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