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| Mass of impure potassium iodate(V) in sample = 0.064 g | Titre value = 17.4 cm3 |
Concentration of sodium thiosulfate solution = 0.010 M
Number of moles of sodium thiosulfate used = 0.010 × 17.4/1000 (Remember to divide by 1000 to convert cm3 into dm3)
The following reaction takes place before the titration:
| KIO3(aq) + 5KI(aq) + 3H2SO4(aq) ⇒ 3I2(aq) + 3H2O(l) + 3K2SO4(aq) |
This shows that KIO3 º 3I2 meaning one mole of iodate(V) produces 3 moles of iodine
The following reaction takes place during the titration:
| 2Na2S2O3(aq) + I2(aq) ⇒ Na2S4O6(aq) + 2NaI(aq) |
This shows that 2Na2S2O3 º I2 meaning two moles of thiosulfate react with 1 mole of iodine
So one mole of iodate produces 3 moles of iodine and each mole of iodine needs two moles of thiosulfate. Putting these two relationships together we find that one mole of iodate needs six moles of thiosulfate KIO3 º 6 Na2S2O3
\ Number of moles of KIO3 in 10 cm3 sample = 0.010 × 17.4/1000 × 1/6
\ Number of moles of KIO3 in original 100 cm3 sample = 0.010 × 17.4/1000 × 1/6 × 10
KIO3 = 214
\ Mass of KIO3 in original sample = 0.010 × 17.4/1000 × 1/6 × 10 × 214 = 0.0621 g
\ Percentage purity of KIO3 in original sample = 0.0621/0.064 × 100 = 97%
Always check that your answer looks reasonable. A percentage purity cannot be more than 100% and is unlikely to be less than about 80%. We wouldn't have done a very good job in making it if it were only 9.7% pure!! If you get an answer like this check your working - have you missed a factor of 10 out somewhere?