Common ion effect: Question bank — Chemistry deep dive

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

Reminder of the symbols used everywhere below:

$s$ = solubility in pure water (mol/L that dissolves).
$s^{'}$ = solubility in the presence of a common ion.
$c$ = concentration of the added common ion.
$K_{s p}$ = solubility product, a temperature-fixed constant.
$Q$ = the ionic product $[M^{+}] [X^{-}]$ computed with whatever concentrations you currently have (not necessarily at equilibrium).

True or false — justify

The common ion effect lowers the value of

K_{s p}

False.

K_{s p}

depends only on temperature. The common ion lowers solubility

s^{'}

by raising one ion's concentration so the other must fall to keep the product at the same

K_{s p}

Adding a common ion always precipitates some extra solid from a saturated solution.

True (for an already-saturated solution). Raising one ion pushes

Q

above

K_{s p}

, so the system is supersaturated and precipitates until

Q = K_{s p}

again — Le Chatelier shifting the dissolution equilibrium left.

For AgCl,

[Ag^{+}]

must equal

[Cl^{-}]

at every equilibrium.

False. They are equal only in pure water. Add NaCl and

[Cl^{-}] ≫ [Ag^{+}]

; add AgNO₃ and

[Ag^{+}] ≫ [Cl^{-}]

. Only the product

[Ag^{+}] [Cl^{-}] = K_{s p}

is forced.

Adding NaNO₃ (no shared ion) to saturated AgCl changes its solubility not at all in an idealized calculation.

True in the simple model. NaNO₃ shares neither Ag⁺ nor Cl⁻, so it does not enter the

K_{s p}

expression. (In reality it slightly increases solubility via ionic-strength/activity effects — the "salt effect", opposite to the common ion effect.)

The common ion effect and the salt (diverse ion) effect push solubility in the same direction.

False. Common ion (shared ion) decreases solubility; the diverse-ion/salt effect (inert electrolyte) increases it slightly by lowering ion activities. They are opposite phenomena.

If you double the common ion concentration

c

, the solubility of an MX salt halves.

True in the approximation

s^{'} = K_{s p} / c

, which is inversely proportional to

c

. Doubling

c

halves

s^{'}

— provided

s^{'} ≪ c

still holds.

For a salt

MX_{2}

, doubling

c

of the

M^{2 +}

common ion halves the solubility.

False. Here

s^{'} = K_{s p} / (4 c) \propto c^{- 1/2}

, so doubling

c

multiplies

s^{'}

1/ 2 \approx 0.71

, not

0.5

. Stoichiometry changes the exponent.

The common ion effect works whether you add the shared cation or the shared anion.

True. Either ion appearing in the

K_{s p}

product will push the equilibrium left. For AgCl, extra Cl⁻ (from NaCl) or extra Ag⁺ (from AgNO₃) both suppress dissolution.

At the exact instant NaCl is added, the solution is momentarily supersaturated.

True. The added Cl⁻ raises

Q

above

K_{s p}

before any AgCl precipitates. The system then relaxes back to

Q = K_{s p}

by depositing solid.

Spot the error

"Since NaCl adds Cl⁻, solubility of AgCl goes down, so

K_{s p}

went down too." — find the flaw.

The conclusion about

K_{s p}

is wrong. Solubility fell but

K_{s p}

is a constant; what happened is

[Cl^{-}]

rose and

[Ag^{+}]

fell so their product is unchanged.

"For PbCl₂ dissolving,

[Cl^{-}] = s^{'}

." — spot the error.

Each formula unit releases two Cl⁻, so

[Cl^{-}] = 2 s^{'}

. Forgetting the stoichiometric coefficient underestimates

[Cl^{-}]

and mis-writes

K_{s p} = [Pb^{2 +}] (2 s^{'})^{2}

"With common ion present I still write

K_{s p} = s^{'} \cdot s^{'} = s^{'2}

." — what's wrong?

The anion concentration is the total

c + s^{'}

, not just

s^{'}

. Correct form is

K_{s p} = s^{'} (c + s^{'})

. Using

s^{'2}

ignores the ion you deliberately added.

s^{'} = K_{s p} / c

is exact, so I never need to check anything." — error?

It is an approximation valid only when

s^{'} ≪ c

(rule of thumb

s^{'} < 5%

c

). If the added-ion concentration is small or

K_{s p}

large, you must solve the full equation — see PbCl₂ where the shortcut overestimated

s^{'}

"Adding AgNO₃ to saturated AgCl increases solubility because you're adding more silver salt." — spot the trap.

Backwards. Ag⁺ is a common ion; extra Ag⁺ pushes the equilibrium left and decreases the amount of AgCl that stays dissolved, precipitating more solid.

K_{s p}

for PbCl₂ is just

[Pb^{2 +}] [Cl^{-}]

." — what's missing?

The Cl⁻ term must be squared:

K_{s p} = [Pb^{2 +}] [Cl^{-}]^{2}

. Each ion is raised to its stoichiometric coefficient in the equilibrium.

"Solid AgCl appears in the

K_{s p}

expression as

[AgCl]

." — flaw?

Pure solids (and pure liquids) have activity

= 1

and never appear in the equilibrium expression.

K_{s p} = [Ag^{+}] [Cl^{-}]

contains only the aqueous ions.

"Since

Q < K_{s p}

after adding some Cl⁻, the solution will precipitate." — error?

Wrong direction. Precipitation requires

Q > K_{s p}

. If

Q < K_{s p}

the solution is unsaturated and more solid would dissolve, not precipitate.

Why questions

Why does the common ion effect follow directly from Le Chatelier's Principle?

Dissolution

MX(s) ⇌ M^{+} + X^{-}

is an equilibrium; adding a product ion is a stress the system relieves by shifting left, consuming ions and depositing solid — exactly Le Chatelier.

Why does solid MX not appear in

K_{s p}

Its concentration (activity) is essentially constant and defined as 1 for a pure solid, so it factors out of the equilibrium constant and only the aqueous ion concentrations remain.

Why is

s^{'} = K_{s p} / c

an inverse relationship for an MX salt?

Because

K_{s p} = s^{'} \cdot c

(with

c + s^{'} \approx c

), so

s^{'}

K_{s p}

divided by

c

. Bigger common-ion load

c

leaves less room for MX to dissolve, hence smaller

s^{'}

Why can the common ion effect separate ions in Qualitative Analysis?

Different salts have very different

K_{s p}

values, so a fixed common-ion concentration precipitates the low-

K_{s p}

salt while the higher-

K_{s p}

one stays dissolved — selective precipitation.

Why does the common ion effect stabilize the pH of a buffer?

The added common ion (e.g. conjugate base A⁻) suppresses ionization of the weak acid HA, pinning

[H^{+}]

near a set value so added acid or base barely shifts it.

Why must you use

2 s^{'}

(not

s^{'}

) for the chloride from PbCl₂, yet still keep

s^{'}

for lead?

Stoichiometry: one dissolved unit yields one Pb²⁺ but two Cl⁻. Concentrations reflect how many of each ion each dissolving unit produces.

Why does adding an inert salt like KNO₃ slightly raise solubility instead of lowering it?

Its ions increase the ionic strength, lowering the effective activities of Ag⁺ and Cl⁻; to keep the activity product at

K_{s p}

, more solid must dissolve. This is the salt effect, not a common ion effect.

Edge cases

What happens to solubility if the common-ion concentration

c

approaches zero?

The formula

s^{'} = K_{s p} / c

blows up, signalling the approximation has broken. As

c \to 0

you smoothly recover the pure-water case

s = K_{s p}

— you must use the full expression

K_{s p} = s^{'} (c + s^{'})

What if the common-ion concentration

c

is comparable to the pure-water solubility

s

The approximation

s^{'} ≪ c

fails and you cannot drop

s^{'}

. Solve the full equation

K_{s p} = s^{'} (c + s^{'})

(a quadratic for MX) rather than the shortcut.

Does the common ion effect help or hurt when

K_{s p}

is very large (a fairly soluble salt)?

A large

K_{s p}

means

s^{'}

stays large and often comparable to

c

, so the neat suppression breaks down and the approximate formula misleads — you must solve exactly (as in PbCl₂).

If you add the common ion but the solution was unsaturated (no solid present), does anything precipitate?

Only if the addition pushes

Q

above

K_{s p}

. If

Q

stays below

K_{s p}

after adding the ion, nothing precipitates — the common ion effect on the solid requires reaching saturation first.

For

MX_{2}

, why is the pure-water solubility

s = 3 K_{s p} /4

rather than

K_{s p}

In pure water

[M^{2 +}] = s

and

[X^{-}] = 2 s

, so

K_{s p} = s (2 s)^{2} = 4 s^{3}

, giving

s = (K_{s p} /4)^{1/3}

. The exponent tracks the total number of ions, not a simple square.

What is the limiting behaviour of

s^{'}

c \to \infty

for an MX salt?

s^{'} = K_{s p} / c \to 0

. Infinite common ion drives solubility toward zero (in the idealized model), though activity effects put a floor on this in reality.

Can the common ion effect ever increase solubility?

Not the common ion effect itself. But at very high ligand/common-ion concentrations, complex-ion formation (e.g.

AgCl_{2}^{-}

) can re-dissolve the solid — a separate equilibrium overriding simple

K_{s p}

suppression.

Connections

Le Chatelier's Principle — the driving principle for every "shift left" answer above.
Solubility Product (Ksp) — the fixed constant that all these traps orbit.
Buffer Solutions — common ion suppression of weak-acid ionization.
Qualitative Analysis — selective precipitation via common ions.
Ionic Equilibrium — the wider setting of coupled equilibria.
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