The $K_{s p}$ of $A g_{2} C r O_{4}, A g C l, A g B r$ and $A g l$ are respectively, $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13},$ . Which of these will precepitate last if $A g N O_{3}$ solution is added to a solution containing equal moles of NaCl, NaBr, NaI and $N a_{2} C r O_{4}$ ?

AgBr

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A g_{2} C r O_{4}

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AgI

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AgCI

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Solution

The correct option is B $A g_{2} C r O_{4}$
The one that has the highest solubility would precipitate out first.

$K_{s p}$ of $A g_{2} C r O_{4}$ would be $4 s^{3}$
$K_{s p}$ of $A g C l, A g B r, A g I$ would be $s^{2}$
Solving for the values of $s$ for all of these compounds, $A g_{2} C r O_{4}$ has the highest value.

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Similar questions

Q. The

K_{s p}

A g_{2} C r O_{4}, A g C l, A g B r

and

A g l

are respectively,

1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13},

. Which of these will precepitate last if

A g N O_{3}

solution is added to a solution containing equal moles of NaCl, NaBr, NaI and

N a_{2} C r O_{4}

Q. The

K_{s p}

A g_{2} C r O_{4}, A g C l, A g B r

and

A g I

are respectively,

1.1 \times 10^{- 12}

1.8 \times 10^{- 10}

5.0 \times 10^{- 13}

and

8.3 \times 10^{- 17}

. Which of the following salts will precipitate last if

A g N O_{3}

solution is added to the solution containing equal moles of

N a C l, N a B r, N a I

and

N a_{2} C r O_{4}

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 $N a_{2} C r O_{4}$ ?