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Common Ion Effect

On adding 0.1...

Question

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

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 $B a S O_{4}$
As the solubility product of $B a S O_{4}$ is lowest, it will get precipitated first upon interaction with $N a_{2} S O_{4}$ as solubility product is inversely proportional to precipitation.

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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 +}] 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} 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 +}], [{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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