Question

Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate first?AgBrCuBrWhat percent of Ag remains in solution at the point when CuBr just begins to precipitate?

Answer 1

AgBr should precipitate first.

The concentration of
`Ag^+` when CuBr just begins to precipitate is,
`1.34* 10^(-6)M`

Percent of
`Ag^+` remains is, 0.0018 %

Explanation :

`K_(sp)` for CuBr is
`4.2* 10^(-8)`

`K_(sp)` for AgBr is
`7.7* 10^(-13)`

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

`CuBr\rightleftharpoons Cu^++Br^-`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Cu^+][Br^-]`

`4.2* 10^(-8)=0.073* [Br^-]`

`[Br^-]=5.75* 10^(-7)M`

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

`AgBr\rightleftharpoons Ag^++Br^-`

The expression for solubility constant for this reaction will be,

`K_(sp)=[Ag^+][Br^-]`

`7.7* 10^(-13)=[Ag^+]* 5.75* 10^(-7)M`

`[Ag^+]=1.34* 10^(-6)M`

Now we have to calculate the percent of
`Ag^+` remains in solution at this point.

Percent of
`Ag^+` remains =
`(1.34* 10^(-6))/(0.073)* 100`

Percent of
`Ag^+` remains = 0.0018 %