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To 2500 ml ...

Question

To $2500 m l 5 \times 10^{- 6} M {C a C l}_{2}$ . How many $m m$ of ${A g N O}_{3}$ should be added to just start precipitation of $A g C l [K_{s p} (A g C l) = 10^{- 10}]$ ?

A

2.5

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B

0.25

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C

10^{- 5}

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D

0.025

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Solution

The correct option is D $10^{- 5}$
for presifitation
$[A g^{+}] [C l^{-}] \geq 10^{- 10}$
for limiting condition
$[A g^{+}] [C l^{-}] = 10^{- 10}$
in $5 \times 10^{- 6} M C a C l_{2} [C l^{-}] = 2 \times 5 \times 10^{- 6}$ (on taking $10.0 %$ dissosiation)
$[C l^{-}] = 10^{- 5}$
$[A g^{+}] = 10^{- 5}$
COMMEN:- question is not correctly Red

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