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TDS to Conductivity ratio? (Read 291 times)

Andres

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TDS to Conductivity ratio?
May 29^th, 2012, 6:54am

Hello,

I have another of those newbe questions, but eveyone has been so helpful to me answering all my previous ones.

I got results from the lab, that the
TDS = 3050 mg/L
Conductivity = 30 000 microS/cm

How likely is this? One of my collegues told me that he read the Conductivity to TDS ratio is usually in the range of 1,3 - 2.

If it is water dependant, then I am talking about a phenolic water stream comming from retorting - had a few questions earlier about that here on the forums.

Best regards, and thank you in advance,
Andres

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David S. Senior Member Offline I have nothing clever to say here. Posts: 373 Gender: Company or Organization: Environmental Express	Re: TDS to Conductivity ratio? Reply #1 - May 29^th, 2012, 8:34am My best guess is that your TDS numbers are biased low. Phenol is a somewhat volatile compound and boils around 180 C. This happens to be the temperature used for TDS determination.
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grrun God Member Offline WaterandWastewa ter.Com is the best! Posts: 3525 Pekin, IL Gender: Company or Organization: Freelance Environmental Engine	Re: TDS to Conductivity ratio? Reply #2 - May 29^th, 2012, 10:43am Don't forget that pH and temperature can affect conductivity. The relationship between conductivity and TDS will vary with the nature of the TDSs. grrun
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Dedalus

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Re: TDS to Conductivity ratio?
Reply #3 - May 29^th, 2012, 3:24pm

There's no hard and fast relationship between TDS and Sp. cond. It depends on the ions present.

It might be instructive to dilute the sample at 4 - 5 different levels and observe the relationship of Sp. cond. to dilution level.

Apart from that, I agree with the other posters. Phenol will fly right up the chimmney at 180 C. However, it's not a very strong electrolyte, either.

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Andres YaBB Newbies Offline I love Water and Wastewater.com Posts: 40 Gender: Company or Organization: Tallinn University of Technol.	Re: TDS to Conductivity ratio? Reply #4 - May 31^st, 2012, 1:23am Thank you all for your most helpful answers
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