The solubility products of $A g C l$ and $A g I$ are $1.1 \times 10^{- 10}$ and $1.6 \times 10^{- 16}$ respectively. If $A g {N O}_{3}$ is added drop by drop the solution containing both chloride and iodide ions, then precipitated first is:

A g I

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A g {N O}_{3}

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A g C l

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Both

A g I

and

A g C l

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Solution

The correct option is A $A g I$
$K_{s p_{A g C l}} = 1.1 \times 10^{- 10} = [S] [S]$
$[A g^{+}]_{A g C l} = (1.1)^{0.5} \times 10^{- 5}$
$K_{s p_{A g I}} = 1.6 \times 10^{- 16} = [S] [S]$
$[A g^{+}]_{A g I} = (1.6)^{0.5} \times 10^{- 8}$
As less amount of $A g^{+}$ is required for the precipitation of $A g I$ so it will precipitate first.

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