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Common Ion Effect

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

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 A $B a {S O}_{4}$
The species having minimum value of $K_{s p}$ will get precipitated first of all because ionic product will exceed the solubility product of such species.

The $K_{s p}$ value is minimum for $B a {S O}_{4} (10^{- 11})$ , so $B a {S O}_{4}$ will get precipitated first amongst all.

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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}]

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})

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}]

Q. The solubility of

P b {S O}_{4}

0.01 M

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

K_{s p}

P b {S O}_{4} = 1.25 \times 10^{- 9}

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