The concentration of arsenic trioxide (As_2O_3) can be experimentally determined

Question

The concentration of arsenic trioxide (As_2O_3) can be experimentally determined via titration with coulometrically generated iodine. To perform the analysis, solid As_2O_3 (MW=197.84 g/mol) is first dissolved in an aqueous sodium bicarbonate solution, forming arsenious acid (As(OH)_3) by the following equilibrium. As_2O_3(s) + 3H_2O(l) reverse arrow 2As(OH)_3(aq) The iodine is coulometrically generated by passing a constant current through the solution which contains potassium iodide (Kl). The arsenious acid in solution is then oxidized by the iodine. Once the reaction has gone to completion, excess generated iodine reacts with a starch indicator, generating a color change and signaling the titration end point. The amount of time it takes to reach the end point is used to determine the amount of AS2O3 in solution. 2I^- right arrow I_2 + 2e^- I_2 + As(OH)_3+ H2O right arrow AsO(OH)_3 + 2H^+ +2I^- An unknown amount of AS_2O_3 was dissolved in 59.00 mL of an aqueous sodium bicarbonate solution, and to this sample, 3.0 g of Kl were added. To reach the titration end point, 554 seconds were required at 44.6 mA. Determine the AS_2O_3 concentration in the original sample and report the value in mg/mL.

Explanation / Answer

As per the equation the oxidation of As(OH)3 requires two electrons per As(OH)3

So from the equation of colulometric titrations

NA= ite / nF

= (0.0446 A)(554 s) / ((2 mol e / mol As(OH)3 )(96487 C / mol e)) = 1.28×10-4 mol of As(OH)3

so we have 1.28 x 10-4/2 moles = 6.4 x 10-5 mol of As2O3 in 59 mL of the solution

MW of As2O3 is 197.841 g/mol = 197.841 g/mol x 6.4 x 10-5 mol = 0.0126 g in 59 mL

12.6 mg in 59 mL so 0.214 mg/mL

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