Question

A 2.892 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.850 grams of KI and 50.00 mL of a 0.00937 M KIO3 solution. The excess I3

Answer 1

Start by writing down reaction equations to try to understand what is happening.

AsCl3 is dissolved in aqueous HCl to inhibit the formation (and precipitation?) of As(OH)3; addition of HCl drives equilibrium to the left:
AsCl3 + H2O ?As(OH)3 + HCl ...(eqn 1).

Addition of KI then forms AsI3. It think this is quite soluble, so perhaps the I3(-) ion forms but does not dissociate into I(-)?
AsCl3 + 3KI(aq)?AsI3 + 3KCl(aq) ....(eqn 2).
Assuming that KI was in excess, this excess of I(-) ions then reacts with iodate ions in the presence of H(+) ions to form iodine, which turns the solution from colourless to brown:
IO3(-)+5I(-)+6H(+)?3I2+3H2O ...(eqn 3).

Finally, I2 (surely not "excess I3-") is titrated against thiosulphate, turning the solution colourless again:
2S2O3(2-)+I2?S4O6(2-)+2I(-) ...(eqn 4).
However, thiosulphate also reacts with dilute HCl, and there may still be some of this in solution, (unless this has been neutralised by the bicarbonate NaHCO3):
Na2S2O3 + 2 HCl ? 2 NaCl + S + SO2 + H2O.



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CALCULATION:

working backwards, 50ml x 0.02000 mol/l = 1.000 mmol of thiosulphate was used in the titration; this reacted with half as much (0.500 mmol) iodine molecules (eqn 4). The iodine was formed from the reaction between 0.500/3=0.16667 mmol of IO3(-) ions and 5x0.500/3=0.83333 mmol of I(-) ions (eqn 3).

The amount of IO3(-) ions which was added earlier in the form of KIO3(aq) was 50ml x 0.00937 mol/l = 0.4685 mmol. This is more than the 0.16667 mmol used to generate the iodine (eqn 3), so iodate was in excess in this reaction and iodide ions were in limited supply, so the 0.83333 mmol of I(-) ions fully used up in (eqn 3) is what was left over from the reaction of KI(aq) with AsCl3 (eqn 2).

Next work out how much KI was used up in reaction with AsCl3 (eqn 2). The 1.850 g of added KI has RMM (relative molecular mass) 39.1+126.9=166.0. So 1.850/166.0=11.4457mmol of I(-) ions were available for this reaction, and 0.83333 mmol was left over afterwards. Therefore 11.4457-0.83333=10.61237 mmol of KI was used in this reaction, and it reacted with 10.61237/3=3.5374 mmol of AsCl3.

AsCl3 has RMM 74.9+3*35.5=181.4, so 3.5374 mmol of AsCl3 has mass3.5374 mmolx181.4g/mol =641.68mg=0.6416g. So the mass % of AsCl3 in the sample is (0.6416/2.892)x100%=22.185%