Question 3 of 4 Sapling Learning lly determined via titration with coulometrical

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Question 3 of 4 Sapling Learning lly determined via titration with coulometrically generated iodine. To perform the analysis, solid As203 (MW 197.84 g/mol) is first dissolved As, O, s 3H, Oli 2As(OH), 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 end point. The amount of time it takes to reach the end point is used to determine the amount of As2O3 in solution. I, 2e Aso (OH),+ 2H 21 An unknown amount of As2O3 was dissolved in 40.00 mL of an aqueous sodium bicarbonate solution, and to this sample, 3.0 g of Kl were added. To reach the titration end point, 704 seconds were required at 51.3 mA. Determine the As203 concentration in the original sample and report the value in mg/mL. Number mg/mL As o, concentration

Explanation / Answer

First find out the amount of I2 generated:

2 I- -----> I2 + 2 e-

Determine the Coulombs of charge delivered:

Current = 51.3 mA = 51.3*10-3 A = 51.3*10-3 C/s

Time = 704 s

Charge delivered = Current*Time = (51.3*10-3 C/s)*(704 s) = 36.1152 C.

Next determine the moles of electrons delivered:

1 mole of electrons = 1 Faraday = 96485 C.

Therefore 36.1152 = (36.1152 C)*(1 mole electrons/96485 C) = 3.743*10-4 mol electrons.

Determine the moles of I2 generated:

1 mole I2 = 2 mole electrons (refer to the equation above).

Therefore, 3.743*10-4 mole electrons = (3.743 mole electrons)*(1 mole I2/2 mole electrons) = 1.8715*10-4 mole I2.

Next determine the moles of arsenious acid formed:

I2 + As(OH)3 + H2O <====> AsO(OH)3 + 2 H+ + 2 I-

1 mol I2 = 1 mol As(OH)3

Therefore, 1.8715*10-4 mol I2 = 1.8715*10-4 mol As(OH)3.

Determine the moles of As2O3:

As2O3 (s) + 3 H2O (l) <====> 2 As(OH)3 (aq)

1 mole As2O3 = 2 moles As(OH)3

Therefore, 1.8715*10-4 mol As(OH)3 = (1.8715*10-4 mol As(OH)3)*(1 mole As2O3/2 mole As(OH)3) = 9.3575*10-5

Determine the mass of As2O3:

Molar mass of As2O3 = 197.84 g/mol.

Therefore, mass of As2O3 in the given sample = (9.3575*10-5 mol)*(197.84 g/mol) = 0.0185128 g = 18.513 mg

The given mass of As2O3 is dissolved in 40.00 mL sodium bicarbonate solution. Therefore, the concentration of As2O3 (mg/mL) = (18.513 mg)/40.00 mL = 0.4628 mg/mL 0.463 mg/mL (ans).

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