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Solubility Product

The Ksp of Ag...

Question

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

A

$A g_{2} C r O_{4}$

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B

AgI

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C

AgCl

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D

AgBr

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Solution

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

Let the concentration of NaCl, NaBr, NaI and $N a_{2} C r O_{4}$ be 1M.

$K_{s p} (A g_{2} C r O_{4}) = [A g^{+}]^{2} [C r O_{4}^{2 -}] = 4 s^{3}$ where s is the solubility and the concentartion of the chromate ion.

$[A g^{+}]_{A g_{2} C r O_{4}} = \sqrt{\frac{K_{s p} (A g_{2} C r O_{4})}{[C r O_{4}^{2 -}]}} = \sqrt{\frac{1.1 \times 10^{- 12}}{(1.1 \times 10^{- 12} / 4)^{1 / 3}}}$

$[A g^{+}]_{A g I} = \frac{K_{s p} (A g I)}{[I^{-}]} = \frac{8.3 \times 10^{- 17}}{1}$

$[A g^{+}]_{A g C l} = \frac{K_{s p} (A g C l)}{[C l^{-}]} = \frac{1.8 \times 10^{- 10}}{1}$

$[A g^{+}]_{A g B r} = \frac{K_{s p} (A g B r)}{[B r^{-}]} = 5.0 \times 10^{- 13}$

$∴$ Solubility of $A g_{2} C r O_{4}$ is the highest it will precipitate last

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