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

Question

The concentration of arsenic trioxide (As_2O_3) can be experimentlly determined via titration with coulometrically generated iodine. To perform the analysis, soild As_2O_3 (megawatt = 197.84 gram per 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) if and only if 2As(OH)_3(aq) The iodine is coulometrically generated by passing a constant current through the solution which contains potassium iodide (KI). 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 on end point. The amount of time it takes to reach the end point is used to determine the amount of As_2O_3 in solution. 2I^- if and only if I_2 + 2e^- I_2 + As(OH)_3 + H_2 O if and only if AsO(OH)_ 3 + 2H^+ + 2I^- An unknown amount of As_2O_3 was dissolved in 60.00 millilitre of an aqueous sodium bicarbonate solution, and to this sample, 3.0 gram of KL were added. To reach the titration end point, 598 second were required at 57.2 milliAmp. Determine the as As_2O_3 concentration in the original sampe and report the value in milligram per millilitre. As_2O_3 concentration =

Explanation / Answer

KI moles = 3.0 / 166 = 0.0181

2 mol I- --------------> 1 mol I2

0.0181 mol I- -----------> 0.0181 /2 mol I2 = 9.036 x10^-3 mol I2

moles of I2 = moles As (OH)3 = 9.036 x10^-3 mol I2

2 mol As(OH)3 = 1 mol As2O3

9.036 x10^-3 mol As (OH)3 = 9.036 x10^-3 /2 mol As2O3

moles of As2O3 = 4.52 x 10^-3

molar mass of As2O3 = 197.84 g/mol

mass of As2O3 = 4.52 x 10^-3 x 197.84 = 0.894 g

mass of As2O3 = 894 mg

for 60 mL

894 /60 mL   = 14.9 mg/mL

answer : 14.9 mg/mL

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