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# How do I prepare a solution with concentration in ppm from a solution with concentration in percent?

I have 100 ml of Clorox containing 5.25% NaOCl and want to make a 1000 ml of total solution (d.i. water and clorox) containing 200 ppm of NaOCl??? How do I start???

Jasmine

 Vocabulary dilution mass percentage parts per million solute solution solvent volume percentage
Solve this problem either by using dilution factors or by setting up a unit conversion calculation.

The quick way (dilution factor method). Think of percentages as 'parts per hundred'. One part per hundred is 10,000 parts per million, so 5.25% NaOCl is 5.25×104 ppm. Preparing a 200 ppm solution requires dilution by a factor of 200 ppm/ 5.25×104 ppm or 1/263. That means that the mass of Chlorox that must be diluted to 1000 mL is 1/263 of 1000 g (the approximate mass of the diluted solution).

The mechanical way (unit conversion method). Take one step at a time. I'll assume the percent and ppm are in terms of masses, not volumes [1].

1. Pick out the target. You must dilute some of the Chlorox by adding water to it. Rephrase the question: How much Chlorox (5.25% NaOCl) should be diluted to 1000 mL to make a 200 ppm NaOCl solution? Find g of Chlorox.

? g of Chlorox

2. List the given information, and connect it with the target. Try to work backwards from the target. You know two things about the Chlorox solution: you have 100 mL, and it's 5.25% NaOCl. The latter means, "5.25 g of NaOCl are found in 100 g of Chlorox". That connects the target with g of NaOCl.

 ? g NaOCl 5.25 g NaOCl = 100 g Chlorox g Chlorox

Now search the given information for something that includes g NaOCl. The dilute solution will contain the same g of NaOCl as the Chlorox. The 200 ppm NaOCl means: there are 200 g of NaOCl in every 1000000 g of dilute solution. That connects g NaOCl with g dilute solution:

 ? g solution 200 g NaOCl= 106g solution g NaOCl 5.25 g NaOCl = 100 g Chlorox g Chlorox

You want 1000 mL of solution, so connect mL solution with g solution. You need the density of the solution to do this, but the problem doesn't include the density. Since the solution is fairly dilute, though, you can assume it will have pretty much the same density as water, 1.00 g solution is about 1 mL of solution:

 1000 mL solution 1.00 g solution= 1 mL solution g solution 200 g NaOCl= 106g solution g NaOCl 5.25 g NaOCl = 100 g Chlorox g Chlorox

Notice that we didn't need the 100 mL of Chlorox given in the problem at all.
3. Do the math. Set up the conversion factors so that all units cancel to give g Chlorox. Use the roadmap we developed in the previous step.
4. Check the answer. You know the Chlorox solution is much more concentrated than the dilute solution, so you expect the amount of Chlorox required to be small. The best way to decide if your answer is correct is to work the problem again using different logic (e. g. the quick solution above.) The next best way is to see if the answer is consistent with the given information.

### Notes

1. Percentage compositions for solutions are ambiguous. "5.25% NaOCl" may mean any of the following:
 percentage type interpretation of "5.25% NaOCl" notation mass/mass 5.25 g of NaOCl are dissolved in every 100 g of solution 5.25 (w/w)% NaOCl mass/volume 5.25 g of NaOCl are dissolvedin every 100 mL of solution 5.25 g NaOCl/100 mL solution volume/volume 5.25 mL of NaOCl are dissolvedin every 100 mL of solution 5.25 (v/v)% NaOCl
For dilute solutions around room temperature, mass and mass/volume percentages agree to two or three figures, when the latter is expressed in g/mL. Volume percentages are convenient for liquid or gaseous solutes. In accurate work you must always specify which type of percentage is being used: the three are not equivalent. It's best to avoid mass/volume percentages entirely.

The calculations assume that 5.25% NaOCl is a mass percentage rather than a volume percentage. A volume percentage is less likely because hypochlorite bleaches are produced by the reaction

Cl2(aq) + 2 NaOH(aq) = NaCl(aq) + NaOCl(aq) + H2O()
and estimating the volume of NaOCl(aq) produced would be much less convenient than estimating its mass.

Author: Fred Senese senese@antoine.frostburg.edu

General Chemistry Online! How do I prepare a solution with concentration in ppm from a solution with concentration in percent?