Answer : The maximum concentration of silver ion is ![5\times 10^{-12}m](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B-12%7Dm)
Solution : Given,
for AgBr = ![5\times 10^{-13}](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B-13%7D)
Concentration of NaBr solution = 0.1 m
The equilibrium reaction for NaBr solution is,
![NaBr(aq)\rightleftharpoons Na^++Br^-](https://tex.z-dn.net/?f=NaBr%28aq%29%5Crightleftharpoons%20Na%5E%2B%2BBr%5E-)
The concentration of NaBr solution is 0.1 m that means,
![[Na^+]=[Br^-]=0.1m](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D%3D%5BBr%5E-%5D%3D0.1m)
The equilibrium reaction for AgBr is,
![AgBr\rightleftharpoons Ag^++Br^-](https://tex.z-dn.net/?f=AgBr%5Crightleftharpoons%20Ag%5E%2B%2BBr%5E-)
At equilibrium s s
The expression for solubility product constant for AgBr is,
![K_{sp}=[Ag^+][Br^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BBr%5E-%5D)
The concentration of
= s
The concentration of
= 0.1 + s
Now put all the given values in
expression, we get
![5\times 10^{-13}=(s)(0.1+s)](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B-13%7D%3D%28s%29%280.1%2Bs%29)
By rearranging the terms, we get the value of 's'
![s=5\times 10^{-12}m](https://tex.z-dn.net/?f=s%3D5%5Ctimes%2010%5E%7B-12%7Dm)
Therefore, the maximum concentration of silver ion is
.