Question

What is the maximum `["CO"_3^(2-)]` that can exist in a saturated solution of AgBr?

AgBr Ksp = 5.4 x 10^-13
Ag2CO3 Ksp = 8.5 x 10^-12

Answer 1

This problem is why you have to always check before or after you do the small `s` approximation to see if it worked... I got `1.28 xx 10^(-4) "M"` the long way, and `"15.74 M"` the short way.

Clearly, the short way didn't agree.

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First, we'll need to find the molar concentration of `"Ag"^(+)` in solution. That gives rise to the common ion effect, which limits how much `"Ag"_2"CO"_3` can dissolve since it also dissolves to give `"Ag"^(+)` in solution.

A saturated `"AgBr"` solution satisfies the solubility product equilibrium:

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

`"I"" "-" "" "" "" "0" "" "" "" "0`
`"C"" "-" "" "" "+s" "" "" "+s`
`"E"" "-" "" "" "" "s" "" "" "" "s`

where `s` is defined to be the solubility in pure water for the ion with a stoichiometric coefficient of `1`.

Note that had the stoichiometric coefficient been `nu ne 1`, the concentration would have changed by `nus`. The solubility product constant expression then gives:

`K_(sp) = 5.4 xx 10^(-13) = ["Ag"^(+)]["Br"^(-)] = s^2`

Thus,

`s = sqrt(5.4 xx 10^(-13)) = 7.35 xx 10^(-7) "M" = ["Ag"^(+)]_1`

So, this becomes our initial concentration of `"Ag"^(+)` while dissolving `"Ag"_2"CO"_3` next. (We can do this because concentration is a state function.)

`"Ag"_2"CO"_3(s) " "rightleftharpoons" " 2"Ag"^(+)(aq) " "+" " "CO"_3^(2-)(aq)`

`"I"" "-" "" "" "" "" "" "7.35 xx 10^(-7)" "" "" "" "0`
`"C"" "-" "" "" "" "" "+2s'" "" "" "" "" "" "+s'`
`"E"" "-" "" "" "" "" "7.35 xx 10^(-7) + 2s'" "" "s'`

where `s'` is defined to be the solubility in ion-containing water for the ion with a stoichiometric coefficient of `1`.

In this case, we CANNOT make the small `s` approximation. If we did... we would get a very unreasonable answer.

`K_(sp) = 8.5 xx 10^(-12) = ["Ag"^(+)]^2["CO"_3^(2-)]`

`= (7.35 xx 10^(-7) + 2s')^2(s')`

`stackrel(?" ")(~~) (7.35 xx 10^(-7))^2s'`

That would have given:

`color(red)(s') = ["CO"_3^(2-)] = (8.5 xx 10^(-12))/(7.35 xx 10^(-7))^2`

`=` `color(red)"15.74 M"`

which makes no sense whatsoever! Concentrated nitric acid is about `"16 M"`. Concentrated hydrochloric acid is about `"12 M"`. Concentrated sulfuric acid is around `"18 M"`.

This unreasonable answer probably arose because the two `K_(sp)` values are very similar and thus the two substances are interacting at a non-negligible level.

Let's NOT do that approximation in this case, and solve for the cubic.

`8.5 xx 10^(-12) = (7.35 xx 10^(-7) + 2s')^2(s')`

`= (5.40 xx 10^(-13) + 4cdot7.35 xx 10^(-7)s' + 4s'^2)s'`

`= 5.40 xx 10^(-13)s' + 2.940 xx 10^(-6)s'^2 + 4s'^3`

The final cubic becomes:

`4s'^3 + 2.940 xx 10^(-6)s'^2 + 5.40 xx 10^(-13)s' - 8.5 xx 10^(-12) = 0`

The numerical solution to this cubic can be done in any modern TI calculator to give a physically reasonable answer of:

`color(blue)(s' = ["CO"_3^(2-)] = 1.28 xx 10^(-4) "M")`

And if we check in `K_(sp)`, we should find that this worked...

`8.5 xx 10^(-12) stackrel(?" ")(=) ["Ag"^(+)]^2["CO"_3^(2-)]`

`= (["Ag"^(+)]_1 + 2s')^2(s')`

`= (7.35 xx 10^(-7) + 2 cdot 1.28 xx 10^(-4))^2(1.28 xx 10^(-4))`

`= 8.44 xx 10^(-12) ~~ 8.5 xx 10^(-12)` `color(blue)(sqrt"")`

Yep, it worked!