A sample of impure potassium iodate (KIO3) weighing 0.50 grams is titrated with a sodium thiosulfate (Na2S2O3) solution. If it takes 35.6 mL of 0.10 M Na2S2O3 to reach the endpoint, calculate the percentage of KIO3 in the sample.
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To calculate the percentage of KIO3 in the sample, we first need to determine the number of moles of KIO3 and Na2S2O3 that reacted.
First, let's calculate the number of moles of Na2S2O3 used:
Volume of Na2S2O3 used = 35.6 mL = 0.0356 L
Molarity of Na2S2O3 = 0.10 M
Number of moles of Na2S2O3 = Molarity × Volume
= 0.10 M × 0.0356 L
= 0.00356 moles
Next, let's determine the balanced chemical equation for the reaction between KIO3 and Na2S2O3. From the equation, we can determine the mole ratio between KIO3 and Na2S2O3.
2Na2S2O3 + KIO3 → Na4S4O6 + 2NaI
From the equation, 2 moles of Na2S2O3 react with 1 mole of KIO3.
Therefore, the number of moles of KIO3 = (Number of moles of Na2S2O3) / 2
= 0.00356 moles / 2
= 0.00178 moles
The molar mass of KIO3 is:
K: 1 × 39.10 g/mol = 39.10 g/mol
I: 1 × 126.90 g/mol = 126.90 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol
Molar mass of KIO3 = 39.10 g/mol + 126.90 g/mol + 48.00 g/mol
= 214.00 g/mol
Finally, let's calculate the percentage of KIO3 in the sample:
Mass of KIO3 in the sample = Number of moles of KIO3 × Molar mass of KIO3
= 0.00178 moles × 214.00 g/mol
= 0.38012 grams
Percentage of KIO3 = (Mass of KIO3 in the sample / Mass of the impure sample) × 100
= (0.38012 grams / 0.50 grams) × 100
= 76.02%
Therefore, the percentage of KIO3 in the impure sample is approximately 76.02%.
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