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Solubility Product

On adding 0.1...

Question

On adding $0.1 M$ solution each of $[A g^{+}], [B a^{2 +}] a n d [C a^{2 +}]$ in a $N a_{2} S O_{4}$ solution, species which precipitated first is: $(K_{s p} (B a S O_{4}) = 10^{- 11}, K_{s p} (C a S O_{4}) = 10^{- 6}, K_{s p} (A g_{2} S O_{4}) = 10^{- 5})$

A

A g_{2} S O_{4}

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B

B a S O_{4}

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C

C a S O_{4}

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D

All of these

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Solution

The correct option is B $B a S O_{4}$
Finding out solubilities of each given compounds:

Solubility of $B a S O_{4}, s_{1} = \sqrt{K_{s p}} = \sqrt{10^{- 11}}$
$= 3.16 \times 10^{- 6}$ $m o l L^{- 1}$

Solubility of $C a S O_{4}, s_{2} = \sqrt{K_{s p}} = \sqrt{10^{- 6}}$
$= 1.0 \times 10^{- 3}$ $m o l L^{- 1}$

Solubility of $A g_{2} S O_{4}$
$∵ 4 (s_{3})^{3} = K_{s p}$
$\Rightarrow$ Solubility of $A g_{2} S O_{4} =_{3} \sqrt{\frac{K_{s p}}{4}}$
$=_{3} \sqrt{\frac{10^{- 5}}{4}} = 1.35 \times 10^{- 2}$ $m o l L^{- 1}$

As $B a S O_{4}$ has least solubility so it will precipitate first.

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Similar questions

Q. On adding 0.1 M solution each of

[A g^{+}], [B a^{2 +}], [C a^{2 +}]

N a_{2} S O_{4}

solution, species first precipitated is:

[K_{s p} (B a S O_{4}) = 10^{- 11}, K_{s p} (C a S O_{4}) = 10^{- 6}

K_{s p} (A g_{2} S O_{4}) = 10^{- 5}]

0.1

M

solution each of

[{A g}^{+}], [{B a}^{2 +}], [{C a}^{2 +}]

{N a}_{2} S O_{4}

solution, species first precipitated is:

[K_{s p} (B a S O_{4}) = 10^{- 11}, K_{s p} (C a S O_{4}) = 10^{- 6} a n d K_{s p} ({A g}_{2} S O_{4}) = 10^{- 5}]

0.1 M

solution each of

[A g^{+}], [B a^{2 +}] a n d [C a^{2 +}]

N a_{2} S O_{4}

solution, species which precipitated first is:

(K_{s p} (B a S O_{4}) = 10^{- 11}, K_{s p} (C a S O_{4}) = 10^{- 6}, K_{s p} (A g_{2} S O_{4}) = 10^{- 5})

0.1 M

solution each of

[{A g}^{+}], [{B a}^{2 +}], [{C a}^{2 +}]

{N a}_{2} {S O}_{4}

solution, species first precipitated is

[K_{s p} B a {S O}_{4} = 10^{- 11}, K_{s p} C a {S O}_{4} = 10^{- 6}, K_{s p} {A g}_{2} {S O}_{4} = 10^{- 5}

]

Q. On adding 0.1 M solution each of

[A g^{+}], [B a^{+ 2}], [C a^{+ 2}]

N a_{2} S O_{4}

, solution, species first precipitated is:

[K_{s p}, B a S O_{4} = 10^{- 11} M, K_{s p} C a S O_{4} = 10^{- 6}, K_{s p} A g_{2} S O_{4} = 10^{- 5}]

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