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

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

If N is normality of H₂O₂, then

N x 25 = 0.3 x 20, $\text{N}_{\text{H}_{2} \text{O}_{2}} = 0 \text{.} 2 4 \text{N}$

$\text{M}_{\text{H}_{2} \text{O}_{2}} = \text{0.12 M}$

N = 0.24

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

=1.334 Vol

Final answer $= 1.334 \times 1000 = 1334$