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An aqueous solution containing 0.10 g `KIO_(3)` (formula weight`=214.0`) was treated with an excess of KI solution the solution was acidified with HCl. The liberated `I_(2)` consumed 45.0 " mL of " thiosulphate solution to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thosulphate solution. |
Answer» We know, `KIO_(3)+5KIrarr3K_(2)O+3I_(2)` i.e, `2I^(5+)+10erarrI_(2)^(0)` `2I^(-)rarrI_(2)^(0)+2e` Now the liberated `I_(2)` reacts with `Na_(2)S_(2)O_(3)` to give `I_(2)+2erarr2I^(-)` `2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+2e` `because` Milli-mole ratio is `I_(2):S_(2)O_(3)::1:2` Thus, milli`-`mole of `Na_(2)S_(2)O_(3) "used"xx(1)/(2)` `=45xxMxx(1)/(2)` ( `M` is molarity of thiosulphate) Also milli-mole ratio of `KIO_(3): I_(2)::1:3` Thus, `((0.1)/(214))/((45M)/(2))=(1)/(3)` `therefore M=(0.1xx1000xx3xx3)/(214xx45)=0.062` |
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