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Topic: calculate chemical volume?  (Read 847 times)

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Offline MCL99

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calculate chemical volume?
« on: May 20, 2021, 08:27:37 AM »
Hi,
I would like to find out how the below was calculated:
- The required concentration of the final solution is 1mg/l as Cl2
- The chemical used: 15% (w/w) sodium hypochlorite
- 1,000,000l of water is used
Answer: 5.6l of chemical (sodium hypochlorite) is needed to satisfy the above water usage.

Look forward to giving me some hints on how it should be solved/approached, thank you in advance.


Offline Borek

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Re: calculate chemical volume?
« Reply #1 on: May 20, 2021, 09:12:50 AM »
Please read the forum rules. You have to show your attempts at answering the question/solving the problem to receive help, it is a forum policy.

There are two parts to this problem: one is a simple dilution, the other is equivalence between hypochlorite and Cl2. Do you know how to approach this kind of calculations?
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Offline MCL99

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Re: calculate chemical volume?
« Reply #2 on: May 20, 2021, 10:48:09 AM »
Hi, thanks for your reply. I forgot to mention in the first post that I was trying to calculate it myself, but with no luck.
 
I was using this equation: C1*V1 = C2*V2, however, I was not sure which constituents I should quote and it still did not work.
For the equivalence between NaOCl and Cl2: did you mean 1 mol Cl2 = 1 mol NaOCl which quals to 2.08 mg/l of NaOCl? If so, I would not know where to apply it either.

Any help will be appreciated. Thanks

Offline Borek

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Re: calculate chemical volume?
« Reply #3 on: May 20, 2021, 11:43:46 AM »
For the equivalence between NaOCl and Cl2: did you mean 1 mol Cl2 = 1 mol NaOCl which quals to 2.08 mg/l of NaOCl?

Close, but no.

NaOCl is in equilibrium with water and NaOH, can you try to write a reaction equation? Once you balance it, you will know what is the molar ratio between Cl2 and NaOCl. This will tell you how much NaOCl is required to give the solution equivalent of 1 mg of Cl2 per L.
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Offline MCL99

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Re: calculate chemical volume?
« Reply #4 on: May 20, 2021, 12:54:43 PM »
I have attached the reaction and calculation, but I am still not getting the 5.6l of chemical needed per 10^6l of water.

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