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Solubility Product

To Ag 2 CrO...

Question

To ${A g}_{2} C r O_{4}$ solution over its own precipitate, $C r O_{4}^{2 -}$ ions are added. This results in:

A

Increase in

{A g}^{+}

concentration

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B

Decrease in

{A g}^{+}

concentration

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C

Increase in solubility product

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D

Shifting of

{A g}^{+}

ions from the precipitate into the solution

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Solution

The correct option is A Decrease in ${A g}^{+}$ concentration
${A g}_{2} C r O_{4} ⇌ {2 A g}^{+} + {C r} O_{4}^{2 -}$
$K_{s p} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$
As $K_{s p}$ remains constant.So on addition of $C r O_{4}^{2 -}$ concentration of $C r O_{4}^{2 -}$ will increase.So this will result in decrease in $A g^{+}$ concentration.

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Similar questions

Q. A solution contains

0.05 m o l . {l i t r e}^{- 1}

{B a}^{2 +}

0.002 m o l . l i t r e

{A g}^{+}

ions. The metals are to be precipitated by addition of chromate ions,

C r O_{4}^{2 -}

. Which ion precipitates first? What percentage of this ion remain in the solution when the second ion begins to precipitate?

K_{s p} ({A g}_{2} C r O_{4}) = 3 \times 10^{12}; K_{s p} (B a C r O_{4}) = 1 \times 10^{- 10}

Q. Silver ions are added to the solution with

[B r^{-}] = [C l^{-}] = [C O_{3}^{2 -}] = [A s O_{4}^{3 -}] = 0.1 M

Which compound will precipitate at the lowest [

A g^{+}

Q. Silver ions are added to the solution with:

[{B r}^{-}] = [{C l}^{-}] = [{C O}_{3}^{2 -}] = [A {s O}_{3}^{2 -}] = 0.1 M

Which compound will precipitate at the lowest

[{A g}^{+}]

?

Q. Concentration of

A g^{+}

ions in a saturated solution of

A g_{2} C r O_{4}

20^{o}

1.5 \times 10^{- 4}

L^{- 1}

20^{0}

C, the solubility product of

A g_{2} C r O_{4}

0.01

A g {N O}_{3}

is added to one litre of a solution which is

0.1 M

{N a}_{2} C r O_{4}

0.005 M

N a I O_{3}

. The concentration of precipitate formed at equilibrium and the concentrations of

{A g}^{+}, {I O}_{3}^{-}

C r O_{4}^{2 -}

K_{s p}

{A g}_{2} C r O_{4}

A g I O_{3}

10^{- 8}

10^{- 13}

respectively) :

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