Solubility product Ksp — common-ion suppression, selective precipitation: Visual walkthrough

This page rebuilds the whole story of the parent topic from a single picture forward. We start with a jar of water, one pinch of a barely-dissolving salt, and nothing else. By the end you will see — not just compute — why dropping in an ion that is already present slams the solubility down by thousands of times.

Every symbol is earned before it is used. If you can count, and you know that "a solid is stuff you can see settled at the bottom," you can follow from line one.

Step 1 — What "dissolving" actually is (the two-way traffic)

WHAT. Picture a pinch of table-salt-like crystal — call the crystal $MA$ — sitting at the bottom of a beaker of water. Some of its particles break off the surface and float away into the water as separate charged particles. In chemistry a charged particle is called an ion. Positive ones we write $M^{+}$ , negative ones $A^{-}$ .

WHY. Nothing here is one-directional. As free ions wander the water, some bump back into the crystal and re-stick. So there are two flows happening at once: crystal → ions (dissolving) and ions → crystal (re-depositing). Look at the two curved arrows in the figure — one leaving the solid, one returning.

PICTURE.

The half-arrow symbol $⇌$ means "both directions at once, balanced": $MA(s) ⇌ M^{+} (a q) + A^{-} (a q)$

$MA(s)$ — the solid crystal (the " $s$ " tag means solid).
$M^{+} (a q), A^{-} (a q)$ — the freed ions floating in water (the " $a q$ " tag means dissolved in water).

Step 2 — Turning the balance into ONE number: $K_{s p}$

WHAT. At equilibrium we multiply the two ion concentrations together. "Concentration" means how many moles of that ion sit in one litre of water; we write it with square brackets, so $[M^{+}]$ = concentration of $M^{+}$ . This product is a fixed number for a given salt at a given temperature.

WHY multiply, and why leave the solid out? The equilibrium constant idea from Chemical Equilibrium andrium Constant says: take products of the freed species, each raised to how many of it appear. The solid itself doesn't get a concentration — a chunk of crystal is a chunk of crystal, its "amount per litre of water" is meaningless, so its activity is defined as $1$ . That leaves only the ions.

$fixed number K_{s p} = how much M^{+} [M^{+}] \times how much A^{-} [A^{-}]$

PICTURE. Think of $K_{s p}$ as the area of a rectangle whose width is $[M^{+}]$ and height is $[A^{-}]$ . The salt "allows" only rectangles of one fixed area. You can make it tall-and-thin or short-and-wide, but the area stays locked.

Step 3 — Solubility $s$ in pure water (the square rectangle)

WHAT. Drop $MA$ into pure water. Let $s$ = how many moles per litre actually dissolve — this is the solubility. Because the formula is one $M^{+}$ per one $A^{-}$ , every dissolved unit gives one of each. So both concentrations equal $s$ : $[M^{+}] = s, [A^{-}] = s .$

WHY. In pure water nothing else supplies these ions — the only source is the crystal itself, and it releases them in a 1-to-1 pair. So width = height. The locked rectangle is a square of side $s$ .

$K_{s p} = s \times s = s^{2} ⟹ s = K_{s p}$

$s^{2}$ — area of the square, side $s$ .
$K_{s p}$ — "which side length gives this area?" That square-root undoes the squaring.

PICTURE. A perfectly square version of Step 2's rectangle.

Step 4 — Now pour in a common ion (widen the rectangle)

WHAT. Into the same beaker, dissolve a different, fully-soluble salt that shares the $A^{-}$ ion — for AgCl that's ordinary $NaCl$ , which floods the water with $Cl^{-}$ . Call the extra amount $C$ . Now, before AgCl dissolves at all, the water already holds $[A^{-}] = C$ .

WHY it's called "common." The $A^{-}$ is common to both salts — one salt shares an ion the other also makes. That shared ion is the common ion.

Here is the key move. The locked area $K_{s p}$ has not changed (same salt, same temperature). But we just forced the height $[A^{-}]$ up to $C$ . To keep the area constant, the width $[M^{+}]$ must shrink.

$K_{s p} = must shrink [M^{+}] \times shoved up to C [A^{-}]$

PICTURE. Take the square from Step 3 and stretch it very tall (height $C$ ). To keep the same area, it becomes razor-thin. That thin width IS the new, tiny solubility.

This "area resists, so the other side gives way" is exactly Le Chatelier's Principle: push on the products, the balance retreats toward the solid.

Step 5 — The suppressed-solubility formula (read the thin width)

WHAT. Let $s^{'}$ = the new solubility once the common ion is present. Each dissolved AgCl still gives one $A g^{+}$ , so $[M^{+}] = s^{'}$ . The chloride is the flood $C$ plus the tiny bit AgCl adds: $[A^{-}] = C + s^{'}$ .

$K_{s p} = [M^{+}] s^{'} \times [A^{-}] (C + s^{'})$

WHY we may drop the $s^{'}$ inside the bracket. $s^{'}$ is fantastically small (the width is nearly zero), while $C$ is a normal lab concentration. Adding a droplet to an ocean changes nothing: $C + s^{'} \approx C$ . Look at the figure — the sliver of extra height is invisible next to $C$ .

$K_{s p} \approx s^{'} C ⟹ s^{'} \approx \frac{K _{s p}}{C}$

$K_{s p}$ — the locked area (top of the fraction).
$C$ — the forced-up common-ion concentration (bottom).
Bigger $C$ ⇒ smaller $s^{'}$ . Solubility is inversely proportional to the common ion.

PICTURE. The rectangle re-drawn with the exact numbers, showing the negligible sliver.

Step 6 — The 1:2 case (the rectangle's height is squared)

WHAT. Some salts break into two of one ion, e.g. $CaF_{2} ⇌ C a^{2 +} + 2 F^{-}$ . Now the locked shape is not width×height but width×height×height, because $[F^{-}]$ appears twice — once for each fluoride the balanced equation makes: $K_{s p} = [C a^{2 +}] [F^{-}]^{2}$

The exponent $2$ on $[F^{-}]$ is literally the coefficient 2 in the equation — two fluorides per formula unit, so the concentration is multiplied by itself.

WHY it still suppresses. Flood the water with $F^{-}$ at concentration $C$ (from NaF). The $[F^{-}]^{2}$ term becomes $C^{2}$ , and to keep $K_{s p}$ locked, $[C a^{2 +}] = s^{'}$ shrinks even harder: $K_{s p} \approx s^{'} C^{2} ⟹ s^{'} \approx \frac{K _{s p}}{C ^{2}}$

Because the flood is squared, the choke is fiercer than the 1:1 case.

PICTURE. A box whose one face is $s^{'}$ and whose square face is $C \times C$ — the volume is the locked $K_{s p}$ .

Step 7 — The edge cases (never leave the reader stranded)

WHAT & WHY — every degenerate scenario:

PICTURE. All four cases as four little rectangles side by side.

The one-picture summary

One salt = one locked rectangle area $K_{s p}$ . Pure water gives the square (side $= s = K_{s p}$ ). Flooding a common ion stretches the rectangle tall, so the other side collapses to a sliver $s^{'} = K_{s p} / C$ . That collapsing sliver is the suppressed solubility — the entire common-ion effect in one image. This same $Q$ -vs- $K_{s p}$ reasoning powers Qualitative Inorganic Analysis and Ionic Equilibria in Solutions.

Recall Feynman retelling — say it to a friend

A barely-dissolving salt sits in water in a tug-of-war: bits break off, bits re-stick, and at balance the product of the two ion concentrations is a fixed number, $K_{s p}$ — think of it as a rectangle whose area is locked. In pure water the two ions come out in equal amounts, so the rectangle is a square and each side is $K_{s p}$ . Now pour in a second salt that shares one of the ions: you've suddenly made that side of the rectangle enormous. But the area is locked, so the other side — which is the amount of our salt that can dissolve — has to shrink to almost nothing. That's why AgCl is 7500× less soluble in salty water: not magic, just "keep the area constant while I stretch one side." If the shared ion appears twice in the formula (like the two fluorides in CaF₂), that side is squared, so the choke is even harsher. And before anything precipitates when you mix two solutions, you just compute the current rectangle area $Q$ and ask: is it bigger than the locked area $K_{s p}$ ? If yes, solid rains out until it fits.

Recall Quick self-test

Why is the solid left out of the $K_{s p}$ expression? ::: Its activity is defined as 1 — a pure solid has no meaningful "concentration per litre of water," so it contributes a factor of 1. In pure water the rectangle is a ________ shape. ::: A square, side $s = K_{s p}$ , because the two ions appear in equal amounts. Adding a common ion at concentration $C$ makes solubility $s^{'} \approx$ ? ::: $K_{s p} / C$ (for a 1:1 salt) — inversely proportional to $C$ . For CaF₂ the flood enters as $C^{2}$ — why the square? ::: The balanced equation makes two F⁻ per unit, so $[F^{-}]$ is raised to the power 2 in $K_{s p}$ . When does $s^{'} = K_{s p} / C$ FAIL? ::: When $C$ is small (not $≫ s^{'}$ ) — then solve the full quadratic $K_{s p} = s^{'} (C + s^{'})$ ; and it never applies at $C = 0$ (use $K_{s p}$ ).