# 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#.

##### 1 Answer

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

#"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

Define a new solubility

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

We assume that

#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

Therefore, the introduction of **decreased the solubility** of