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Question

In basic solution, CrO42 oxidises S2O32 to form Cr(OH)4 and S2O42. How many mL (nearest integer) of 0.154 M CrO42 are required to react with 40.0 mL of 0.246 M S2O32?
[Hint : 0.0154 M =0.154×3 N CrO42 and 0.246 M =0.246×8 N S2O32]

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Solution

The reaction is as follows:
8CrO24+3S2O236SO24+8Cr(OH)4
The normality of 0.154 M CrO4 is 0.154×3 N.
Similarly, the normality of 0.246 M S2O32 solution is 0.246×8 N.
N1V1=N2V2
V×0.154×3=0.246×8×40
V=170 mL.
153764_160811_ans.png

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