Common ion effect: Worked examples — Chemistry deep dive

#chemistry/equilibrium #solubility

Intuition Why this page exists

The parent note gave you the machinery: $K_{s p}$ , the "add a common ion, solubility drops" story, and two examples. But real problems come in flavours — different salt shapes, the common ion being the cation vs the anion, the approximation working vs breaking, weird edge cases like "what if there's no common ion at all" or "what if the common ion concentration is so tiny it does nothing". This page walks every one of those cells so you never meet a scenario you haven't seen.

If a symbol here feels unfamiliar, it was built in the parent note and in Solubility Product (Ksp) — every quantity below is either $K_{s p}$ (the fixed number nature assigns each solid), a concentration written $[\cdot]$ in mol/L, or a solubility $s$ = "how many moles of the solid dissolve per litre".

Notation used on this page

Definition The three symbols you will meet everywhere below

$K_{s p}$ — the solubility product: the fixed equilibrium constant for a solid dissolving, equal to the product of its ion concentrations at saturation (built in Solubility Product (Ksp)).
$s$ — the solubility in pure water: how many moles of the solid dissolve per litre when nothing else is in solution.
$s^{'}$ — the solubility in the presence of a common ion: how many moles of the same solid dissolve per litre once a common ion is already present. It is a different number from $s$ , which is exactly why we give it its own prime.
$c$ — the initial common-ion concentration: the amount of the shared ion (mol/L) that a soluble salt puts into solution before the sparingly soluble solid re-adjusts. In the generic 1:1 shortcut $s^{'} = K_{s p} / c$ , this $c$ is that supplied concentration.

The scenario matrix

Before any numbers, let us list every kind of problem this topic can throw at you. Read the table as "these are all the doors; each worked example below opens one." (Recall: $c$ = the initial concentration of the common ion supplied by a soluble salt; $s^{'}$ = solubility of the sparingly soluble solid in its presence.)

Cell	What makes it distinct	Which example
A. 1:1 salt, no common ion	The baseline $s = K_{s p}$ case	Ex 1
B. 1:1 salt, common ion (anion)	Add X⁻, use $s^{'} = K_{s p} / c$	Ex 2
C. 1:1 salt, common ion (cation)	Add M⁺ instead — symmetry check	Ex 3
D. 1:2 salt, no common ion	Stoichiometry ⇒ $s = 3 K_{s p} /4$	Ex 4
E. 1:2 salt, common ion, approx VALID	$s^{'} ≪ c$ genuinely holds	Ex 5
F. 1:2 salt, common ion, approx FAILS	Must solve the cubic exactly	Ex 6
G. Degenerate: common ion so small it's negligible	Limiting behaviour $c \to 0$ recovers cell A	Ex 7
H. Real-world word problem	Will a precipitate even form? ( $Q$ vs $K_{s p}$ )	Ex 8
I. Exam twist: percent-suppression	"By what factor did solubility drop?"	Ex 9

The two axes of this matrix are salt shape (1:1 like AgCl, or 1:2 like PbCl₂) and which ion is common / whether the approximation survives. Cover every row and you have covered the topic.

Figure s01 — Solubility $s^{'}$ of AgCl (vertical axis, log scale) plotted against the initial common-ion concentration $c$ (horizontal axis, log scale). The blue curve is a hyperbola: for tiny $c$ it sits flat at the pure-water value $s = 1.34 \times 1 0^{- 5}$ M (green arrow, cell G), then plunges through the steep-drop region (orange arrow, cells B and C), reaching ~1500× suppression at $c = 0.02$ M (red arrow, cell I). One picture holds most of the matrix.

The figure above plots solubility of AgCl against added common-ion concentration $c$ . Notice the shape: a hyperbola — as soon as $c$ climbs past the pure-water value, solubility plunges. The flat left region (tiny $c$ ) is cell G; the steep-drop region is cells B/C; the far-right is cell I's "huge suppression".

Cell A — 1:1 salt, no common ion (the baseline)

Worked example Ex 1: AgCl in pure water

Given $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ . Find its solubility $s$ .

Forecast: guess the order of magnitude — is $s$ nearer $1 0^{- 5}$ or $1 0^{- 10}$ ? (Take the square root in your head: $1 0^{- 10} = 1 0^{- 5}$ .)

Write the dissolving equilibrium. $AgCl(s) ⇌ Ag^{+} + Cl^{-}$ Why this step? Every solubility problem starts by writing what breaks into what, so we know how many of each ion appear.
Let $s$ = moles of AgCl dissolving per litre. Then $[Ag^{+}] = s$ and $[Cl^{-}] = s$ . Why this step? One formula unit gives one Ag⁺ and one Cl⁻, so both concentrations equal $s$ .
Plug into $K_{s p} = [Ag^{+}] [Cl^{-}]$ : $K_{s p} = s \cdot s = s^{2} ⟹ s = K_{s p} = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M .$ Why this step? With no other source of either ion, $K_{s p}$ is just $s^{2}$ — the simplest case.

Verify: plug back: $(1.34 \times 1 0^{- 5})^{2} = 1.80 \times 1 0^{- 10}$ ✓. Units: (mol/L)² = M², correct for a 1:1 salt.

Cell B — 1:1 salt, common anion

Worked example Ex 2: AgCl in 0.10 M NaCl

Given same $K_{s p} = 1.8 \times 1 0^{- 10}$ , now in 0.10 M NaCl (supplies Cl⁻, so here $c = 0.10$ M).

Forecast: the common ion is Cl⁻ at 0.10 M — that is far bigger than the $1.34 \times 1 0^{- 5}$ of pure water. Guess: does $s^{'}$ drop by a factor of thousands?

Total $[Cl^{-}] = 0.10 + s^{'}$ , and $[Ag^{+}] = s^{'}$ . Why this step? Chloride now has two sources: 0.10 M from NaCl plus $s^{'}$ from dissolving AgCl.
Approximate $0.10 + s^{'} \approx 0.10$ (we bet $s^{'}$ is tiny). Why this step? If the guess holds it turns a quadratic into one line of algebra.
$K_{s p} = s^{'} (0.10) ⟹ s^{'} = \frac{1.8 \times 1 0 ^{- 10}}{0.10} = 1.8 \times 1 0^{- 9} M .$ Why this step? With $[Cl^{-}]$ pinned at the common-ion value $c = 0.10$ , the $K_{s p}$ expression $s^{'} \cdot c$ solves for $s^{'}$ directly by dividing — this is the generic shortcut $s^{'} = K_{s p} / c$ in action.

Verify: is $s^{'} ≪ 0.10$ ? $1.8 \times 1 0^{- 9}$ is $1.8 \times 1 0^{- 6} %$ of 0.10 — yes. Suppression factor $= \frac{1.34 \times 1 0 ^{- 5}}{1.8 \times 1 0 ^{- 9}} \approx 7.4 \times 1 0^{3}$ . About 7500× less soluble ✓.

Cell C — 1:1 salt, common cation (symmetry check)

Cell D — 1:2 salt, no common ion

Cell E — 1:2 salt, common ion, approximation VALID

Cell F — 1:2 salt, common ion, approximation FAILS

Worked example Ex 6: PbCl₂ in 0.010 M Pb(NO₃)₂

Given same $K_{s p}$ ; common ion is Pb²⁺ at 0.010 M (so $c = 0.010$ M).

Forecast: here the common ion is the unsquared one and it is small (0.010). Guess: might $s^{'}$ come out comparable to 0.010, breaking the shortcut?

$[Pb^{2 +}] = 0.010 + s^{'}$ , $[Cl^{-}] = 2 s^{'}$ . Why this step? Lead has the extra source now; chloride comes only from dissolving PbCl₂, hence $2 s^{'}$ .
Naive approximation $0.010 + s^{'} \approx 0.010$ : $K_{s p} = (0.010) (2 s^{'})^{2} = 0.010 \cdot 4 s^{'2} ⟹ s^{'2} = \frac{1.7 \times 1 0 ^{- 5}}{0.04} = 4.25 \times 1 0^{- 4}, s^{'} = 0.0206 M .$ Why this step? We deliberately run the usual shortcut (drop $s^{'}$ next to $c = 0.010$ ) so its failure becomes visible — the result 0.0206 turns out bigger than $c$ , which is the alarm bell that the $s^{'} ≪ c$ assumption is false.
Solve the exact cubic instead: $4 s^{'2} (0.010 + s^{'}) = 1.7 \times 1 0^{- 5} .$ Iterating (try $s^{'} = 0.0135$ ): $4 (0.0135)^{2} (0.010 + 0.0135) = 4 (1.8225 \times 1 0^{- 4}) (0.0235) = 1.71 \times 1 0^{- 5}$ ✓. Why this step? When the approximation fails you keep the full $(c + s^{'})$ factor and hunt for the root numerically.

Answer: $s^{'} \approx 1.35 \times 1 0^{- 2}$ M.

Verify: plug $s^{'} = 0.0135$ : $4 (0.0135)^{2} (0.0235) = 1.71 \times 1 0^{- 5} \approx K_{s p}$ ✓. The lesson: always test $s^{'} < 5%$ of $c$ before trusting the shortcut.

Cell G — Degenerate limit: common ion → 0

Worked example Ex 7: AgCl in 1.0×10⁻⁸ M NaCl (essentially none)

Given $K_{s p} = 1.8 \times 1 0^{- 10}$ ; a whisper of common ion, $c = 1.0 \times 1 0^{- 8}$ M.

Forecast: the common ion is far smaller than pure-water solubility ( $1.34 \times 1 0^{- 5}$ ). Guess: does the answer basically fall back to Ex 1?

Total $[Cl^{-}] = c + s^{'} = 1.0 \times 1 0^{- 8} + s^{'}$ , $[Ag^{+}] = s^{'}$ . Why this step? We must not assume $c$ dominates — here it doesn't.
Since $c ≪ s^{'}$ we instead drop $c$ : $c + s^{'} \approx s^{'}$ , giving $K_{s p} = s^{'} \cdot s^{'} = s^{'2} ⟹ s^{'} = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5} M .$ Why this step? In the limit $c \to 0$ the formula must recover the pure-water result — a sanity check that our whole framework is continuous.

Verify: $s^{'} = 1.34 \times 1 0^{- 5}$ , and indeed $c = 1 0^{- 8}$ is $0.075%$ of that — negligible ✓. So the common-ion effect switches off when $c$ is tiny; the hyperbola in the figure is flat there.

Cell H — Real-world word problem: will it even precipitate?

Worked example Ex 8: Mixing to test for a precipitate

Given Mix equal volumes of $2.0 \times 1 0^{- 4}$ M AgNO₃ and $2.0 \times 1 0^{- 4}$ M NaCl. Does AgCl precipitate? ( $K_{s p} = 1.8 \times 1 0^{- 10}$ .)

Forecast: mixing equal volumes halves each concentration. Guess: is the product still above $K_{s p}$ ?

After mixing (double the volume), each ion halves: $[Ag^{+}] = [Cl^{-}] = \frac{2.0 \times 1 0 ^{- 4}}{2} = 1.0 \times 1 0^{- 4} M .$ Why this step? Dilution: same moles, twice the litres ⇒ half the concentration. Forgetting this is the classic word-problem error.
Compute the reaction quotient $Q = [Ag^{+}] [Cl^{-}]$ : $Q = (1.0 \times 1 0^{- 4}) (1.0 \times 1 0^{- 4}) = 1.0 \times 1 0^{- 8} .$ Why this step? $Q$ is " $K_{s p}$ evaluated at the current concentrations". Comparing $Q$ to $K_{s p}$ tells us the direction (this is Le Chatelier's Principle made quantitative — see Qualitative Analysis).
Compare: $Q = 1.0 \times 1 0^{- 8} > K_{s p} = 1.8 \times 1 0^{- 10}$ . Why this step? $Q > K_{s p}$ means the solution is supersaturated ⇒ precipitation occurs until $Q$ falls to $K_{s p}$ .

Verify: ratio $Q / K_{s p} = 1.0 \times 1 0^{- 8} /1.8 \times 1 0^{- 10} = 55.6 > 1$ ✓ — yes, AgCl precipitates.

Cell I — Exam twist: percent / factor of suppression

Worked example Ex 9: "By what factor does solubility drop?"

Given AgCl: solubility in pure water vs in 0.020 M NaCl. State the suppression factor. ( $K_{s p} = 1.8 \times 1 0^{- 10}$ .)

Forecast: you already know both formulas — $s = K_{s p}$ and $s^{'} = K_{s p} / c$ . Guess: the factor should be $K_{s p} / (K_{s p} / c) = c / K_{s p}$ .

Pure water: $s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}$ M. Why this step? Baseline from cell A.
In NaCl ( $c = 0.020$ M): $s^{'} = \frac{K _{s p}}{c} = \frac{1.8 \times 1 0 ^{- 10}}{0.020} = 9.0 \times 1 0^{- 9}$ M. Why this step? Cell B shortcut, valid since $9.0 \times 1 0^{- 9} ≪ 0.020$ .
Suppression factor: $\frac{s}{s ^{'}} = \frac{1.34 \times 1 0 ^{- 5}}{9.0 \times 1 0 ^{- 9}} \approx 1.49 \times 1 0^{3} .$ Why this step? The exam wants a ratio, not the raw solubilities — reading the question precisely.

Verify: cross-check with the derived shortcut $c / K_{s p} = 0.020/1.34 \times 1 0^{- 5} = 1.49 \times 1 0^{3}$ ✓. Solubility drops about 1500-fold.

Recall

Recall When may I use

$s^{'} = K_{s p} / c$ instead of the full quadratic? Only when the dissolved amount is negligible next to the common ion ::: when $s^{'} < 5%$ of $c$ ; otherwise keep the full $(c + s^{'})$ (or $(c + 2 s^{'})$ ) factor and solve exactly (Ex 6).

Recall For a 1:2 salt

$MX_{2}$ in pure water, what is $s$ in terms of $K_{s p}$ ? $s = 3 K_{s p} /4$ ::: because $K_{s p} = s (2 s)^{2} = 4 s^{3}$ .

Recall How do I decide if a precipitate forms on mixing two solutions?

Compute $Q$ at the diluted concentrations and compare ::: if $Q > K_{s p}$ it precipitates; if $Q < K_{s p}$ it stays dissolved; $Q = K_{s p}$ is exactly saturated.

Connections

Solubility Product (Ksp) — every $Q$ -vs- $K_{s p}$ comparison lives here.
Le Chatelier's Principle — the why behind the leftward shift.
Buffer Solutions — same algebra applied to weak-acid ionization.
Qualitative Analysis — selective precipitation via common ions (Ex 8's logic).
Ionic Equilibrium — the broader home of all these simultaneous equilibria.