Question

Question #52bf5

Answer 1

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`