Homework Answers
Here HCl is used to make the solution acidic.
Moles of KI = moles of I-(aq) initially taken = 1.820 g / 166.0 g/mol = 0.010964 mol
Moles of IO3-(aq) initially taken = C*V = 0.00871 mol/L*50.00 mL * (1L/1000 mL) = 4.355*10-4 mol
KI and KIO3 is a source of I2 according to the following reaction:
5I-(aq) + IO3-(aq) + 6H+(aq) ----> 3I2(aq) + 3H2O(l)
Since the moles of I-(aq) is very high in comparison to the moles of IO3-(aq), the later is the limiting reactant. Hence the moles of IO3-(aq) decides how much product, I2(aq) is formed.
=> Moles of I2(aq) formed = 4.355*10-4 mol IO3-(aq) * [3 mol I2(aq) / 1 mol IO3-(aq)] = 0.0013065 mol I2(aq)
Now I2(aq) converts to I3-(aq) that oxidizes As3+(aq) to As5+(aq) according to the following equation:
As3+(aq) +I3-(aq) ----> As5+(aq) + 3I-(aq)
Moles of S2O32-(aq) titrated with excess I3-(aq) = C*V = 0.02000 mol/L * *50.00 mL * (1L/1000 mL) = 0.001 mol
Excess I3-(aq) oxidizes S2O32-(aq) to S4O62-(aq) according to the following equation:
I3-(aq) + 2S2O32-(aq) ----> S4O62-(aq) + 3I-(aq)
Hence moles of excess I3-(aq) remained = 0.001 mol S2O32-(aq) * [1 mol I3-(aq) / 2 mol S2O32-(aq)] = 0.0005 mol I3-(aq)
=> Moles of I3-(aq) actually reacted with As3+(aq) = 0.0013065 mol - 0.0005 mol = 0.0008065 mol I3-(aq)
=> Moles of As3+ in unknown sample =0.0008065 mol I3-(aq) * [1 mol As3+(aq) /1 mol I3-(aq)] = 0.0008065 mol As3+(aq)
Hence moles of AsCl3 = 0.0008065 mol
=> Mass of AsCl3 = 0.0008065 mol * (181.28g/mol) = 0.1462 g
=> mass percent = (0.1462 g / 1.749 g)*100 = 8.36% (Answer)
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