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0.168 g of impure, solid sodium iodate(V) was weighed out and made up to exactly 100 cm3 in a volumetric flask. A 20.0 cm3 sample of this solution was pipetted into a conical flask. 10 cm3 of 1 M sulfuric acid (an excess) and 20 cm3 of 0.1 M potassium iodide solution (an excess) was added. This produced iodine according to the following equation:
|
IO3-(aq) + 5I-(aq) + 6H+(aq) ⇒ 3I2(aq) + 3H2O(l) |
This iodine was then titrated with 0.100M sodium thiosulfate solution. The titres were 9.70 cm3, 9.30 cm3 and 9.40 cm3.
What was the purpose of the added sulfuric acid?
What was the percentage purity of the solid sodium iodate?
The sulfuric acid provides the 6H+ for the reaction to produce the iodine.
Average titre (ignoring 9.70 cm3) = (9.30 + 9.40)/2 = 9.35 cm3
Number of moles of Na2S2O3 = 9.35/1000 × 0.100
|
2Na2S2O3(aq) + I2(aq) ⇒ Na2S4O6(aq) + 2NaI(aq) |
I2 º2Na2S2O3
\number of moles of I2 in 20.0 cm3 = ½ × 9.35/1000 × 0.100
This iodine was produced by this reaction:
|
IO3-(aq) + 5I-(aq) + 6H+(aq) ⇒ 3I2(aq) + 3H2O(l) |
IO3- º 3I2
\number of moles of IO3- in 20.0 cm3 = 1/3 × ½ × 9.35/1000 × 0.100
\number of moles of IO3- in 100.0 cm3 = 5 × 1/3 × ½ × 9.35/1000 × 0.100
NaIO3 = 198
\Mass of NaIO3 in 100.0 cm3 = 5 × 1/3 × ½ × 9.35/1000 × 0.100 × 198 = 0.154 g
\Percentage purity of solid sodium iodate(V) = 0.154/0.168 × 100 = 91.8%