x millimole of ${\mathrm{KIO}}_3$ react completely in acid medium with y millimole of KI to give ${\text{I}}_2$ 

quantitatively. If z millimole of hypo are required for complete titration against the ${\text{I}}_2$  produced, then

${\text{i)\hspace{0.17em}KIO}}_{\text{3}}{\text{\hspace{0.17em}+\hspace{0.17em}5KI\hspace{0.17em}+\hspace{0.17em}H}}_{\text{2}}\text{O\hspace{0.17em}}\to {\text{\hspace{0.17em}3I}}_{\text{2}}\text{\hspace{0.17em}+\hspace{0.17em}6KOH}$

${\text{ii)\hspace{0.17em}I}}_{\text{2}}{\text{\hspace{0.17em}+\hspace{0.17em}2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\to {\text{\hspace{0.17em}S}}_{\text{4}}{\text{O}}_{\text{6}}^{\text{2-}}{\text{\hspace{0.17em}+\hspace{0.17em}2I}}^{-}$

${\text{1\hspace{0.17em}I}}_{\text{2}}\to {\text{\hspace{0.17em}2S}}_{\text{2}}{\text{O}}_{\text{3}}^{\text{2-}}\text{\hspace{0.17em}?\hspace{0.17em}}\to \text{\hspace{0.17em}z}$

$\therefore {\text{\hspace{0.17em}moles\hspace{0.17em}of\hspace{0.17em}I}}_{\text{2}}\text{\hspace{0.17em}formed\hspace{0.17em}from\hspace{0.17em}}\left(\text{i}\right)\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{z}}{\text{2}}$

$\text{From\hspace{0.17em}}\left(\text{i}\right){\text{\hspace{0.17em}1\hspace{0.17em}KIO}}_{\text{3}}\to {\text{\hspace{0.17em}3I}}_{\text{2}}\text{;\hspace{0.17em}x\hspace{0.17em}}\to \frac{\text{z}}{\text{2}}$

$\frac{\text{z}}{\text{2}}\text{\hspace{0.17em}=\hspace{0.17em}3x\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}z\hspace{0.17em}=\hspace{0.17em}6x;\hspace{0.17em}5KI\hspace{0.17em}}\to {\text{\hspace{0.17em}1\hspace{0.17em}KIO}}_{\text{3}}$

$\text{y\hspace{0.17em}}\to \text{\hspace{0.17em}x}$

$\Rightarrow \text{\hspace{0.17em}5x\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}}\Rightarrow \frac{\text{6y}}{\text{5}}\text{\hspace{0.17em}=\hspace{0.17em}z\hspace{0.17em}}\Rightarrow \text{\hspace{0.17em}6y\hspace{0.17em}=\hspace{0.17em}5z}$