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

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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?

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

##### 1 Answer

After further clarification, here's what the questioner was actually looking for. Suppose

The concentration given was an initial concentration, so **we are not at equilibrium yet**. Instead, we'll establish a solubility equilibrium by adding

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) = 1.8xx10^(-10) = x(0.0040 + x)#

Before doing anything, what if

To a good approximation, we could actually cross out

#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 **decreased**. More specifically, you can put

This illustrates the **common ion effect**, in which an ion belonging to the solute (such as