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Common Ion Effect

100 mL of a s...

Question

$100 m L$ of a saturated solution of $A g_{2} S O_{4}$ is added to $250 m L$ of a saturated solution of $P b C r O_{4}$ . Will any precipitate form and if so what?

Given that : $K_{s p}$ for $A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}$ and $P b S O_{4}$ are $1.4 \times 10^{- 5},$ $2.4 \times 10^{- 12},$ $2.8 \times 10^{- 13}$ and $1.6 \times 10^{- 8}$ respectively.

A

A g_{2} C r O_{4}

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B

P b C r O_{4}

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C

A g_{2} S O_{4}

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D

P b S O_{4}

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Solution

The correct option is A $A g_{2} C r O_{4}$
For $A g_{2} S O_{4} ⇌ 2 A g^{\oplus} + S O_{4}^{2 -}$
$2 S$ $S$

$K_{s p} = 4 S^{3}$ or $S = \sqrt[3]{\frac{K_{s p}}{4}} = \sqrt[3]{\frac{1.4 \times 10^{- 5}}{4}} = 1.52 \times 10^{- 2} M$

For $P b C r O_{4} ⇌ P b^{2 +} + C r O_{4}^{2 -}$

$K_{s p} = S_{1}^{2}$ or $S_{1} = \sqrt{K_{s p}} = \sqrt{2.8 \times 10^{- 13}} = 5.29 \times 10^{- 7} M$

In solution, conentration. of each ion can be given as:

Thus, $[A g^{\oplus}] = \frac{2 S \times 100}{350} = \frac{2 \times 1.52 \times 10^{- 2} \times 100}{350}$ $= 0.869 \times 10^{- 2} M$

$[S O_{4}^{2 -}] = \frac{S \times 100}{350} = \frac{1.52 \times 10^{- 2} \times 100}{350} = 0.43 \times 10^{- 2}$

$[P b^{2 +}] = \frac{S_{1} \times 250}{350} = \frac{5.29 \times 10^{- 7} \times 250}{350} = 3.78 \times 10^{- 7}$

$[C r O_{4}^{2 -}] = \frac{S_{1} \times 250}{350} = \frac{5.29 \times 10^{- 7} \times 250}{350} = 3.78 \times 10^{- 7}$

It is the evident that,
$[A g^{\oplus}] [C r O_{4}^{2 -}] = (0.869 \times 10^{- 2})^{2} \times (3.78 \times 10^{- 7})$
$= 2.85 \times 10^{- 11} > K_{s p} (A g_{2} C r O_{4})$

The product is greater than $K_{s p}$ of $A g_{2} C r O_{4}$ and thus, $A g_{2} C r O_{4}$ will precipitate out.

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Similar questions

100.0

mL of a saturated solution of

A g_{2} S O_{4}

250.0

mL of saturate solution of

P b C r O_{4}

. Will any precipitate from and if so what?

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4},

P b S O_{4}

1.4 \times 10^{- 5},

2.4 \times 10^{- 12},

2.8 \times 10^{- 13}

1.6 \times 10^{- 8}

Q. 100 mL of a clear saturated solution of

A g_{2} S O_{4}

is added to 250 mL of a clear saturated solution of

P b C r O_{4}

. What is the precipitate formed? Given,

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}

P b S O_{4}

1.4 \times 10^{- 5}, 2.4 \times 10^{- 12}, 2.8 \times 10^{- 13}

1.6 \times 10^{- 5}

Q. 100 mL of a clear saturated solution of

A g_{2} S O_{4}

is added to 250 mL of a clear saturated solution of

P b C r O_{4}

. What is the precipitate formed? Given,

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}

P b S O_{4}

1.4 \times 10^{- 5}, 2.4 \times 10^{- 12}, 2.8 \times 10^{- 13}

1.6 \times 10^{- 5}

100.0 m L

of a clear saturated solution of

A g_{2} S O_{4}

250.0 m L

of a clear saturated solution of

P b C r O_{4}

. Will any precipitate form and if so what? Given,

K_{s p}

A g_{2} S O_{4}, A g_{2} C r O_{4}, P b C r O_{4}

1.4 \times 10^{- 5}, 2.4 \times 10^{- 12}, 2.8 \times 10^{- 13}

1.6 \times 10^{- 8}

Q. A mixture of

H_{2} O

A g C l

shaken to obtain saturated solution. The solid is filtered and to

100 m L

100 m L

0.03 M

N a B r

is added, will a precipitate formed?

K_{s p} A g C l = 10^{- 10}, K_{s p} A g B r = 5 \times 10^{- 13}

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