A 2.304 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 1.500 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?

Question

A 2.304 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 1.500 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?

Thanks

Answer 1

To find the mass percent of arsenic trichloride in the original sample, we need to understand the steps involved in the given procedure and analyze the data provided.

1. Start with the given sample: We have a 2.304 gram sample containing an unknown amount of arsenic trichloride and the rest inerts.

2. Dissolve the sample: The sample is dissolved into a NaHCO3 and HCl aqueous solution.

3. Add KI and KIO3 solutions: 1.500 grams of KI and 50.00 mL of a 0.00912 M KIO3 solution are added to the sample solution.

4. Excess I3– titration: The excess I3– (formed by the reaction of KI and KIO3) is titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution.

Now, let's consider the chemical equations involved:

1. The reaction between KI and KIO3:
KI + KIO3 + HCl -> I3– + KCl + H2O

2. The reaction between I3– and Na2S2O3:
2Na2S2O3 + I3– -> Na2S4O6 + 2NaI

Based on these equations, we can conclude that one mole of I3– is equivalent to one mole of arsenic trichloride (AsCl3). Therefore, we need to determine the number of moles of I3– and use that to find the number of moles of AsCl3.

Let's calculate the number of moles of I3– using the titration data:

Volume of Na2S2O3 solution used = 50.00 mL = 0.05000 L
Molarity of Na2S2O3 solution = 0.02000 M

Number of moles of Na2S2O3 = Molarity x Volume
Number of moles of Na2S2O3 = 0.02000 M x 0.05000 L = 0.00100 moles

Since one mole of I3– is produced per mole of Na2S2O3, the number of moles of I3– is also 0.00100 moles.

Now, we can use the stoichiometric relationship between AsCl3 and I3– to find the number of moles of AsCl3.

According to the equation: AsCl3 + 3I3– -> AsI3 + 3Cl–

It is evident that one mole of AsCl3 reacts with 3 moles of I3–. Therefore, the moles of AsCl3 are also 0.00100 moles.

To calculate the mass percent of AsCl3 in the original sample, we need to find the mass of AsCl3 and divide it by the total mass of the sample.

The molar mass of AsCl3 = 181.24g/mol

Mass of AsCl3 = Number of moles x Molar mass
Mass of AsCl3 = 0.00100 moles x 181.24 g/mol = 0.18124 grams

Now, we can calculate the mass percent:

Mass percent of AsCl3 = (Mass of AsCl3 / Total mass of the sample) x 100%

Total mass of the sample = 2.304 grams

Mass percent of AsCl3 = (0.18124 grams / 2.304 grams) x 100% = 7.88%

Therefore, the mass percent of arsenic trichloride in the original sample is approximately 7.88%.