An excess of acidified solution of KI is added to a 25 mL H2O2 solution. The iodine liberated requires 20 mL of 0.3NNa2S2O3 solution. Calculate the volume strength of H2O2:
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I2+2Na2S2O3→Na2S4O6+2NaI 20 ml of 0.3M(=0.3N)Na2S2O3 will react with (1000)(2)(20)(0.3) moles of I2 which is 0.003 moles or 0.006 equivalents. Thus there are 0.006 equivalents in 25 ml of H2O2, hence, its normality is 0.24 N and thus, the volume strength is 0.24×5.6=1.344.