:

__1.75 gal water__= 1.75

1.00 gal solution

*When preparing these solutions, multiple the volume ofsolution by 1.75 to find the volume of water to be added.*

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82 + 8 =

Dilution Formula In Environmental Engineering