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Water samples (X and Y) from two different sources were brought to the laboratory for the measurement of dissolved oxygen (DO) using modified Wrinklier method. Samples were transferred to 300 ml BOD bottles. 2 ml of MnSO4 solution and 2 ml of alkaliodide-azide reagent were added to the bottles and mixed. Sample X developed a brown precipitate, whereas sample Y developed a white precipitate. In reference to these observations, the correct statement is
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- Both the samples were devoid of DO
- Sample X was devoid of DO while sample Y contained DO
- Sample X contained DO while sample Y was devoid of DO
- Both the samples contained DO
Correct Option: C
When O2 is present:
| Mn + 2OH- | O2 → MnO2↓(brown) + H2O | |
| 2 |
When O2 is absent
| Mn + 2OH- | O2 → MnO2↓(white) | |
| 2 |
Therefore sample X contain DO and develop brown precipitate Sample Y does not contain DO, thus producing white precipitate
