Answer : The maximum concentration of silver ion is [tex]5\times 10^{-12}m[/tex]
Solution : Given,
[tex]K_{sp}[/tex] for AgBr = [tex]5\times 10^{-13}[/tex]
Concentration of NaBr solution = 0.1 m
The equilibrium reaction for NaBr solution is,
[tex]NaBr(aq)\rightleftharpoons Na^++Br^-[/tex]
The concentration of NaBr solution is 0.1 m that means,
[tex][Na^+]=[Br^-]=0.1m[/tex]
The equilibrium reaction for AgBr is,
[tex]AgBr\rightleftharpoons Ag^++Br^-[/tex]
At equilibrium s s
The expression for solubility product constant for AgBr is,
[tex]K_{sp}=[Ag^+][Br^-][/tex]
The concentration of [tex]Ag^+[/tex] = s
The concentration of [tex]Br^-[/tex] = 0.1 + s
Now put all the given values in [tex]K_{sp}[/tex] expression, we get
[tex]5\times 10^{-13}=(s)(0.1+s)[/tex]
By rearranging the terms, we get the value of 's'
[tex]s=5\times 10^{-12}m[/tex]
Therefore, the maximum concentration of silver ion is [tex]5\times 10^{-12}m[/tex].
