Question

Calculate the value of the equilibrium constant, Kc, for the reaction CuBr(s) + Br−(aq) ↽⇀ CuBr−2(aq) Kc= ? The solubility product constant, Ksp, for CuBr is 6.27×10−9 and the overall formation constant, Kf (????2), for CuBr2− is 8.0×105.

Answer 1

Answer : The equilibrium constant for the reaction is
`5.0* 10^(-3)`

Explanation :

The given equilibrium reaction is:

`CuBr(s)+Br^-(aq)\rightleftharpoons [CuBr_2]^-(aq)`
`K_c=?`

The dissociation reaction will be:

`CuBr(s)\rightleftharpoons Cu^+(aq)+Br^-(aq)`
`K_(sp)=6.27* 10^(-9)`

The formation reaction will be:

`Cu^+(aq)+2Br^-(aq)\rightleftharpoons [CuBr_2]^-(aq)`
`K_f=8.0* 10^(5)`

Thus, the value of equilibrium constant will be:

`K_c=K_(sp)* K_f`

Now put all the given values in this expression, we get:

`K_c=(6.27* 10^(-9))* (8.0* 10^(5))`

`K_c=5.0* 10^(-3)`

Therefore, the equilibrium constant for the reaction is
`5.0* 10^(-3)`