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