Question

1. An unknown sample is analyzed for iodide by oxidizing the iodide ion to lodate ion, according to the unbalanced reaction Br2 +Br +10;: The iodate formed is then precipitated as Ba(10), dried and then weighed. a. Balance the first reaction (the redox reaction) b. In the analysis of a sample with a mass of 2.97 g. 0.08231 g of barium iodate was recovered. Determine the in the sample

Answer 1

Ans 1 :

a)

The balanced reaction is given as :

3Br₂ + I^- + 3H₂O = 6Br^- + IO₃^- + 6H⁺

b)

Mol of barium iodate = mass / molar mass

= 0.08231 g / 487.15 g/mol = 1.69 x 10^-4 mol

Each mol has compound has two mole iodate ions that are formed by 2 mol KI

so mol KI = 2 x 1.69 x 10^-4 mol = 3.38 x 10^-4 mol

Mass KI = mol x molar mass

= 3.38 x 10^-4 mol x 166.0028 g/mol

= 0.056 g

% KI = ( mass KI / mass sample) x 100

= ( 0.056 g / 2.97 g ) x 100

= 1.89 %