Question #52bf5

1 Answer
MeneerNask
Jun 23, 2015

If you look at the concentrations, you will see that #Br^-# is greatly in excess, about a factor of 1 to 1000.

Explanation:

So the tiny bit of #Br^-# that reacts with the #Ag^+# will hardly make a dent in the concentration (less than 0,1% = third decimal).
Of course the concentration is halved (twice the amount of liquid).
We may take [Br^-] to be #1.0*10^-3# after the reaction.

If we now set up the solubility product:

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

Solving:
#[Ag^+]*[1.0*10^-3]=5.0*10^-13->#

#[Ag^+]=(5.0*10^-13)/(1.0*10^-3)=5.0*10^-10M#

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