SiS Sensoren Instrumente Systeme GmbH


Selections of the standard and the titer in dependence from the bottle volume, the burette volume and the expected maximum DO concentration

1. There is a dependency in the needed concentration of the titer Tn from the used bottle volumes Vbot(ml), the expected DO concentration (ml/l NTP) and the burette volume VBurette (µl). You should use a concentration of the titer, so that with the largest bottle Vbotmax and with the highest expected concentration DOmax, one single burette volume would be sufficient for the titration including a let say 10% reserve for the over-titration needed to find the end point. The calculation formula is:

  DOmax × 4 × VBotmax
Tn
  0.9 × VBurette × 22.39

2. You should select the normality of the standard Sn and volume of the standard VS (ml) product, so that you need one half to 90% of the burette volume for the standardisation of the titer. The calculation formula is:

Tn × 0.5 × VBurette   Tn × 0.9 × VBurette

≤ Sn × Vs
1000   1000

3. Example:

Burette volume 1000 µl, max bottle volume 150 ml and max DO concentration of 12 ml/l. The result for the titernormality is

  12 × 4 × 150  
Tn
= 0.36
  0.9 × 1000 × 22.39  

The result for the product of standard normality and standard volume is

  0.36 × 0.5 × 1000   0.36 × 0.9 × 1000  
0.18 =
≤ Sn × Vs
= 0.324
  1000   1000  

Suppose you use a calibrated pipette of volume VS = 10 ml, then the normality Sn of your standard should be between 0.018 and 0.0324. Observe that the normality is molarity × 6 with KIO3 and molarity × 12 with KH(IO3)2. Suppose you decide for a normality of 0.03, then you have to dissolve 1.0700 g KIO3 or 0.9748 g KH(IO3)2 and make up to exactly 1000 ml by distilled water. For the titer (Tn = 0.36) you need 89.3457 g of Na2S2O3·5H2O.

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