# 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 $4 {x}^{3}$ etc but equal to the ksp?

Jan 22, 2017

After further clarification, here's what the questioner was actually looking for. Suppose $4.0 \times {10}^{- 3}$ $\text{M}$ of ${\text{Ag}}^{+}$ was already in solution. Let's find the concentration of ${\text{Cl}}^{-}$ needed to precipitate $\text{AgCl} \left(s\right)$, if ${K}_{s p} = 1.8 \times {10}^{- 10}$.

The concentration given was an initial concentration, so we are not at equilibrium yet. Instead, we'll establish a solubility equilibrium by adding ${\text{Cl}}^{-}$.

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.

${\text{AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" ""0.0040 M"" "" ""0.0000 M}$
$\text{C"" "-" "" "" ""+x M"" "" "" ""+x M}$
$\text{E"" "-" "" ""(0.0040 + x) M"" "" ""x M}$

So, ${K}_{s p}$ is now expressed as:

${K}_{s p} = 1.8 \times {10}^{- 10} = x \left(0.0040 + x\right)$

Before doing anything, what if ["Ag"^(+)] = "0 M"? That would give $1.8 \times {10}^{- 10} = {x}^{2}$, or $\textcolor{g r e e n}{x = \left[{\text{Cl}}^{-}\right] = 1.34 \times {10}^{- 5}}$ $\textcolor{g r e e n}{\text{M}}$. This is without the starting $\left[{\text{Ag}}^{+}\right]$.

To a good approximation, we could actually cross out $x$ in the addition, since ${K}_{s p} \text{<<} {10}^{- 5}$. So:

$1.8 \times {10}^{- 10} \approx 0.0040 x$

$\implies x = \textcolor{b l u e}{\left[{\text{Cl}}^{-}\right] = 4.5 \times {10}^{- 8}}$ $\textcolor{b l u e}{\text{M}}$

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

To compare, if we just solved the full quadratic:

$1.8 \times {10}^{- 10} = 0.0040 x + {x}^{2}$

$\implies {x}^{2} + 0.0040 x - 1.8 \times {10}^{- 10}$

To save time, Wolfram Alpha gives:

$\implies \textcolor{g r e e n}{x = 4.49995 \times {10}^{- 8}}$ $\textcolor{g r e e n}{\text{M}}$

which is hardly any different.

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

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