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Reducing Water pH with Acid.


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#1 Dewald42

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Posted 6 December 2016 - 06:57 AM

From The word go I have to emphasise that I am not a chemist, but I need the help of you guys that has the knowhow.

We have a RO plant on our site and most of our water goes through it (dairy Factory).

However, when one section of the RO is on CIP, we bleed from the normal water into the supply to the factory. So the water is not always 100% RO water.

In the RO plant we correct the pH to be within 7.4 to 8.0 as this is the requirement for most of our products.

The problem is that we need a pH of approx 7.0 for one specific product. To achieve this we batch 11000 lites and correct the pH by adding acid by hand.

In the last few weeks we had more trouble to achieve this.

 

We have the following acid on site:

HCL 0.1M

HCL 0.5M

Nitric 30%

 

Volume is 11000 litres.

 

Which acis would be best to acieve this with the minimal amount needed?

If you must give a broad indication, how much of each acid would be needed if the initial pH is for example 7.8?

Is there a way of calculating (please not too complicated) approx how much to add if we have the volume and initial pH?

 

Thanks in advance

 

 


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#2 AbstractDreamer

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Posted 6 December 2016 - 07:57 AM

I'm not a chemist or anyone with knowhow.

A few of my thoughts:

 

  • Isn't HCL dangerous and toxic, especially if your product is for consumption or entry into the food chain?
  • You don't have much neutralising to do, isn't something like acetic acid (vinegar) safer to use?
  • The minimal amount by volume is dependent on the concentration of your acid and the PH for that concentration.  So surely you can discount HCL 0.1M in favour of HCL 0.5M?
  • When you measured it by hand, did it not occur to you to note down how much you put in, so you would at least have a ball figure to work with on the next dilution?

 

EDIT:

 

Apparently HCL is safe according to SCOGS  :-)

 

http://www.fda.gov/F...S/ucm260426.htm

 

"There is no evidence in the available information on hydrochloric acid that demonstrates or suggests reasonable grounds to suspect a hazard to the public when it is used at levels that are now current or that might reasonably be expected in the future."


Edited by AbstractDreamer, 6 December 2016 - 08:46 AM.

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#3 Dewald42

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Posted 6 December 2016 - 09:09 AM

We did, but it has come to light that they were not issued with the same acis every time. they would sometimes get 0.1M, and other times 0.5M.

then they would get frustrated and add Nitric as it is 30% and they think it will work better.

So I have volemes added, but I do not know which acid was used.

This is to gain information to put measures in place to standardise and correct, and the query was merely to have an idication of what to expect.


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#4 John Cuthber

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Posted 6 December 2016 - 11:08 AM

I have bad news for you: there is no way to reliably answer the question given just the pH of the incoming water.

Do you have other information about the feedwater?


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#5 AbstractDreamer

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Posted 6 December 2016 - 11:14 AM

http://www.sciencefo...rochloric-acid/

 

This might help, just substitute in your numbers.

 

You might need to find the Molarity of your Nitric solution.


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#6 CharonY

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Posted 6 December 2016 - 03:57 PM

That won't help much unless, (as John said) you get some readings on e.g. carbonate content of the water. 


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#7 AbstractDreamer

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Posted 8 December 2016 - 09:37 PM

Well like i said I don't know much about this topic.   But it seems if you only have one unknown variable like carbonate content of the water (whatever that is), then you should be able to solve the problem, if this unknown variable is consistent throughout the water.

 

You take a measured sample of your feedwater, say 10 litres.

You thoroughly mix in a measured sample of one of your acids.

You measure the resulting PH of the water+acid solution.

From that you can reverse calculate the carbonate content of the feedwater source.

 

It will be an estimate, the larger your sample of feedwater, or the more accurate your PH measurement, then the more accurate your answer.

 

But I'm just guessing here.   Listen to the experts.  Can't be done.


Edited by AbstractDreamer, 8 December 2016 - 09:42 PM.

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#8 StringJunky

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Posted 8 December 2016 - 09:54 PM

Would it not work if you take a test litre sample of water, add the acid until neutrality is reached, note the amount of acid then multiply that by the volume of water to be treated - does it work that way? I assume the main volume of water might be known or fixed within reasonable limits before treating; could the tank/reservoir water  level  be calibrated to a mark that denotes a known volume?


Edited by StringJunky, 8 December 2016 - 09:56 PM.

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#9 AbstractDreamer

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Posted 8 December 2016 - 10:00 PM

That's kinda my thinking.  That would work if all the OP wants is to get the water at PH 7

 

But there was another question like which acid requires least amount.  So calculation is required.

 

But from a practical point of view, I'm with you.


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#10 Dewald42

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Posted 9 December 2016 - 04:05 AM

That's kinda my thinking.  That would work if all the OP wants is to get the water at PH 7
 
But there was another question like which acid requires least amount.  So calculation is required.
 
But from a practical point of view, I'm with you.


I hear.

We will start with some experiments and see if it works.
Next week we have about 12 batches to make, so we will do a titration with estimate, and we will get an actual value fot the big batch.

I will post the results.
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#11 AbstractDreamer

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Posted 9 December 2016 - 06:54 AM

So doing some reading:

 

http://chem.libretex...ns/The_pH_Scale

 

This tells me that temperature affects PH.  How much I don't know, but if you keep it constant it should cancel out in measurements.  Also at the bottom are some maths and solutions that might help.

 

http://chem.libretex...h_a_Strong_Acid

 

This sounds like what you have to do.  Need to take care of the equivalance point.  The examples provided are reversed for your case.   Eg1 is where you are given the volume and molarity of the base and the molarity of acid and have to calculate the PH of both separately and together,  and volume of acid.  Your case, you have the molarity of your acid, and can measure both volumes and both PH (separately and together) to obtain molarity of the base (your feedwater).  I cant see a second variable that would make this impossible, but I'm not familiar with the calculations.  Good luck.


Edited by AbstractDreamer, 9 December 2016 - 07:05 AM.

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#12 Dewald42

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Posted 9 December 2016 - 07:41 AM

Very nice article.

Thanks.

 

Will post results


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#13 AbstractDreamer

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Posted 9 December 2016 - 07:43 AM

After mixed solution is in equilibrium, and pH=7.0, and temperature for all solutions are unchanged:

 

 Mol_{acid}*Volume_{acid} = Mol_{base}*Volume_{feedwater}

 

If you overshoot the pH and make the solution acidic, you will need an base agent to neutralise, and recalculate the volume and check temperature.  Like sodium hydroxide.  Don't overshoot, just redo with new batch.  Do several batches of different feedwater volumes, for assurance and confidence in results.

 

"The pH of the final solution of titration changes as a result of the concentration of the standard solution (your acid). Ideally, if the titration has been done precisely and accurately, the final solution of the titration process should be neutralized and have a pH of 7.0. However, this is not always the case. The pH of the final solution often fluctuates depending upon the concentration of the unknown solution and the standard solution that is being added"

 

So the fluctuation will be based on the concentration of base in your feedwater, which thankfully is low, and the concentration of your acid (so use the low concentration HCL).

 

So then you have the Mol of your feedwater.

 

http://chem.libretex...on_Fundamentals

 

I might have missed something this looks too easy.


Edited by AbstractDreamer, 9 December 2016 - 08:25 AM.

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#14 hypervalent_iodine

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Posted 9 December 2016 - 09:27 AM

AbstractDreamer,

It is not that simple if you are not just working with strong acids and bases, as is the case here since there is carbonate present. The equivalence point in this case is not pH 7, so you cannot use that math to work out a volume.
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#15 AbstractDreamer

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Posted 9 December 2016 - 12:33 PM

Yep i realised that later when looking at the formulae.   Please correct me if im wrong.

 

The equivalent point is NOT what OP needs to use or calculate though hopefully.   The equivalent point of equal moles of acid and base in solution would probably end up much more acidic.   OP probably wont be able to derive back to Kb, but shouldnt need to.  

 

Would the carbonate not just react with the acid and give off CO2 and salt residue and water?  It might be a 2-stage reaction, but it should reach equilibrium eventually.  As neither the base analyte (ph8.0) or the acid (0.1M HCL) is strong , there should be smaller fluctuations, but might take a bit longer to complete.

eg Sodium carbonate

 

Na2CO3+ HCl→ NaHCO3 + NaCl

NaHCO3+ HCl → NaCl+ CO2(g) + H2O

 

The volume can be measured physically.  Just have to hit pH7.0  pretty much the same as when they did it by hand, except measure the analyte volume and measure the titrate volume, and use just one type of titrate.

 

Another reason why multiple batches of different volumes should be used, to derive any proportionality constant based on volume of reaction, such as gases released.

 

I was only trying to help.  Sorry if Im wrong and misled anyone.  Though cant be much worse than their previous methods.


Edited by AbstractDreamer, 9 December 2016 - 12:43 PM.

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