Visual walkthrough — Common ion effect

#chemistry/equilibrium #solubility #le-chateliers-principle

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Step 1 — What "dissolving" even means

WHAT. Picture a lump of a solid salt sitting at the bottom of a glass of water. Call the salt $\text{MX}$ — a made-up name where $\text{M}$ is the positive part (the cation) and $\text{X}$ is the negative part (the anion). Water molecules chip ions off the surface: some ${\text{M}}^{+}$ and ${\text{X}}^{-}$ float away, while others bump back onto the solid and re-attach.

WHY. Before any formula, we must agree on the physical event: dissolving is two things happening at once — ions leaving the solid, and ions rejoining it. That two-way traffic is the whole story.

PICTURE. In the figure below, purple arrows point OFF the solid (ions escaping) and orange arrows point ONTO the solid (ions returning).

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Step 2 — Equilibrium: the two arrows match

WHAT. Wait a while. The rate of ions leaving the solid slows down (the surface has less exposed) and the rate of ions returning speeds up (more ions now floating). At some moment the two rates become equal. We write this balance with a double arrow:

$\underset{\text{solid salt}}{\underbrace{\text{MX(s)}}}\rightleftharpoons \underset{\begin{array}{c}\text{cation}\\ \text{in water}\end{array}}{\underbrace{{\text{M}}^{+}\text{(aq)}}}+\underset{\begin{array}{c}\text{anion}\\ \text{in water}\end{array}}{\underbrace{{\text{X}}^{-}\text{(aq)}}}$

Reading the symbols: $\text{(s)}$ means solid, $\text{(aq)}$ means dissolved in water (aqueous), and $\rightleftharpoons$ means "goes both ways and is now balanced."

WHY. Equilibrium is the frozen snapshot where nothing net changes. Only at this snapshot can we write a fixed relationship between the ion amounts — that is what makes the algebra possible. This balance point is governed by Le Chatelier's Principle.

PICTURE. The two arrows are now the same length — leaving traffic equals returning traffic.

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Step 3 — Naming amounts: solubility $s$

WHAT. We need a number for "how much dissolved." Let

$s=\text{moles of MX that dissolve in one litre of water (mol/L)}.$

Because each $\text{MX}$ unit splits into exactly one ${\text{M}}^{+}$ and one ${\text{X}}^{-}$, if $s$ units dissolve then:

$[{\text{M}}^{+}]=s,[{\text{X}}^{-}]=s.$

The square brackets $[]$ are shorthand for "concentration of, in mol/L."

WHY. We can't compare "pure water" with "common-ion water" until we have a single measuring stick. $s$ is that stick — the height of the dissolved-ion bar.

PICTURE. Two equal bars, one for each ion, both of height $s$.

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Step 4 — The rule the ions must obey: $K_{sp}$

WHAT. Experiments show that at equilibrium the product of the two ion concentrations is a fixed number for each salt at a given temperature. We name it the solubility product:

$K_{sp}=\underset{\text{cation amount}}{\underbrace{[{\text{M}}^{+}]}}\times \underset{\text{anion amount}}{\underbrace{[{\text{X}}^{-}]}}$

The solid itself does not appear (its "concentration" is fixed at 1 by convention — a pure solid has no dilution to vary).

Plug in Step 3 ($[{\text{M}}^{+}]=[{\text{X}}^{-}]=s$):

$K_{sp}=s\cdot s=s^2⟹\boxed{s=\sqrt{K_{sp}}}$

WHY a product and not a sum? Because equilibrium constants always multiply the amounts of things being made (products of the reaction), each raised to how many form. That multiplication is what stays constant when you push the system. See Solubility Product (Ksp).

WHY the square root? We asked "what number times itself gives $K_{sp}$?" — that is exactly what ​ answers. It undoes the squaring in $s^2$.

PICTURE. The two bars of height $s$ form a square of area $s^2$; that shaded area is locked at the value $K_{sp}$.

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Step 5 — The intruder: adding a common ion

WHAT. Now pour in a fully soluble salt, $\text{NaX}$, at concentration $c$ mol/L. It dumps its ${\text{X}}^{-}$ instantly and completely into the water — before the $\text{MX}$ has a chance to react. So the water's ${\text{X}}^{-}$ shelf suddenly jumps up by $c$. The shared ion here is ${\text{X}}^{-}$: the common ion.

WHY. This is the "stress" in Le Chatelier's Principle. We flooded the right-hand side of the equation with extra ${\text{X}}^{-}$. The system must respond to relieve that stress — and Step 6 shows how.

PICTURE. The ${\text{X}}^{-}$ bar rockets up to height $c$ (magenta block), while the ${\text{M}}^{+}$ bar hasn't moved yet.

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Step 6 — The push-back: equilibrium shifts left

WHAT. Too much ${\text{X}}^{-}$ means the returning arrow now wins: excess ${\text{X}}^{-}$ grabs ${\text{M}}^{+}$ and slams it back onto the solid as $\text{MX}$. This continues until the product $[{\text{M}}^{+}][{\text{X}}^{-}]$ is back down to $K_{sp}$.

Let $s^{\prime }$ = the new, smaller solubility. Now the amounts are:

$[{\text{M}}^{+}]=\underset{\text{only from dissolved MX}}{\underbrace{s^{\prime }}},[{\text{X}}^{-}]=\underset{\text{from NaX}}{\underbrace{c}}+\underset{\text{from dissolved MX}}{\underbrace{s^{\prime }}}$

Both ion sources of ${\text{X}}^{-}$ pile into the same shelf, so we add them: $c+s^{\prime }$. Put this into the locked rule:

$K_{sp}=\underset{[{\text{M}}^{+}]}{\underbrace{s^{\prime }}}\cdot \underset{[{\text{X}}^{-}]}{\underbrace{(c+s^{\prime })}}$

WHY. $K_{sp}$ never changes (same salt, same temperature). But we just forced $[{\text{X}}^{-}]$ up to $\approx c$. For the product to stay the same, $[{\text{M}}^{+}]=s^{\prime }$ must shrink — that shrinking IS the common ion effect.

PICTURE. The equation is a see-saw: the ${\text{X}}^{-}$ side went up, so the ${\text{M}}^{+}$ side must come down to keep the product (area) constant at $K_{sp}$.

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Step 7 — The clean formula (and when it's allowed)

WHAT. Because the common ion crushed solubility, the new $s^{\prime }$ is tiny — far smaller than $c$. So the extra $s^{\prime }$ riding on the ${\text{X}}^{-}$ shelf is a rounding error:

$c+s^{\prime }\approx c(\text{valid when }{s}^{\prime }\ll c)$

Drop it and the algebra collapses:

$K_{sp}\approx s^{\prime }\cdot c⟹\boxed{s^{\prime }=\frac{K_{sp}}{c}}$

Compare the two results side by side:

$s=\sqrt{K_{sp}}\text{(pure water)}⟶s^{\prime }=\frac{K_{sp}}{c}\text{(common ion)}$

WHY the approximation? It turns an ugly quadratic into one division. But it is a loan — you must pay it back by checking $s^{\prime }$ really is small. Rule of thumb: $s^{\prime }$ should be under $5\%$ of $c$.

PICTURE. Solubility vs. added $c$ falls as a hyperbola — double $c$, halve $s^{\prime }$. The curve's steep plunge near the origin is the effect at work.

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Step 8 — The degenerate case: when the shortcut breaks

WHAT. The approximation fails if $s^{\prime }$ turns out comparable to $c$. This bites hardest for salts like ${\text{PbCl}}_2$ that release two anions:

${\text{PbCl}}_2\text{(s)}\rightleftharpoons {\text{Pb}}^{2+}+2{\text{Cl}}^{-}$

With $K_{sp}=1.7\times 10^{-5}$ and $c=0.01$ M of added ${\text{Pb}}^{2+}$, the naive formula gives $s^{\prime }\approx 0.0206$ M — bigger than $c$ itself. That is nonsense for a "suppressed" solubility, so we must solve the full cubic:

$K_{sp}=(0.01+s^{\prime })(2s^{\prime }{)}^2=0.04s^{\prime 2}+4s^{\prime 3}$

which yields $s^{\prime }\approx 0.0135$ M.

WHY. Whenever the salt's stoichiometry raises the dissolved term to a power ($(2s^{\prime }{)}^2$ here), $s^{\prime }$ can be large enough that dropping it lies to you. The lesson: always re-check $s^{\prime }\ll c$; if it fails, keep the full expression.

PICTURE. Two bars for the true answer vs. the naive answer — the naive one overshoots the fence.

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The one-picture summary

The whole derivation on one canvas: the locked area $K_{sp}$ (Step 4), the intruder raising $[{\text{X}}^{-}]$ (Step 5), the see-saw pushing $[{\text{M}}^{+}]$ down (Step 6), and the hyperbola of falling solubility (Step 7).

Recall Feynman retelling — the walkthrough in plain words

A pinch of salt sits half-dissolved in water: for every ion that floats off the solid, another ion drifts back and sticks. When those two flows match, we're "balanced," and nature enforces one rule — the product of the two floating ion amounts is a fixed number, $K_{sp}$. In clean water both ions have the same height $s$, so $s\times s=K_{sp}$ and $s=\sqrt{K_{sp}}$. Now cheat: pour in a different salt that shares one of the ions. That ion's pile suddenly towers up. But the product must stay $K_{sp}$! So the other ion has to shrink to compensate — meaning less of our original salt can stay dissolved. That's the common ion effect: $s^{\prime }=K_{sp}/c$, a much smaller number. One catch — if the salt spits out two ions at once, or if the added amount is small, the tidy division lies, and you must solve the full equation and always double-check the answer stayed small.

Recall Quick self-test

Why does $[{\text{X}}^{-}]$ equal $c+s^{\prime }$ and not just $c$? ::: Because both the added $\text{NaX}$ (gives $c$) and the tiny bit of dissolved $\text{MX}$ (gives $s^{\prime }$) drop ${\text{X}}^{-}$ into the same solution; they add on the same shelf. What does the square root in $s=\sqrt{K_{sp}}$ undo? ::: The squaring in $K_{sp}=s^2$; it answers "what number times itself gives $K_{sp}$?" When is $s^{\prime }=K_{sp}/c$ NOT safe to use? ::: When $s^{\prime }$ is not much smaller than $c$ (over $\sim 5\%$), e.g. multi-ion salts like ${\text{PbCl}}_2$ — then solve the exact equation.

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Connections

• Le Chatelier's Principle — the shift-left push in Step 6 is a direct application.
• Solubility Product (Ksp) — the locked area of Step 4.
• Buffer Solutions — same common-ion logic applied to weak acids.
• Qualitative Analysis — using a common ion to precipitate selectively.
• Ionic Equilibrium — the broader setting.
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