Answer : The equilibrium constant for the reaction is [tex]5.0\times 10^{-3}[/tex]
Explanation :
The given equilibrium reaction is:
[tex]CuBr(s)+Br^-(aq)\rightleftharpoons [CuBr_2]^-(aq)[/tex] [tex]K_c=?[/tex]
The dissociation reaction will be:
[tex]CuBr(s)\rightleftharpoons Cu^+(aq)+Br^-(aq)[/tex] [tex]K_{sp}=6.27\times 10^{-9}[/tex]
The formation reaction will be:
[tex]Cu^+(aq)+2Br^-(aq)\rightleftharpoons [CuBr_2]^-(aq)[/tex] [tex]K_f=8.0\times 10^{5}[/tex]
Thus, the value of equilibrium constant will be:
[tex]K_c=K_{sp}\times K_f[/tex]
Now put all the given values in this expression, we get:
[tex]K_c=(6.27\times 10^{-9})\times (8.0\times 10^{5})[/tex]
[tex]K_c=5.0\times 10^{-3}[/tex]
Therefore, the equilibrium constant for the reaction is [tex]5.0\times 10^{-3}[/tex]
