Question

To a 25 mL H₂O₂ solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H₂O₂ solution and report your answer by multiplying it with 1000.

Answer 1

Answer: 1334

Meq of H₂O₂ = Meq of I₂ = Meq of Na₂S₂O₃.

If N is normality of H₂O₂, then

N x 25 = 0.3 x 20, ${\text{N}}_{{\text{H}}_2{\text{O}}_2}=0\text{.}24\text{N}$

${\text{M}}_{{\text{H}}_2{\text{O}}_2}=\text{0.12 M}$

N = 0.24

Volume strength = 0.12 $\times \text{11.2}$

=1.334 Vol

Final answer $=1.334\times 1000=1334$