Question

How much `KBr` should be added to `1` `L` of `0.05` `M` `AgNO_3` solution just to start precipitation of `AgBr`? `K_(sp)` of `AgBr` = `5` x `10^-13`.

Answer 1

`K_(sp)` means that `[Ag^+]*[Br^-]=5*10^-13`,
because the 'sp' suffix stands for 'solution product'.

Explanation:

Concentration of `Ag^+=0.05mol//L`
(one mole of `Ag^+` per mole of `AgNO_3`)
So `[Ag^+]=5*10^-2`

Now we plug in what we know:
`(5*10^-2)*[Br^-]=5*10^-13`
`[Br^-]=(cancel5*10^-13)/(cancel5*10^-2)=1*10^-11mol//L`

If you have the `KBr` in a not very diluted form, you may assume that you'll only need a drop or two, and that the `1L` won't change significantly.
So you'll need `1*10^-11mol` of `KBr` and this can easily be converted to grams (or rather milligrams).

Extra :
One drop more and the solution will turn cloudy white, and turn dark on exposure to sunlight -- that's what classic photography is based upon.