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Reaction Quotient

50.0 ml of Ag...

Question

50.0ml of $A g N O_{3}$ solution of 0.05 M is mixed with 50.0ml of 0.01M $K B r O_{3}$ solution. Will the precipitate of $A g B r O_{3}$ be formed?

Given : $K_{s p}$ of $A g B r O_{3}$ = $6.0 \times 10^{- 5}$

Precipitate formed

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No precipitate

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Ready to get precipitate

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none

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Solution

The correct option is A
Precipitate formed

Since 50ml of $A g N O_{3}$ and 50ml of $K B r O_{3}$ are mixed together, the total volume becomes 100ml.

The concentrations of $A g N O_{3}$ and $K B r O_{3}$ are exactly halved.

$∴$ Conc. Of $A g^{+}$ = $\frac{0.05}{2}$ = $0.025 M$

conc. Of $B r O_{3}$ = $\frac{0.01}{2}$ = $0.005 M$

Ionic product = $[A g^{+}] [B r O_{3}^{-}]$ = $2.5 \times 10^{- 2} \times 5.0 \times 10^{- 3}$

= $12.5 \times 10^{- 5}$ = $1.25 \times 10^{- 4}$

Ionic product exceeds solubility product.

Hence precipitation results.

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50.0ml of AgNO3 solution of 0.05 M is mixed with 50.0ml of 0.01M KBrO3 solution. Will the precipitate of AgBrO3 be formed? Given : Ksp of AgBrO3 = 6.0 × 10−5

50.0ml of $A g N O_{3}$ solution of 0.05 M is mixed with 50.0ml of 0.01M $K B r O_{3}$ solution. Will the precipitate of $A g B r O_{3}$ be formed?

Given : $K_{s p}$ of $A g B r O_{3}$ = $6.0 \times 10^{- 5}$