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))`

Explanation

`{:(,Na_(2)CO_(3)+,2AgCIhArr,2NACI+,Ag_(2)CO_(3)),("Millimole added",7.5,"excess",0,0),("Milli-mole left",(7.5-a),"excess",2a,a):}`
Given `[CI^(-)]=(0.0026)/(35.5)=7.32xx10^(-5)`
Also conc. Of `CI^(-)` formed `= ("Milli-mole")/("Volume"("in mL"))=(2a)/(5)`
`:. (2a)/(5)=(0.0026)/(35.5)`
`:. a= 1.83xx10^(-4)"milli-mole"`
`:.` "Milli-mole of" `Na_(2)CO_(3)` left in 5 mL
`=7.5-1.83xx10^(-4)=7.5`
or `[CO_(3)^(2-)]=(7.5)/(5)`
Now `K_(SP_(Ag_(2CO_(3))))=[Ag^(+)]^(2)[CO_(3)^(2-)]`
`:. [Ag^(+)]^(2)=(8.2xx10^(-12))/(7.5//5)=5.46xx10^(-12)`
`:, [Ag^(+)]=2.34xx10^(-6)`
`:. K_(SP)` for `AgCI=[Ag^(+)][CI^(-)]=2.34xx10^(-6)xx(0.0026)/(35.5)`
`K_(SP)=1.71xx10^(-10)`