25 mL of silver nitrate solution (1 M) is a | vedclass

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(C) The reaction is: $AgNO_3 + KI ightarrow AgI(s) + KNO_3$.Initial moles of $AgNO_3 = 25 \ mL 	imes 1 \ M = 25 \ mmol$.Initial moles of $KI = 25 \

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$Ag^{+}$ and $I^{-}$ both

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(C) The reaction is: $AgNO_3 + KI ightarrow AgI(s) + KNO_3$.Initial moles of $AgNO_3 = 25 \ mL 	imes 1 \ M = 25 \ mmol$.Initial moles of $KI = 25 \ mL 	imes 1.05 \ M = 26.25 \ mmol$.Since $AgNO_3$ is the limiting reagent,all $Ag^{+}$ ions react with $I^{-}$ to form $AgI$ precipitate.After the reaction,$25 \ mmol$ of $AgI$ is formed,$25 \ mmol$ of $K^{+}$ and $NO_3^{-}$ remain as spectator ions,and $1.25 \ mmol$ of $I^{-}$ remains in excess.$AgI$ is a sparingly soluble salt,so it exists in equilibrium: $AgI(s) ightleftharpoons Ag^{+}(aq) + I^{-}(aq)$.Due to the common ion effect from the excess $I^{-}$,the concentration of $Ag^{+}$ becomes extremely small.Since $I^{-}$ is present in excess $(1.25 \ mmol)$,the ion present in the smallest quantity is $Ag^{+}$.

$25 \ mL$ of silver nitrate solution $(1 \ M)$ is added dropwise to $25 \ mL$ of potassium iodide $(1.05 \ M)$ solution. The ion$(s)$ present in very small quantity in the solution is/are $........$.

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