A solution containing a mixture of $0.05 M N a C I$ and 0.05 M Nal is taken. ( $K_{s p}$ of $A g C l = 10^{- 10}$ and $K_{s p}$ of $A g I = 4 \times 10^{- 16}$ ). When $A g N O_{3}$ is added to such a solution

A g C I

and

A g I

will be precipitate together

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First

A g C l

will be precipitated

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First

A g I

will be precipitated

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neither

A g I

nor

A g C l

precipitate

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Solution

The correct option is C First $A g I$ will be precipitated
According to given data, $[A g^{+}]$ needed for precipitation of
$A g C l = \frac{10^{- 10}}{5 \times 10^{- 2}} = 2 \times 10^{- 9} M$
$A g I = \frac{4 \times 10^{- 16}}{5 \times 10^{- 2}} = 8 \times 10^{- 15} M$
Thus, minimum $[A g^{+}]$ required is for $A g I$ and so it will be precipitated first.

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Similar questions

Q. A solution containing a mixture of 0.05 M NaCI and 0.05 M Nal is taken. (

K_{s p}

A g C l = 10^{- 10}

and

K_{s p}

A g I = 4 \times 10^{- 16}

). When

A g N O_{3}

is added to such a solution

The $K_{s p}$ of $A g_{2} C r O_{4}, A g C l, A g B r$ and AgI are respectively, $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13}, 8.3 \times 10^{- 17}$ . Which one of the following salts will precipitate last if $A g N O_{3}$ solution is added to the solution containing equal moles of NaCl, NaBr, Nal and $N a_{2} C r O_{4}$ ?