The $K_{s p}$ of $A g_{2} C r O_{4}, A g C l, A g B r AgI$ 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 $N a C l, N a B r, N a I and N a_{2} C r O_{4}$ ?

A g C l

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A g B r

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A g_{2} C r O_{4}

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A g I

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Solution

The correct option is C $A g_{2} C r O_{4}$
For $A g_{2} C r O_{4}$ ,

$K_{s p} = \frac{(2 S)^{2} S}{1 - S}, S << 1$
$= 1.1 \times 10^{- 12} = 4 S^{3} \Rightarrow S = 6.5 \times 10^{- 5}$

For $A g C l$
$A g C l \to A g^{+} + C l^{-}$
$S S$
$K_{s p} = S^{2}$
$1.8 \times 10^{- 10} = S^{2}$
$S = 1.34 \times 10^{- 5}$

Similarly for $A g B r$ and $A g I$ is $7.1 \times 10^{- 7}$ and $9.1 \times 10^{- 9}$ respectively. Since, solubility of $A g_{2} C r O_{4}$ is the highest, it will precipitate last.
Hence, option (c) is correct.

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Similar questions

Q. The

K_{s p}

A g_{2} C r O_{4}

, AgCl, AgBr 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, NaI 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 $N a_{2} C r O_{4}$ ?