Question

A 5.0 cm³ solution of H₂O₂ liberates 0.50 g of iodine from an acidified K1 solution. Calculate the strength of H₂O₂ solution in terms of volume strength at STP.

Answer 1

(a) 2K1 + H₂SO₄ + H₂O → K₂SO₄ + 2H₂O + I₂

From the above equation, H₂O₂ s I₂ (both are required)

34g of H₂O₂ = 254g of I₂

∴ 0.508 g of I₂ will be liberated from H₂O₂ = \(\frac{34}{254}\) x 0.508 = 0.068g

(b) The decomposition of H₂O₂ occurs as

2H₂O₂ = 2 x 34 = 68g → 2H₂O + O₂ 22400 cm³ at NTP

∴ 0.068 of H₂O upon decomposition will give O₂ = \(\frac{22400}{68}\) x 0.068 = 22.4ml

(c) Now 5.0 cm³ of H₂O₂ solution gives O₂ = 22.4cm³ at NTP

∴ 1.0 cm³ of H₂O₂ solution gives O₂ = \(\frac{22.4}{5}\) = 4.48 cm³ at NTP

Thus volume strength of give H₂O₂ solution = 4.48