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In the process
$\large {I_2} + \,{I^ - } \rightleftharpoons \,I_3^ -$
(in aq medium), initially there are 2 mole I₂ & 2 mole I^- . But at equilibrium, due to addition of AgNO_3(aq), 1.75 mole yellow ppt is obtained. K_C for the process is (V_flask=1 dm³) nearly

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0.08

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answer is A.

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Detailed Solution

\large {I_2}\left( {aq} \right) + {I^ - }\left( {aq} \right) \rightleftharpoons I_3^ - \left( {aq} \right)

Initial moles of I₂ = 2

Initial moles of I ^- = 2

Volume of the vessel = 1 litre

Yellow ppt formed by the addition of AgNO₃ is AgI

Number moles of AgI precepitated at equilibrium = 1.75

number of moles of I^- at equilibrium = 1.75

Number of moles of I^- reacted = (2 -1.75) = 0.25

According to stoichiometry

1 mole of I₂ reacts with 1 mole of I^-

x mole of I₂ reacts with 0.25 moles of I^-

Similarly moles of

formed due to the reaction of 0.25 moles each of I₂ & I^- will 0.25

\large {K_C} = \frac{{\left[ {I_3^ - } \right]}}{{\left[ {{I_2}} \right]\left[ {{I^ - }} \right]}}

\large {K_C} = \frac{{0.25}}{{1.75 \times 1.75}}

\large {K_C} = \frac{{\left( {\frac{1}{4}} \right)}}{{\left( {\frac{7}{4}} \right)\left( {\frac{7}{4}} \right)}}

\large \boxed{{K_C} \approx 0.08{M^{ - 1}}}

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