Question

Suppose 6.54g of potassium bromide is dissolved in 50.mL of a 0.70 M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the potassium bromide is dissolved in it.

Answer 1

Answer:

[$Br^{-}$]=0.4M

Explanation:

When you add bromide to a silver nitrate solution, silver bromide can be produced. It depends of the reactants concentration.

$Br^{-} + Ag^{+}$⇄$AgBr$ Kps for this reaction is $K= 5x10^{-13}$, 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.

$[Br^{-}]=6.54gKBr.\frac{1molKBr}{119gKBr}.\frac{1molBr^{-} }{1molKBr} .\frac{1}{0.05L} =1.10M$

$Qps=[Br^{-}][Ag^{+}]$=$0.70x1.10=0.77$

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.

$[Br^{-}]=\frac{(\frac{1.1mol}{L}.0.05L-\frac{0.70mol}{L}.0.05L}{0.05L} =0.4M$