Question

`K_(sp)` with different starting solubility for each ion. Is the solution not at equilibrium anymore?

Pretend that there is a solution with ions in equilibrium with the solid.
Is the whole thing still at equilibrium if the solubility of the ions are different (like ksp=xy instead of `x^2` or `4x^3` etc but equal to the ksp?

Answer 1

After further clarification, here's what the questioner was actually looking for. Suppose `4.0xx10^(-3)` `"M"` of `"Ag"^(+)` was already in solution. Let's find the concentration of `"Cl"^(-)` needed to precipitate `"AgCl"(s)`, if `K_(sp) = 1.8xx10^(-10)`.

The concentration given was an initial concentration, so we are not at equilibrium yet. Instead, we'll establish a solubility equilibrium by adding `"Cl"^(-)`.

Normally an ICE Table is not required for solubility problems, but for this scenario it helps to make one to see what's going on.

`"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)`

`"I"" "-" "" "" ""0.0040 M"" "" ""0.0000 M"`
`"C"" "-" "" "" ""+x M"" "" "" ""+x M"`
`"E"" "-" "" ""(0.0040 + x) M"" "" ""x M"`

So, `K_(sp)` is now expressed as:

`K_(sp) = 1.8xx10^(-10) = x(0.0040 + x)`

Before doing anything, what if `["Ag"^(+)] = "0 M"`? That would give `1.8xx10^(-10) = x^2`, or `color(green)(x = ["Cl"^(-)] = 1.34xx10^(-5))` `color(green)("M")`. This is without the starting `["Ag"^(+)]`.

To a good approximation, we could actually cross out `x` in the addition, since `K_(sp) "<<" 10^(-5)`. So:

`1.8xx10^(-10) ~~ 0.0040x`

`=> x = color(blue)(["Cl"^(-)] = 4.5xx10^(-8))` `color(blue)("M")`

`=> 0.0040 + x = color(blue)(["Ag"^(+)] ~~ "0.0040 M")`

To compare, if we just solved the full quadratic:

`1.8xx10^(-10) = 0.0040x + x^2`

`=> x^2 + 0.0040x - 1.8xx10^(-10)`

To save time, Wolfram Alpha gives:

`=> color(green)(x = 4.49995xx10^(-8))` `color(green)("M")`

which is hardly any different.

Anyways, the result is that the solubility of `"Cl"^(-)` decreased. More specifically, you can put `ul(0.336%)` the amount of `"Cl"^(-)` in and still precipitate `"AgCl"`, which is quite a drastic decrease in solubility.

This illustrates the common ion effect, in which an ion belonging to the solute (such as `"Ag"^(+)` in `"AgCl"(s)`) is already in solution, decreasing the solubility of the other ion that would have been dissociated from the solute (such as `"Cl"^(-)` in `"AgCl"(s)`).