Arsenic(III) oxide is determined by iodimetry: H3AsO3 + I2 + H2O -> H3AsO4 + 2I- + 2H+ (in a buffered medium). 0.099 g of As2O3 (molar mass 198) is dissolved to give arsenite and titrated with 0.0500 M I2. The volume of iodine solution needed is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx186

Arsenic(III) oxide is determined by iodimetry: $H X_{3} AsO X_{3} + I X_{2} + H X_{2} O H X_{3} AsO X_{4} + 2 I X^{-} + 2 H X^{+}$ (in a buffered medium). 0.099 g of $As X_{2} O X_{3}$ (molar mass 198) is dissolved to give arsenite and titrated with 0.0500 M $I X_{2}$ . The volume of iodine solution needed is:

mmol

As X_{2} O X_{3}

= 0.099/198 x 1000 = 0.50, giving 1.0 mmol

H X_{3} AsO X_{3}

. Each arsenite reacts with one

I X_{2}

, so mmol

I X_{2}

= 1.0. Volume = 1.0/0.0500 = 20 mL.