[shinghan]

I measured 0.7965 g of an unknown and dissolved it with ~50 mL of 1M NaOH. It was then transferred to a 500 mL volumetric flask and then ~10 mL of 3M H2SO4 was added to the flask. Water was added to the mark. For the titration, ~10 mL of the sample was used and done in triplicate.

The unknown containing As(III) is oxidized to As(V) by iodine (tri-iodide) which is generated at the anode from a solution containing potassium iodide (KI) containing iodide. The tri-iodide is consumed as long as there is arsenic in the solution. The excess tri-iodide will react with starch indicating the endpoint.

The blank was prepared and titrated.

10 mL of the unknown was added to the blank and the analysis was carried out. After the first titration, the second one was added to the same blank and titration was carried out. The same occurred for the third sample.

As2O3 + H2O --> 2 H3AsO3    (by dissolution)

Iodide is generated at the anode:

2 I- --> I2 + 2 e-

The overall reaction is:

I2 + H2O + H3AsO3 --> 3 I- + H2AsO4- + 3 H+

Two electrons for each mole of H3AsO3 gives us four electrons for one mole of As2O3



What I don't really understand is how to calculate the % As2O3 in my sample. For one of my runs, I get:

Time (t) = 299.7 s
Voltage start (E) = 1.129 V
Voltage endpoint (E) = 1.131 V
Resistance (R) = 100 Ohms

I think my equation for the moles of As2O3 = (I)(t)/4(F) Where current (I)=(E)/(R). Then I would find the mass of As2O3 = (moles of As2O3)(molar mass of As2O3). I'm not sure if I'm in the right track or not. I would use the endpoint voltage right?

I tried: (I) = (E)/(R)
               = (1.131 V)/(100 Ohms)
               = 0.01131 A

moles of As2O3 = (I)(t)/4(F)
                      = (0.01131 A)(299.7)/4(96485.3399 C mol-1)
                      = 8.782699536 e-6 mol

mass of As2O3 = (moles of As2O3)(molar mass of As2O3)
                     = (8.782699536 e-6 mol)(197.8422 g mol-1)
                     = 1.737588598 e-3 g

% As2O3 = (1.737588598 e-3 g/0.7965 g) x 100%
             = 0.218%

This is totally wrong because the range of the %As2O3 is from 8.0% - 15.0%

[Borek]

mass of As2O3 = (moles of As2O3)(molar mass of As2O3)
                     = (8.782699536 e-6 mol)(197.8422 g mol-1)
                     = 1.737588598 e-3 g

This is mass of As₂O₃ in 10 mL, isn't it?

[shinghan]

I understood what you meant and forgot to convert to back to the 500 mL volumetric flask.

So from:

moles of As2O3 = (I)(t)/4(F)
                      = (0.01131 A)(299.7)/4(96485.3399 C mol-1)
                      = 8.782699536 e-6 mol

concentration of As2O3 = (8.782699536 e-6 mol)/(0.01 L)
                                = 8.782699536 e-4 mol/L

CV = CV
C (volumetric flask) = (8.782699536 e-4 mol/L)(0.01 L)/0.500 L
                           = 1.756539907 e-5 mol/L

mass of As2O3 = (1.756539907 e-5 mol/L)(0.500 L)(197.8422 g/mol)
                     = 0.34376835 g

% yield = (0.34376835 g)/(0.7965 g) x 100%
           = 43.16%

Im still getting the wrong answer. Not sure where I went wrong.