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A sample of AgCI was treated with 5.00mL...

A sample of AgCI was treated with `5.00mL` of `1.5M` `Na_(2)CO_(3)` solubility to give `Ag_(2)CO_(3)` . The remaining solution contained `0.0026g of CI^(-)` per litre. Calculate the solubility product of AgCI. `(K_(SP)for Ag_(2)CO_(3)=8.2xx10^(-12))`

A

`1.1x10^(-2)`

B

`1.71x10^(-11)`

C

`1.71x10^(-10)`

D

`1.32x10^(-9)`

Text Solution

Verified by Experts

The correct Answer is:
C
`[Ag^(+)]=sqrt((K_(sp))/([CO_(3)^(2-)]))=2.34xx10^(-6)m`
`K_(spAgCl)=[Ag^(+)][Cl^(-1)=1.71xx10^(-10)`
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