Answer:
[[tex]Br^{-}[/tex]]=0.4M
Explanation:
When you add bromide to a silver nitrate solution, silver bromide can be produced. It depends of the reactants concentration.
[tex]Br^{-} + Ag^{+}[/tex]⇄[tex]AgBr[/tex] Kps for this reaction is [tex]K= 5x10^{-13}[/tex], this constant indicates if the reaction can happen spontaneously or not. in this case, a small value of Kps means that precipitate can be formed.
Now we need to know if our conditions are enough to form that precipitate.
[tex][Br^{-}]=6.54gKBr.\frac{1molKBr}{119gKBr}.\frac{1molBr^{-} }{1molKBr} .\frac{1}{0.05L} =1.10M[/tex]
[tex]Qps=[Br^{-}][Ag^{+}][/tex]=[tex]0.70x1.10=0.77[/tex]
Due to Qsp>Ksp precipitate will be formed. notice that reaction will go on until silver has been consumed completely, silver is limiting reagent.
moles of bromide remainig =total moles of bromide-moles of bromide that reacts
moles of bromide that reacts are the same number of moles from silver beacause in reaction the mole ratio is 1:1.
[tex][Br^{-}]=\frac{(\frac{1.1mol}{L}.0.05L-\frac{0.70mol}{L}.0.05L}{0.05L} =0.4M[/tex]
