${\mathrm{I}}_2\left(\mathrm{aq}\right)+{\mathrm{I}}^{-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{I}}_3^{-}\left(\mathrm{aq}\right)$ We started with 1 mole of l₂ and 0.5 mole of ${\mathrm{I}}^{-}$ in one liter flask.  After equilibrium is reached, excess of Ag NO ₃ gave 0.25 mole of yellow precipitate. Equilibrium constant is :

           ${\mathrm{I}}_2(\mathrm{aq}.) + {\mathrm{I}}^{-}(\mathrm{aq}.) \rightleftharpoons {\mathrm{I}}_3^{-}(\mathrm{aq}.)$ 

t  =0 ; 1 mole   0.5 mole           0

t_eq. ;   1-x          0.5-x                 x

$$\begin{gathered} {\mathrm{AgNO}}_3 \mathrm{reacts} \mathrm{with} {\mathrm{I}}^{-}\mathrm{and} \mathrm{forms} \mathrm{precipitate}. \\ \mathrm{i}.\mathrm{e}. , {\mathrm{AgNO}}_3+{\mathrm{I}}^{-}⟶\mathrm{AgI}\left(\mathrm{s}\right)+{\mathrm{NO}}_3^{-}(\mathrm{aq}.) \\ \Rightarrow (0.5-\mathrm{x})=0.25 \\ \Rightarrow \mathrm{x}=0.25 \\ {\mathrm{k}}_{\mathrm{c}}=\frac{\left[{\mathrm{I}}_3^{-}\right]}{\left[{\mathrm{I}}_2\right]\left[{\mathrm{I}}^{-}\right]}=\frac{0.25}{0.75\times 0.25}=1.33 \end{gathered}$$