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Author: Subject: NaOCl
ChemyNooby
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[*] posted on 11-5-2009 at 15:08
NaOCl


I'm trying to prepare a 250ml solution of 4.2% NaOCl from the standard 5.25% and 6%. Please let me know if my calculations are correct. Thanks!


From 5.25%

To prepare 250 mL of a 4.2 % v/v solution of NaOCl , Measure 185.19 mL of NaOCl and dilute up to 250 mL with solvent.


From 6%

To prepare 250 mL of a 4.2 % v/v solution of NaOCl , Measure 162.04 mL of NaOCl and dilute up to 250 mL with solvent.



What is the density of the 5.25 and 6% solution? Is it about 1.1-1.5g/cm3?
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Magpie
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[*] posted on 11-5-2009 at 15:43


Are the known solutions stated as volume% or weight %? Life just becomes so much easier when you just stay with weight%. Then you don't have to have densities and you can weigh everything out.

If you must deal with volumes then you are going to have to have the densities with the same degree of accuracy that you expect to get your answers. CRC might have these for NaOCl.
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ChemyNooby
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[*] posted on 11-5-2009 at 16:08


Yep, it's 5.25 and 6% by weight. So, if I change my calculation from mL to gm then I would be correct?!

[Edited on 12-5-2009 by ChemyNooby]
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Ozone
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[*] posted on 11-5-2009 at 16:23


The difference in density between a 5.25 and 6 % (v/v:mad:) is not a big deal when considering that the concentration given for NaOCl is bosh by the time you get it. The material needs to be standardized before a dilution is made. The only reliable way I know to do this involves a spectrophotometric titration vs H2O2 standardized via iodometric titration with standard thiosulfate. the details of this are posted elsewhere on this site.

Once that is known (it might be 2.3 %, for example), you can go ahead and use your C1V1=C2V2. Keep in mind that diluted solutions are less stable, so use them quick...or just finish your homework.

Cheers,

O3




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Magpie
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[*] posted on 11-5-2009 at 17:21



from 5.25wt%:

(0.0425)(250g) = W(0.0525)
W = (0.0425)(250g)/0.0525
W = 202.4g

So, for 250g of this solution add 202.4g of the 5.25% solution to (250g-202.4g) = 47.6g of water.

It is a similar calculation for the use of 6wt% NaOCl.

[Edited on 12-5-2009 by Magpie]
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ChemyNooby
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[*] posted on 11-5-2009 at 22:34


Thanks! :)
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ChemyNooby
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[*] posted on 12-5-2009 at 22:11


Ozone, etc. I was wondering what you said about the potency diminishing. How likely is it for 5.25% to do so? I take it that stronger solutions (like 12%) do this at a much greater rate?
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ChemyNooby
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[*] posted on 13-5-2009 at 18:41


How stable is 5.25% ?? Anyone??
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Ozone
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[*] posted on 14-5-2009 at 15:35


Over time, -OCl disproportionates:

2OCl- -->2Cl- + O2
3OCl- --> 2Cl- + ClO3-

This is accelerated by heat and light.

Opening and closing the bottle also admits dust, which leads to the reduction of the -OCl.

Underneath my sink, a jug of bleach lasts <1 year, afterwhich, it no longer smells of HOCl/Cl2.

Elsewhere on this site can be found methods for measuring the -OCl and HOCl content of bleach at various pH.

Cheers,

O3

[Edited on 14-5-2009 by Ozone]




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ChemyNooby
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[*] posted on 15-5-2009 at 18:08


Thanks again! :D
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ChemyNooby
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[*] posted on 16-5-2009 at 19:10


5.25% doesn't exists. How much water do I add to 6% to get 250ml/gm of a 4.2% bleach solution? Please help me out asap as I'm building a nuclear weapon and don't want to kill myself. :D :mad:
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[*] posted on 19-5-2009 at 16:34


damnit, i had to create a new account. can borth accounts be merged or something?! i have some pms i think. :(
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Magpie
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[*] posted on 19-5-2009 at 18:02


from 6.0wt%:

(0.0425)(250g) = W(0.060)
W = (0.0425)(250g)/0.060
W = 177g

So, for 250g of this solution add 177g of the 6.0% solution to (250g-177g) = 73g of water.

If you have no scale then just measure everything by volume and assume 1g = 1ml. This is a good assumption for water and dilute aqueous solutions.






The single most important condition for a successful synthesis is good mixing - Nicodem
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chemynooby2
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[*] posted on 19-5-2009 at 19:05


Thanks! :D:D:D

I R Done bothering people on here. ;)
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