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

Answer :

AgBr should precipitate first.

The concentration of $Ag^+$ when CuBr just begins to precipitate is, $1.34\times 10^{-6}M$

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

Explanation :

$K_{sp}$ for CuBr is $4.2\times 10^{-8}$

$K_{sp}$ for AgBr is $7.7\times 10^{-13}$

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,

$K_{sp}=[Cu^+][Br^-]$

$4.2\times 10^{-8}=0.073\times [Br^-]$

$[Br^-]=5.75\times 10^{-7}M$

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,

$K_{sp}=[Ag^+][Br^-]$

$7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M$

$[Ag^+]=1.34\times 10^{-6}M$

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

Percent of $Ag^+$ remains = $\frac{1.34\times 10^{-6}}{0.073}\times 100$

Percent of $Ag^+$ remains = 0.0018 %