[lablackey]

Let me start by saying I really want to get better with calculations involving dilution factors. I've always been bad at them, and if anyone knows of a reference to consult I'd appreciate it.

Please bear with me 

Anyway, we're doing a survey of sulfate concentration at various points in our process and part of that is doing the conventional titration for sulfur dioxide concentration.  But since the sample type varies from starch filtrate, to light steep water, to steeped corn, I need to break down the calculation and figure out the proper dilution factors.

For example, if I modify CRA method E-67 for measuring the sulfur dioxide content of corn syrup to a sample of steeped corn I still use 50 grams of steeped corn, add 100 mL water - blend and get the liquid off - of that liquid I take 10 mL and titrate with a standard iodide solution.

Then
ppm SO₂ =
Sample titer - blank titer (mL) x Normality Iodide x 0.032 x 10⁶
all divided by sample weight

The method notes that the 0.032 is the milliequivalent weight of SO₂ =

64.07
2x1000

My question is - how/where do I account for the 100 mL dilution and taking 10 mL out?


The next sample might be a liquid and I'll just take 10 mL of that and dilute with 100 mL water.  Do I just replace the sample weight with sample volume?

[Mr Peanut]

Let's consider an example.

Say the normality of the Iodine is 0.1N (eq/l)


A = 50 grams of sample to 100 mls DI water

Say: 10 mls of A takes 5 ml of titrant

5 ml of titrant has 0.005 X 0.1 equivalents of SO2 = 0.0005 eq

But you had 100/10 times as much in your original solution (10X dilution)

So you have 0.005 eq in A

your original sample weighed 50 grams and all of it was in the original A solution

So you have 0.005eq/50grams AKA 5meq/50 grams

5 meq of SO2 X 32 mg-SO2/meq = 160 mg

So you have 160 mg SO2/50 grams

= 3200 mg/kg (ppm).

[lablackey]

Thank you, Mr. Peanut. If you wouldn't mind, could you help me check my math one more time?


50 grams of corn in 100 mL water, centrifuged and took 10mL of the liquid - added 100 mL more water

Ran through the usual treatement and titrated with 0.02N iodine = 4.4 mL titrant

So 0.0044 L titrant x 0.02 eq/L = 8.8 x 10^-5 eq SO₂

But I diluted twice. So do I actually have 8.8 milliequivalents?

If so, 8.8 meq SO₂ x 32mg SO2/meq = 281.6 mg SO2/50 grams = 5.632 mg/g = 5632 ppm

That seems ridiculously high, so I'm going to assume I screwed up the double dilution portion. 

[Borek]

My question is - how/where do I account for the 100 mL dilution and taking 10 mL out?

Calculate amount of SO₂ in 10 mL, then multiply by 10 - or more precisely, by 100 mL/10 mL. That will give you amount in the original sample, which have to be converted to ppm.

For serious and accurate work you should use not 100&10 but precise volumes of glass used (see http://www.titrations.info/volumetric-glass-calibration for more detailed explanation).

50 grams of corn in 100 mL water, centrifuged and took 10mL of the liquid - added 100 mL more water

Added 100 mL more, or filled up to 100 mL?

Assuming you filled up to - that means your 100 mL contains 1/10th of the initial amount. You are taking 1/10th of that for titration - so you titrate 1/100th of the initial amount.

Ran through the usual treatement and titrated with 0.02N iodine = 4.4 mL titrant

So 0.0044 L titrant x 0.02 eq/L = 8.8 x 10^-5 eq SO₂

But I diluted twice. So do I actually have 8.8 milliequivalents?

Yes, 100 x 8.8 x 10^-5 eq.

If so, 8.8 meq SO₂ x 32mg SO2/meq = 281.6 mg SO2/50 grams = 5.632 mg/g = 5632 ppm

That seems ridiculously high, so I'm going to assume I screwed up the double dilution portion.

I have not checked numbers thoroughly, but your approach seems OK to me and numbers seem to land in correct ballparks.

[lablackey]

I went over the matter at length with a co-worker and we came up with a calculation he seems satisfied, but I am still uneasy

I'm going to try to articulate things a little better now that I am calmer about the whole mess.

The method I am working from was designed for corn syrup.  It calls for weighing out 50-100 grams of syrup and adding 100 mL water to dilute.   For the titration, 10 mL of the corn syrup solution is transferred to a flask and another 100 mL of water is added.  Treat then titrate.

the calculation is given as ppm SO₂ =
Sample titer - blank titer (mL) x Normality Iodide x 0.032 x 10⁶
sample weight (g)

Of course I'm not working with corn syrup.  I have two kinds of samples: various process liquids and corn.

To titrate the process liquids I simply draw off 10 mL and dilute with 100 mL of water. Treat & titrate with 0.02N

For my SO₂ calculation

Sample titer - blank titer (mL) x 0.02N Iodine x 0.032 x 10⁶
10 mL of sample

My understanding is that I do not need to worry about the 100 mL of water  because the method only calls for it to make seeing the titration endpoint easier

Next is the corn itself

This time I weigh out 50 grams of steeped corn, add 100 mL water and use a blender to puree it.  The mixture has to be centrifuged to separate the solid matter and I pour off the liquid fraction into a beaker.  From that beaker I draw 10 mL out and add 100 mL of water as usual.  Treat and titrate. 

For this, my co-worker has suggested that the calculation be:

Sample titer - blank titer (mL) x 0.02N Iodine x 0.032 x 10⁶
50 grams sample

AND multiply the whole thing by 100/10 = 10

Am I accounting for the initial 100 mL of water and drawing out 10 mL in this case because it is more of an extraction than solution preparation?

BTW, Borek, I saw your point about using good glassware and normally I would, but this could eventually become something tested by our quality technicians and no way do they use volumetric glassware.

[Borek]

This time I weigh out 50 grams of steeped corn, add 100 mL water and use a blender to puree it.  The mixture has to be centrifuged to separate the solid matter and I pour off the liquid fraction into a beaker.  From that beaker I draw 10 mL out and add 100 mL of water as usual.  Treat and titrate. 

For this, my co-worker has suggested that the calculation be:

Sample titer - blank titer (mL) x 0.02N Iodine x 0.032 x 10⁶
50 grams sample

AND multiply the whole thing by 100/10 = 10

Yes and no.

Yes - that's not different from what you did earlierand what seemed to be correct then.

No - there is a possible serious flaw in this approach. You can't be sure that separated solid doesn't contain part of the SO₂. Could be that's the case, if all SO₂ moved to the liquid phase.

Also you are missing () in your formulas, but let's assume they are there.

BTW, Borek, I saw your point about using good glassware and normally I would, but this could eventually become something tested by our quality technicians and no way do they use volumetric glassware.

That was more for awareness then to say you should do it this way, uncalibrated glass means errors in the 1/1000 range.

[lablackey]

No - there is a possible serious flaw in this approach. You can't be sure that separated solid doesn't contain part of the SO2. Could be that's the case, if all SO2 moved to the liquid phase.

For how the information is to be used, I think it's all right to assume all the SO₂ moves to the liquid phase.  There are no plans to use a more rigorous extraction.

Also you are missing () in your formulas, but let's assume they are there.

Oops, sorry.

(Sample titer - blank titer) x 0.02N Iodine x 0.032  x (10) x 10⁶
50 grams sample

Does the calculation successfully account for the dilution?

[Borek]

Yes, 100/10 is the coefficient that allows you calculation of mass of SO₂ in the whole sample (assuming extraction was 100%).

[lablackey]

Awesome.  Thank you for your help.