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: 8CrO2−4+3S2O2−3→6SO2−4+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.