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 T_{n} from the used bottle volumes V_{bot}(ml), the expected DO concentration (ml/l NTP) and the burette volume V_{Burette} (µl). You should use a concentration of the titer, so that with the largest bottle V_{botmax} and with the highest expected concentration DO_{max}, 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:
DO_{max} × 4 × V_{Botmax} | |
T_{n} ≥ | |
0.9 × V_{Burette} × 22.39 |
2. You should select the normality of the standard S_{n} and volume of the standard V_{S} (ml) product, so that you need one half to 90% of the burette volume for the standardisation of the titer. The calculation formula is:
T_{n} × 0.5 × V_{Burette} | T_{n} × 0.9 × V_{Burette} | |
≤ S_{n} × V_{s} ≤ | ||
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 | ||
T_{n} ≥ | = 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 = | ≤ S_{n} × V_{s} ≤ | = 0.324 | ||
1000 | 1000 |
Suppose you use a calibrated pipette of volume V_{S} = 10 ml, then the normality S_{n} of your standard should be between 0.018 and 0.0324. Observe that the normality is molarity × 6 with KIO_{3} and molarity × 12 with KH(IO_{3})_{2}. Suppose you decide for a normality of 0.03, then you have to dissolve 1.0700 g KIO_{3} or 0.9748 g KH(IO_{3})_{2} and make up to exactly 1000 ml by distilled water. For the titer (T_{n} = 0.36) you need 89.3457 g of Na_{2}S_{2}O_{3}·5H_{2}O.
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