Answer:

The following procedure was followed to determine the percentage purity of KIO3 in an impure potassium iodate sample. The mass of the given sample was measured. It was 0.305g. The sample was completely dissolved in deionized water in a beaker. Then the solution was placed in a 250cm^3 volumetric flask and deionized water was added up to the volumetric point. 25.0cm^3 of the solution was taken separately in volumetric flasks and more KI and H2SO4 were added to it. The solution was then titrated with a Na2S2O3 solution of concentration 0.022moldm^-3. The volume of Na2S2O3 consumed at the endpoint of the simulation was 26.20cm^3. 1.Calculate the percent purity of KIO3 in this sample.

Answer:

To calculate the percent purity of KIO3 in the sample, we need to determine the number of moles of KIO3 and Na2S2O3 involved in the reaction.

First, we calculate the number of moles of Na2S2O3:

Moles of Na2S2O3 = Concentration × Volume

Moles of Na2S2O3 = 0.022 mol/dm^3 × 26.20 cm^3 / 1000 cm^3/dm^3

Moles of Na2S2O3 = 0.0005764 mol

Next, we use the balanced chemical equation to determine the stoichiometry between KIO3 and Na2S2O3:

2Na2S2O3 + KIO3 + H2SO4 → Na2SO4 + K2SO4 + H2O + I2

From the balanced equation, we can see that 1 mole of KIO3 reacts with 2 moles of Na2S2O3. Therefore, the number of moles of KIO3 can be calculated as follows:

Moles of KIO3 = (Moles of Na2S2O3) / 2

Moles of KIO3 = 0.0005764 mol / 2

Moles of KIO3 = 0.0002882 mol

Now, we calculate the mass of KIO3 in the sample:

Mass of KIO3 = Moles of KIO3 × Molar mass of KIO3

Mass of KIO3 = 0.0002882 mol × (39.10 g/mol + 126.90 g/mol + 16.00 g/mol × 3)

Mass of KIO3 = 0.0002882 mol × 214.10 g/mol

Mass of KIO3 = 0.0618 g

Finally, we calculate the percent purity of KIO3:

Percent purity = (Mass of pure KIO3 / Mass of sample) × 100

Percent purity = (0.0618 g / 0.305 g) × 100

Percent purity = 20.26%

Therefore, the percent purity of KIO3 in the sample is approximately 20.26%.