The concentration of arsenic trioxide (AS2O3) can be experimentally determined v

Question

The concentration of arsenic trioxide (AS2O3) can be experimentally determined via titration with coulometrically generated iodine. To perform the analysis, solid AS2O3 (MW=197.84 g/mol) is first dissolved in an aqueous sodium bicarbonate solution, forming arsenious acid (As(OH)3(aq) by the following equilibrium. As2O3(s) + 3H2O(l) 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^- I2 + 2e^- I2+ As(OH)3+ H2O AsO(OH)3 + 2H^+ 2I^- An unknown amount of AS2O3 was dissolved in 45.00 mL of an aqueous sodium bicarbonate solution, and to this sample, 3.0 g of Kl were added. To reach the titration end point, 728 seconds were required at 47.1 mA. Determine the AS2O3 concentration in the original sample and report the value in mg/mL.

Explanation / Answer

Q(charge)= I x t

= 47.1x 10-3A x 728s

= 34.28 C

1 mol e- --------------96500C

x= 3.55x10-4 mol e- -----------34.28 C

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2mol e- -------------- 1 mol I2

3.55x10-4 mol e- ------- x= 1.77 x 10-4 mol I2

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1 mol I2 ------ 1 mol As(OH)3 -----> mol As(OH)3= 1.77 x 10-4 mol

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1 mol As2O3--------2 mol As(OH)3

x=8.88 x 10-5 mol As2O3 ----------1.77 x 10-4 mol As(OH)3

mass= mol x MW

mass= 8.88 x 10-5 mol x 198g/mol

= 0.0175 g

= 17.5 mg

Concentration= 17.5 mg/45.00 mL= 0.389 mg/ml

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