Solubility product Ksp — common-ion suppression, selective precipitation: Question bank

This page is a misconception minefield walkthrough. Every item below targets a specific way students misread solubility equilibria. Read the prompt, say your answer out loud before revealing, then check the reasoning — the answer side always explains why, never just "yes/no".

Prerequisites live in the parent note (Hinglish version here) and lean on Le Chatelier's Principle, Chemical Equilibrium andrium Constant, Ionic Equilibria in Solutions, and — for the pH traps below — pH and pOH Calculations.

Definition The five symbols this page keeps reusing

Before the traps, pin down every symbol so no reveal is opaque. Consider a salt $M_{x} A_{y} (s) ⇌ x M^{n +} (a q) + y A^{m -} (a q)$ .

$x, y$ = the stoichiometric coefficients: how many cations ( $x$ ) and anions ( $y$ ) one formula unit releases when it dissolves. For AgCl, $x = y = 1$ ; for $CaF_{2}$ , $x = 1, y = 2$ ; for $Ca_{3} (PO_{4})_{2}$ , $x = 3, y = 2$ .
$s$ = the solubility in pure water, in mol/L: how many formula units of the salt dissolve per litre before saturation. Each dissolved unit gives $x$ cations and $y$ anions, so $[M^{n +}] = x s$ and $[A^{m -}] = y s$ .
$s^{'}$ = the solubility in the presence of a common ion: the same quantity but measured in a solution that already contains one of the salt's ions. It is (almost always) much smaller than $s$ .
$C$ = the concentration of the already-present common ion (e.g. from added NaCl or NaF).
$Q$ = the reaction quotient / ionic product: the same product of ion concentrations as $K_{s p}$ , but evaluated at the current, possibly non-equilibrium concentrations — $Q = [M^{n +}]_{now}^{x} [A^{m -}]_{now}^{y}$ . It equals $K_{s p}$ only at saturation.

True or false — justify

A smaller

K_{s p}

always means a smaller solubility

s

False. The conversion

s = (K_{s p} / x^{x} y^{y})^{1/ (x + y)}

depends on stoichiometry, so a 1:2 salt with a smaller

K_{s p}

can still be more soluble than a 1:1 salt with a larger

K_{s p}

— you can only compare

s

directly for salts of the same ion count.

Adding NaCl to a saturated AgCl solution changes the value of

K_{s p}

for AgCl.

False.

K_{s p}

depends only on temperature. Adding a common ion shifts the position of equilibrium (less AgCl dissolves) but the product

[Ag^{+}] [Cl^{-}]

still equals the same

K_{s p}

At equilibrium,

[Ag^{+}] = [Cl^{-}]

for AgCl in every solution.

False. That equality holds only in pure water. In 0.1 M NaCl,

[Cl^{-}] \approx 0.1

while

[Ag^{+}]

drops to

\sim 1 0^{- 9}

— very unequal, yet their product is still

K_{s p}

The concentration of solid

M_{x} A_{y}

appears in the denominator of the equilibrium expression.

False. A pure solid has activity 1 (it is in its standard state and its "concentration" does not change), so it is folded into the constant and never written — this is exactly why

K_{s p}

has only the ion terms.

If you double the amount of undissolved AgCl solid sitting at the bottom, the solubility doubles.

False. More solid does not change any concentration — the solid's activity stays 1. Solubility depends on

K_{s p}

and the ions already present, not on how much solid is heaped up.

For

CaF_{2}

K_{s p} = [Ca^{2 +}] [F^{-}]

False. The balanced equation is

CaF_{2} ⇌ Ca^{2 +} + 2 F^{-}

, so the fluoride coefficient 2 becomes an exponent:

K_{s p} = [Ca^{2 +}] [F^{-}]^{2}

A supersaturated solution has

Q > K_{s p}

and is thermodynamically stable.

False.

Q > K_{s p}

means the solution is over the tolerated ion product and is metastable — it will precipitate (sometimes only after a nucleation trigger) until

Q

falls back to

K_{s p}

Common-ion suppression works even if the added ion is not one of the salt's own ions.

False. Suppression requires a shared (common) ion so Le Chatelier can push the dissolution equilibrium backward. An unrelated spectator ion does not appear in the

K_{s p}

expression and so cannot shift it directly.

The solubility of

CaF_{2}

is the same in neutral water and in strong acid.

False.

F^{-}

is the conjugate base of the weak acid HF, so acid protonates it (

F^{-} + H^{+} \to HF

), removing free fluoride; the equilibrium dissolves more solid to replace it, so solubility rises as pH falls.

Adding aqueous

NH_{3}

to solid AgCl can only decrease its dissolution.

False.

NH_{3}

forms the complex

[Ag(NH_{3})_{2}]^{+}

, pulling free

Ag^{+}

out of solution; the dissolution equilibrium responds by dissolving more AgCl, so solubility actually increases.

Spot the error

K_{s p} = 1.8 \times 1 0^{- 10}

for AgCl, so its solubility is

1.8 \times 1 0^{- 10}

mol/L."

Error:

K_{s p}

is the product

s^{2}

, not

s

itself. Solubility is

s = 1.8 \times 1 0^{- 10} = 1.34 \times 1 0^{- 5}

mol/L — five orders of magnitude larger than the raw

K_{s p}

value.

"For

CaF_{2}

with solubility

s

, I set

K_{s p} = s \cdot s^{2} = s^{3}

because there are three ions."

Error:

[F^{-}] = 2 s

, not

s

. The correct substitution is

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

; forgetting the factor of 2 undercounts fluoride and gives a wrong

s

"AgCl in 0.10 M NaCl: I set

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

and solved

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

Error: the chloride is dominated by NaCl. Correct setup is

[Cl^{-}] = 0.10 + s^{'} \approx 0.10

and

[Ag^{+}] = s^{'}

, giving

K_{s p} = s^{'} (0.10)

— the common ion is added, not equal to

s^{'}

"Mixing 100 mL of 0.010 M

AgNO_{3}

with 100 mL of NaCl, I use

[Ag^{+}] = 0.010

M in

Q

Error: mixing doubles the total volume, halving each concentration. You must use the diluted value

[Ag^{+}] = 0.0050

M before computing

Q

; forgetting dilution overstates

Q

and can wrongly predict precipitation.

"I checked

s^{'} ≪ C

and it failed, so I'll just use the approximate answer anyway."

Error: if the approximation fails you must solve the full quadratic

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

. Ignoring a failed check gives a numerically wrong solubility, sometimes by a large factor.

Q < K_{s p}

means a precipitate is about to form."

Error: it's the opposite.

Q < K_{s p}

is unsaturated — the solution can still dissolve more salt. Precipitation requires

Q > K_{s p}

"Solubility of

CaF_{2}

in 0.010 M NaF: I forgot to square the fluoride, so

s^{'} = K_{s p} /0.010

Error: fluoride enters squared. Correct expression is

K_{s p} = s^{'} (0.010)^{2}

, so

s^{'} = K_{s p} / (0.010)^{2} = K_{s p} /1 0^{- 4}

— dropping the square inflates the answer by a factor of 0.010.

K_{s p}

CaCO_{3}

is fixed, so its solubility can't depend on pH."

Error:

K_{s p}

is fixed, but

CO_{3}^{2 -}

is protonated by acid to

HCO_{3}^{-}

H_{2} CO_{3}

; removing free carbonate makes

Q

drop below

K_{s p}

so more solid dissolves — pH changes solubility without changing

K_{s p}

Why questions

Why does the pure-solid activity of 1 let us drop the solid from the equilibrium expression?

Because a pure solid is in its standard state — its "amount per unit of itself" is fixed and unchanging, so it contributes a constant factor of 1 that is absorbed into

K_{s p}

rather than tracked as a variable.

Why is solubility inversely proportional to common-ion concentration in the simple

s^{'} \approx K_{s p} / C

case?

Because

K_{s p}

is fixed: if the common-ion term

C

in the product goes up, the other ion (which equals

s^{'}

) must go down to keep the product constant, giving

s^{'} \propto 1/ C

Why can two salts with the same

K_{s p}

have different solubilities?

Because

s

depends on stoichiometry through

s = (K_{s p} / x^{x} y^{y})^{1/ (x + y)}

; the exponents and

x^{x} y^{y}

factor differ between, say, a 1:1 and a 1:2 salt, so equal

K_{s p}

does not force equal

s

Why does selective precipitation work — why does one ion drop out before another?

Because each target salt reaches its own

K_{s p}

threshold at a different added-reagent concentration; you slowly raise the precipitating ion until

Q

exceeds the smaller

K_{s p}

first, precipitating that salt while the other stays dissolved.

Why do we use

Q

(not

K_{s p}

) to decide whether precipitation happens?

Q

is computed from the actual current concentrations, which may be far from equilibrium; comparing this instantaneous ion product to the fixed threshold

K_{s p}

tells you which direction the system must move.

Why does the common-ion effect not lower solubility "to zero"?

Because

K_{s p}

is a nonzero constant — no matter how large

C

gets,

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

stays positive. Some salt always dissolves; suppression only makes

s^{'}

very small, never exactly zero.

Why is raising temperature usually (but not always) increasing

K_{s p}

Because most dissolution is endothermic, so by Le Chatelier heat acts like a reactant and higher T shifts equilibrium toward dissolved ions; for the rare exothermic-dissolution salts the opposite holds and

K_{s p}

falls with T.

Why does lowering the pH increase the solubility of salts of weak-acid anions (like

F^{-}

CO_{3}^{2 -}

S^{2 -}

Because

H^{+}

protonates the anion into its weak acid, draining free anion out of the dissolution equilibrium; to keep

[M^{n +}] [A^{m -}]^{y}

K_{s p}

the system must dissolve more solid, so acid boosts solubility.

Why does adding a complexing agent (like

NH_{3}

CN^{-}

) make a "very insoluble" salt dissolve?

The ligand ties up the free metal ion in a complex (e.g.

[Ag(NH_{3})_{2}]^{+}

), so

[M^{n +}]

falls,

Q

drops below

K_{s p}

, and dissolution continues — the apparent solubility can be thousands of times the pure-water value even though

K_{s p}

never changed.

Edge cases

What is the solubility of a salt in a solution where the common-ion concentration

C = 0

Then there is no suppression — you are back in pure water, so

s

comes from the full

K_{s p} = x^{x} y^{y} s^{x + y}

relation, not from

s^{'} \approx K_{s p} / C

(which would blow up and is undefined at

C = 0

What happens to

Q

at the exact instant two solutions are mixed but before any solid forms?

Q

is computed from the freshly diluted concentrations of both ions; if that

Q > K_{s p}

the system is momentarily supersaturated and will precipitate, if

Q \leq K_{s p}

nothing forms.

If you add so much common ion that

C + s^{'}

can no longer be approximated by

C

, what must change?

Nothing about the physics, but the algebra: you must keep the full expression

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

(or with exponents for non-1:1 salts) and solve the resulting quadratic/cubic rather than using the shortcut.

Consider a salt so soluble it is not sparingly soluble — does the

K_{s p}

framework apply cleanly?

Not really. For highly soluble salts the ion concentrations are large, activities deviate strongly from concentrations, and the simple

K_{s p} = [ion]^{x} [ion]^{y}

(which assumes ideal dilute behaviour) becomes inaccurate.

What is

[Ag^{+}]

in a saturated AgCl solution to which you add a huge excess of NaCl?

It becomes vanishingly small:

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

, and as

[Cl^{-}]

grows,

[Ag^{+}]

shrinks toward (but never reaches) zero — the basis of driving a metal ion almost entirely out of solution.

But if that "huge excess" of chloride is really extreme, what surprising thing can happen to AgCl solubility?

At very high

[Cl^{-}]

, chloride acts as a ligand and forms soluble complexes like

[AgCl_{2}]^{-}

; solubility, having been driven down, can turn around and rise again — a competition between common-ion suppression and complex formation.

When two salts share a common ion and both are near saturation, what happens if you add more of that shared ion?

Both equilibria feel the same pressure — the shared ion appears in both

Q

expressions, so both salts are suppressed simultaneously and whichever crosses its

K_{s p}

first begins precipitating.

At the boundary

Q = K_{s p}

exactly, is anything precipitating?

No net change: the solution is exactly saturated and at equilibrium, so any solid present coexists in dynamic balance with the ions — dissolution and precipitation rates are equal, no net solid forms or dissolves.

A metal sulfide has a tiny

K_{s p}

; does that alone guarantee it precipitates from any acidic sulfide solution?

No — in acid,

S^{2 -}

is largely protonated to

HS^{-}

H_{2} S

, so free

[S^{2 -}]

(and hence

Q

) is throttled by pH; this pH control of sulfide is exactly what lets qualitative analysis separate metal groups selectively.

Recall Fast self-test

Which comparison decides precipitation, $s$ vs $K_{s p}$ or $Q$ vs $K_{s p}$ ? ::: $Q$ vs $K_{s p}$ — $Q$ uses the actual (diluted) concentrations, precipitation when $Q > K_{s p}$ . For $Ca_{3} (PO_{4})_{2} ⇌ 3 Ca^{2 +} + 2 PO_{4}^{3 -}$ , write $K_{s p}$ in $s$ using the master formula. ::: $x = 3, y = 2$ , so $K_{s p} = (3 s)^{3} (2 s)^{2} = 27 \cdot 4 s^{5} = 108 s^{5}$ . What two side reactions can make a salt more soluble than its $K_{s p}$ suggests? ::: Protonation of a weak-acid anion (low pH) and complex-ion formation of the metal — both drain a free ion and pull the dissolution equilibrium forward.