A 1.147 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 1.690 grams of KI and 50.00 mL of a 0.00821 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?
To determine the mass percent of arsenic trichloride in the original sample, we need to understand the series of chemical reactions involved and the stoichiometry of these reactions.
Here's a step-by-step guide to solving this problem:
1. Calculate the number of moles of KI added:
moles of KI = mass of KI / molar mass of KI
Given: mass of KI = 1.690 grams
The molar mass of KI is potassium (K) with atomic mass 39.10 g/mol and iodine (I) with atomic mass 126.90 g/mol.
2. Calculate the number of moles of KIO3 added:
moles of KIO3 = volume of KIO3 solution (in liters) x molarity of KIO3 solution
Given: volume of KIO3 solution = 50.00 mL = 0.05000 L
Given: molarity of KIO3 solution = 0.00821 M
3. Determine the limiting reagent between KI and KIO3:
KI and KIO3 are reacted together in a well-known redox reaction. The limiting reagent will be the one that produces the least amount of I3– ions.
The balanced equation for the reaction is:
5 KI + KIO3 + 3 HCl -> 3 I3– + 3 KCl + 3 H2O
By comparing the stoichiometric coefficients, you can determine which reactant limits the formation of I3– ions. The ratio of KI to KIO3 is 5:1.
Calculate the number of moles of I3– formed from each reactant:
moles of I3– from KI = moles of KI x (3 moles of I3– / 5 moles of KI)
moles of I3– from KIO3 = moles of KIO3 x (3 moles of I3– / 1 mole of KIO3)
The reactant that produces fewer moles of I3– will be the limiting reagent.
4. Calculate the number of moles of Na2S2O3 used in the titration:
moles of Na2S2O3 = volume of Na2S2O3 solution (in liters) x molarity of Na2S2O3 solution
Given: volume of Na2S2O3 solution = 50.00 mL = 0.05000 L
Given: molarity of Na2S2O3 solution = 0.02000 M
5. Calculate the number of moles of I3– reacted with Na2S2O3:
The balanced equation for the reaction between I3– and Na2S2O3 is:
I3– + 2 S2O32– -> 3 I– + S4O62–
By comparing the stoichiometric coefficients, you can determine the ratio of moles of Na2S2O3 to moles of I3–
moles of I3– = moles of Na2S2O3 x (1 mole of I3– / 2 moles of Na2S2O3)
6. Calculate the number of moles of I3– originally present:
moles of I3– originally present = moles of I3– from KI + moles of I3– from KIO3
7. Calculate the number of moles of AsCl3 originally present:
In the reaction between AsCl3 and I3–, they react at a 1:2 molar ratio.
So, moles of AsCl3 = moles of I3– originally present / 2
8. Calculate the mass of AsCl3 in the original sample:
mass of AsCl3 = moles of AsCl3 x molar mass of AsCl3
The molar mass of AsCl3 is:
Arsenic (As): 74.92 g/mol
Chlorine (Cl): 35.45 g/mol (x3 for three chlorine atoms)
9. Calculate the mass percent of AsCl3:
mass percent of AsCl3 = (mass of AsCl3 / mass of sample) x 100
Given: mass of the sample = 1.147 grams
By following these steps and performing the necessary calculations, you should be able to determine the mass percent of arsenic trichloride in the original sample.
Note: I don't use I^3-; it's the I2 that does the job.
As^3+ + I2 ==> 2I^- + As^5+
Then mols I2 initially = 0.050L x 0.0821 = 0.004105
mols I2 titrated with S2O3^2- = 0.050 x 0.02 = 0.001 which means 0.0005 mols I2 were used.
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
mols I2 used for the As/I2 rxn = 0.004105-0.0005 = 0.003605
Then g As = mols As x atomic mass As
%As = (g As/g sample)*100 = ?