Question

The percentage of arsenic in a sample can be determined by titration with iodine. In the determination, all of the arsenic (As) in the sample is converted to HAsO2 followed by titration with iodine (I2) solution. The chemical equation for the reaction between HAsO2 and I2 is:

HAsO2 (aq) + I2 (aq) + 2 H2O (l) → H3AsO4 (aq) + 2 H+ (aq) + 2 I − (aq)

In an actual analysis, a 0.7034 g sample is dissolved and the arsenic (As) converted to HAsO2. Then the sample is titrated to the end point with 0.05225 M I2 solution. If 32.21 mL of the I2 solution were required to reach the end point, calculate the percentage of arsenic (as As) in the sample. (20 points)

Answer 1

The balanced equation is

HAsO₂ (aq) + I₂ (aq) + 2 H₂O (l) → H₃AsO₄ (aq) + 2 H⁺ (aq) + 2 I ⁻ (aq)

Volume of I₂ solution = 32.21 mL

Number of moles of I₂ = M*V = 0.05225 M * 32.21 mL = 1.6829725 mmol

According to balanced equaiton,

1 mole of I₂ reacts with 1 moles of HAsO₂

So, 1.6829725 mmol of I₂ would react with 1.6829725 mmol of HAsO₂

Number of moles of HAsO₂ = 1.6829725 mol

1 mole of HAsO₂ contains 1 mole of As

So,

Number of moles of As = moles of HAsO₂ = 1.6829725 mmol = 0.001683 mol

Molecular weight of As = 74.92 g/mol

Mass of As present in the sample = moles * mol.wt. =0.001683 mol * 74.92 g/mol = 0.126088 g

Mass percent of As present in the sample = mass of As / mass of the sample * 100

Mass percent of As present in the sample = 0.126088 / 0.7034 * 100 = 17.93 %

Mass percent of As present in the sample = 17.93 %