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0.3 g of KI is dissolved in 25 " mL of " water. After adding to this solution double its volume of concentration HCl, a solution of KIO_(3) is graduaaly added with stirring. Iodine is liberated as first but redissolved. It is observed that 24.1 " mL of " iodate solution is just sufficient to dissolve the iodine. If the iodate solution contains 0.8 g per 100 mL formulate the reaction that has taken place.

Answer» <html><body><p></p>Solution :0.3 of KI <a href="https://interviewquestions.tuteehub.com/tag/contains-11473" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINS">CONTAINS</a> `(0.3)/(166)=1.807xx10^(-3)` " mol of "iodide ion, 24.1 " mL of " iodate <a href="https://interviewquestions.tuteehub.com/tag/salution-3004965" style="font-weight:bold;" target="_blank" title="Click to know more about SALUTION">SALUTION</a> contains `(24.1)/(100)xx(0.8)/(214)=9.01xx10^(-4)` mol. It is clear from these figures that 2 moles of `I^(ɵ)` ions react with 1 <a href="https://interviewquestions.tuteehub.com/tag/mole-1100299" style="font-weight:bold;" target="_blank" title="Click to know more about MOLE">MOLE</a> of iodate ion. In thus <a href="https://interviewquestions.tuteehub.com/tag/process-11618" style="font-weight:bold;" target="_blank" title="Click to know more about PROCESS">PROCESS</a> `I^(ɵ)` is oxidised and `IO_(3)^(ɵ)` is reduced to form some common species in which the oxidation number of iondine is x (say). <br/> Thus. <br/> `2(x-(-1)=(5-x)` <br/> i.e., `2x+2=5-x` <br/> `therefore3x=3impliesx=+1` <br/> In presence of excess HCl, this would correspond to the formation of <a href="https://interviewquestions.tuteehub.com/tag/icl-496585" style="font-weight:bold;" target="_blank" title="Click to know more about ICL">ICL</a>. `therefore` the reaction is formulated as follows, <br/> `2I^(ɵ)+IO_(3)^(ɵ)+3Cl^(ɵ)+6H^(o+)to3ICl+3H_(2)O`</body></html>


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