Question

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`.

Answer 1

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?