0.2475 g of pure $As X_{2} O X_{3}$ (198 g/mol) is dissolved in alkali, neutralised with bicarbonate, and titrated with an iodine solution; the arsenic(III) is oxidised to arsenic(V) and $I X_{2}$ goes to $I X^{-}$ . The end point is reached at 24.5 mL. Find the molarity of the iodine solution (three decimal places).

Numerical answer type, answer in M, tolerance 0.0006.

Answer: 0.102 M

As X_{2} O X_{3}

= 0.2475/198 = 1.25 mmol, containing 2.50 mmol As(III). Each As changes by 2 e

^{-}

, so n-factor of

As X_{2} O X_{3}

= 4, giving 5.00 meq, i.e. 2.50 mmol

I X_{2}

(n = 2). Molarity = 2.50/24.5 = 0.102 M.