The solubility of AgBr with solubility product 5.0×10−13 at 298 K in 0.1 M NaBr solution would be

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The solubility of $A g B r$ with solubility product $5.0 \times 10^{- 13}$ at 298 K in 0.1 M $N a B r$ solution would be

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$5 \times 10^{- 14} M$

$5 \times 10^{- 12} M$

$7 \times 10^{- 6} M$

$5 \times 10^{- 6} M$

answer is B.

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Detailed Solution

Solution: Let, the solubility of $A g B r$ be $S m o l / L .$
$A g B r ⇌ A g^{+} + B r^{-}$
Hence, $[A g^{+}] [B r^{-}] = 5 \times 10^{- 13}$
Given that, $[B r] = 0.1$
$K_{s p} = [A g^{+}] [B r^{-}]$
$5.0 \times 10^{- 13} = [A g^{+}] \times 0.1$
Or $[A g^{+}] = (5 \times 10^{- 13}) / 0.1 = 5 \times 10^{- 12} M$
It means solubility in NaBr is $5 \times 10^{- 12}$ .