A 1.857 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 1.540 grams of KI and 50.00 mL of a 0.00912 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?
The idea here is that I2 reacts with As and oxidizes it to As^5+, then you titrate the excess I2 wit S2O3^2- to see how much of the I2 initially there was used.
As^3+ + I2 ==> As^5+ + 2I^-
1 mol As = 1 mol I2.
The I2 came from the reaction of
5I^- + IO3^- + 6H^+ ==> 3I2 + 3H2O
Then I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
1 mol I2 = 1 mols S2O3^2-
mmols IO3^- = 50 x 0.00912 = 0.456
mmols I2 initially = 0.456 x 3 = about 1.37
mmols S2O3^2- = 50 x 0.02 = 1 mmol or 1/2 mmols I2; therefore,
1.368-0.500 = about 0.8 mmol I2 used in the As step or 8.68E-4 mols AsCl3
Convert that to grams AsCl3 and
%AsCl3 = (mass AsCl3/1.857g sample)*100 = ?
To find the mass percent of arsenic trichloride in the original sample, we need to determine the number of moles of arsenic trichloride and the total moles of the sample.
Step 1: Calculate the number of moles of KI (potassium iodide) added.
Given:
Mass of KI = 1.540 grams
Molar mass of KI = 166.0028 g/mol
Number of moles of KI = Mass of KI / Molar mass of KI
Number of moles of KI = 1.540 g / 166.0028 g/mol
Step 2: Calculate the number of moles of KIO3 (potassium iodate) added.
Given:
Volume of KIO3 solution = 50.00 mL = 50.00 cm^3
Concentration of KIO3 solution = 0.00912 M
Number of moles of KIO3 = Concentration of KIO3 solution x Volume of KIO3 solution
Number of moles of KIO3 = 0.00912 mol/L x 0.05000 L
Step 3: Calculate the number of moles of I3- (triiodide) formed.
Since KI reacts with KIO3 to form I3-, the number of moles of I3- formed would be equal to the number of moles of KI added.
Number of moles of I3- formed = Number of moles of KI
Step 4: Calculate the number of moles of Na2S2O3 (sodium thiosulfate) used in the titration.
Given:
Volume of Na2S2O3 solution = 50.00 mL = 50.00 cm^3
Concentration of Na2S2O3 solution = 0.02000 M
Number of moles of Na2S2O3 = Concentration of Na2S2O3 solution x Volume of Na2S2O3 solution
Number of moles of Na2S2O3 = 0.02000 mol/L x 0.05000 L
Step 5: Calculate the number of moles of I3- that reacted with Na2S2O3.
The reaction between I3- and Na2S2O3 is a 1:2 ratio. So, the moles of I3- reacted would be half the moles of Na2S2O3 used.
Number of moles of I3- reacted = (1/2) x Number of moles of Na2S2O3
Step 6: Calculate the number of moles of AsCl3 (arsenic trichloride) in the original sample.
Since I3- is produced by the reaction between AsCl3 and excess KI, the number of moles of AsCl3 can be determined based on the moles of I3- reacted.
Number of moles of AsCl3 = Number of moles of I3- reacted
Step 7: Calculate the mass percent of AsCl3 in the original sample.
Mass percent of AsCl3 = (Number of moles of AsCl3 x Molar mass of AsCl3) / Mass of the original sample
Now let's perform the calculations:
Molar mass of KI = 166.0028 g/mol
Number of moles of KI = 1.540 g / 166.0028 g/mol = 0.009287 mol
Concentration of KIO3 solution = 0.00912 M
Volume of KIO3 solution = 50.00 mL = 0.05000 L
Number of moles of KIO3 = 0.00912 mol/L x 0.05000 L = 0.000456 mol
Number of moles of I3- formed = Number of moles of KI = 0.009287 mol
Concentration of Na2S2O3 solution = 0.02000 M
Volume of Na2S2O3 solution = 50.00 mL = 0.05000 L
Number of moles of Na2S2O3 = 0.02000 mol/L x 0.05000 L = 0.001000 mol
Number of moles of I3- reacted = (1/2) x Number of moles of Na2S2O3 = (1/2) x 0.001000 mol = 0.000500 mol
Number of moles of AsCl3 = Number of moles of I3- reacted = 0.000500 mol
Molar mass of AsCl3 = 181.2816 g/mol (Calculated based on the atomic masses of arsenic and chlorine)
Mass percent of AsCl3 = (Number of moles of AsCl3 x Molar mass of AsCl3) / Mass of the original sample
Mass percent of AsCl3 = (0.000500 mol x 181.2816 g/mol) / 1.857 g
Mass percent of AsCl3 = 0.0488 x 100
Mass percent of AsCl3 = 4.88%
Therefore, the mass percent of arsenic trichloride in the original sample is 4.88%.
To determine the mass percent of arsenic trichloride in the original sample, we need to analyze the chemical reactions and use stoichiometry calculations.
Let's break down the steps:
1. Start by finding the amount of iodine generated from the reaction between KI and KIO3:
- KI + KIO3 → KIO3 + I2
To do this, we need to determine the limiting reagent (the reactant that is completely consumed). We will compare the moles of KI and KIO3 to find the limiting reagent.
- Moles of KI = mass of KI / molar mass of KI
= (1.540 g) / (166 g/mol) = 0.00928 mol
- Moles of KIO3 = volume of KIO3 solution (L) x molarity of KIO3 solution (mol/L)
= (50.00 mL) x (0.00912 mol/L) / 1000 mL/L = 0.000456 mol
Since KI has more moles than KIO3, KIO3 is the limiting reagent, and all of it will react. Therefore, 0.000456 moles of I2 are produced.
2. Calculate the amount of thiosulfate (Na2S2O3) reacted with the excess iodine:
- I2 + 2S2O3^2- → 2I^- + S4O6^2-
We need to find the moles of Na2S2O3 that reacted with the known amount of iodine:
- Moles of Na2S2O3 = volume of Na2S2O3 solution (L) x molarity of Na2S2O3 solution (mol/L)
= (50.00 mL) x (0.02000 mol/L) / 1000 mL/L = 0.00100 mol
This means that 0.00100 moles of Na2S2O3 reacted with the iodine.
3. Calculate the moles of iodine that reacted with the Na2S2O3:
- From the balanced equation, we know that 1 mole of I2 reacts with 2 moles of Na2S2O3.
- Therefore, the moles of I2 reacted = 0.00100 mol Na2S2O3 x (1 mol I2 / 2 mol Na2S2O3) = 0.000500 mol
4. Calculate the moles of arsenic trichloride (AsCl3) in the original sample.
- From the balanced equation for the reaction between iodine and arsenic trichloride, we are given that:
- 2 moles of I2 react with 1 mole of AsCl3.
So the moles of AsCl3 = 0.000500 mol I2 x (1 mol AsCl3 / 2 mol I2) = 0.000250 mol
5. Calculate the mass of arsenic trichloride in the original sample:
- Mass of AsCl3 = moles of AsCl3 x molar mass of AsCl3
= 0.000250 mol x (181.28 g/mol) = 0.04532 g
6. Calculate the mass percent of arsenic trichloride in the original sample:
- Mass percent = (mass of arsenic trichloride / mass of original sample) x 100%
= (0.04532 g / 1.857 g) x 100% ≈ 2.44%
Therefore, the mass percent of arsenic trichloride in the original sample is approximately 2.44%.