How do you find the solubility of silver bromide in 0.01 M NaBr(aq)? The solubility product of silver bromide is #7.7 xx 10^-13#.

When finding the solubility of a compound in a solution, it is always easier to find its solubility in pure water first.

Define solubility as #s#, so that with the 1:1 stoichiometry in the reaction

#"AgBr"(s) rightleftharpoons "Ag"^(+)(aq) + "Br"^(-)(aq)#,

the solubility product constant is

#K_(sp) = 7.7 xx 10^(-13) = ["Ag"^(+)]["Br"^(-)]#

#= s*s = s^2#.

Thus, we have that the solubility in pure water is:

#s = sqrt(K_(sp)) = 8.775 xx 10^(-7)# #"M"# #"Br"^(-)# or #"Ag"^(+)#

In #"NaBr"(aq)#, what we have is a new initial concentration of #"Br"^(-)#, in which the concentration of #"Br"^(-)# is 1:1 with that of #"NaBr"#, and induces a skewed equilibrium upon being introduced into the previous aqueous solution.

Define a new solubility #s'# so that for the following ICE table we have:

#"AgBr"(s) rightleftharpoons "Ag"^(+)(aq) + "Br"^(-)(aq)#,

#"I"" "--" "" ""0 M"" "" "" ""0.01 M"#
#"C"" "--" "+s'" "" "" "+s'#
#"E"" "--" "" "s'" "" "" ""(0.01 + s') M"#

Notice how the equilibrium previously had #K_(sp) = s*s = s^2#. This one instead is:

#K_(sp) = 7.7 xx 10^(-13) = s'(0.01 + s')#

We assume that #s'# is small compared to #0.01# as #K_(sp) "<<" 10^(-5)#. Thus, we have a simplified expression:

#7.7 xx 10^(-13) ~~ 0.01s'#

#=> color(blue)(s' = 7.7 xx 10^(-11))# #color(blue)("M")#

The true answer would have been from solving

#0 = (s')^2 + 0.01s' - 7.7 xx 10^(-13)#

which would give an identical answer because the percent dissociation is much less than #5%#.

Therefore, the introduction of #"NaBr"# into solution decreased the solubility of #"Br"^(-)# ions by a factor of #~~11396#. Does that fall in line with Le Chatelier's principle?