Question

$I_2+I^{-}\rightleftharpoons I_3{}^{-}$. This reaction is set up in aqueous medium. We start with $1$ mole of $I_2$ and $0.5$ mol of $I^{-}$ in $1$ L flask. After equilibrium, the excess of $AgN{O}_3$ gave $0.25$ moles of yellow ppt. Then the equilibrium constant is:

• A. $1.33$
• B. $2.66$
• C. $2.00$
• D. $3.00$

Answer 1

The correct option is A $1.33$

$I^{-}+AgN{O}_3\to AgI\downarrow +N{O}_3^{-}$

$0.25$ moles of yellow precipitate corresponds to $0.25$ moles of iodide ions at equilibrium.

$0.5-0.25=0.25$ moles of iodide ions have reacted with $0.25$ moles of iodine to form $0.25$ moles of iodate ions.

$1-0.25=0.75$ moles of iodine remains at equilibrium.

Since total volume is $1L$, the number of moles is equal to molar concentration.

$K=\frac{\left[I_3^{-}\right]}{\left[I_2\right]\left[I^{-}\right]}=\frac{0.25}{0.75\times 0.25}=1.33$