Respuesta :
Answer :
AgBr should precipitate first.
The concentration of [tex]Ag^+[/tex] when CuBr just begins to precipitate is, [tex]1.34\times 10^{-6}M[/tex]
Percent of [tex]Ag^+[/tex] remains is, 0.0018 %
Explanation :
[tex]K_{sp}[/tex] for CuBr is [tex]4.2\times 10^{-8}[/tex]
[tex]K_{sp}[/tex] for AgBr is [tex]7.7\times 10^{-13}[/tex]
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:
[tex]CuBr\rightleftharpoons Cu^++Br^-[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Cu^+][Br^-][/tex]
[tex]4.2\times 10^{-8}=0.073\times [Br^-][/tex]
[tex][Br^-]=5.75\times 10^{-7}M[/tex]
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
[tex]AgBr\rightleftharpoons Ag^++Br^-[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ag^+][Br^-][/tex]
[tex]7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M[/tex]
[tex][Ag^+]=1.34\times 10^{-6}M[/tex]
Now we have to calculate the percent of [tex]Ag^+[/tex] remains in solution at this point.
Percent of [tex]Ag^+[/tex] remains = [tex]\frac{1.34\times 10^{-6}}{0.073}\times 100[/tex]
Percent of [tex]Ag^+[/tex] remains = 0.0018 %
