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#.

1 Answer
MeneerNask
Jan 30, 2016

#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.

Related questions
See all questions in Ksp
Impact of this question
5904 views around the world
You can reuse this answer
Creative Commons License