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In an analysis, 200 ml of the sample is oxidized by 10 ml of a 0.04167 m of Cr2O72-. The excess Cr2O72- consumes 4 ml of 0.25 n Fe(NH4)2(SO4)26H2O. How many equivalents per liter of organic matter is present in the sample?
Accepted Answer
Assuming that the Cr2O72- is quantitatively reduced to Cr3+ in acidic solution:
2 Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O
The number of equivalents of Cr2O72- used is:
10 mL × 0.04167 M × 2 equivalents/mole = 0.008334 equivalents
The number of equivalents of Fe(NH4)2(SO4)26H2O used is:
4 mL × 0.25 N × 6 equivalents/mole = 0.006 equivalents
Therefore, the number of equivalents of organic matter in the sample is:
0.008334 equivalents - 0.006 equivalents = 0.002334 equivalents
The concentration of organic matter in the sample is:
0.002334 equivalents / 0.2 L = 0.01167 equivalents/L
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