Chemistry: Atoms First 2e · Equilibria of Other Reaction Classes

Precipitation and Dissolution

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

Many salts are only slightly soluble in water, yet even "insoluble" salts dissolve a little — the solid sits in equilibrium with a tiny concentration of its ions. For silver chloride:

$AgCl (s) ⇌ Ag + (aq) + Cl - (aq)$

The solubility product constant, K_sp, quantifies this equilibrium:

$K sp = [Ag +][Cl -] = 1.8 \times 10 -10 at 25 ° C$

The solid itself is left out of the expression because its activity is constant. From K_sp you can compute the molar solubility — moles of salt that dissolve per liter of — and, going the other way, the reaction quotient Q tells you whether a solution is unsaturated (more solid can dissolve), saturated, or supersaturated (precipitation is expected). Two factors dominate practical use: the , which suppresses solubility when a shared ion is already present, and , which exploits differences in K_sp to separate ions.

Why this matters

Precipitation chemistry surrounds us. Kidney stones form when calcium and oxalate concentrations in urine exceed the K_sp of calcium oxalate. Limestone caves, stalactites, and scale in pipes are precipitation–dissolution systems, and acid rain accelerates limestone dissolution by consuming carbonate ion. Photographic film relies on the low K_sp values of silver halides. In the lab, precipitation separates ions (qualitative analysis), powers gravimetric analysis, and underlies water softening and metal removal in environmental cleanup. Understanding K_sp also sets up the next topics, which explain why "insoluble" solids can still dissolve when acids or complexing agents are present.

The college version

Core Concepts

The dissolution equilibrium and K_sp

For a slightly soluble salt A_xB_y:

$A x B y (s) ⇌ x A y+ (aq) + y B x- (aq) K sp = [A y+] x [B x-] y$

The exponents are the stoichiometric coefficients and the pure solid is excluded. Compare K_sp values directly only for salts with the same stoichiometry — an AB salt and an A₂B salt with the same numerical K_sp are not directly comparable.

From K_sp to molar solubility

If s is the molar solubility, then for an AB salt (AgCl):

$K sp = s \cdot s = s 2$

For an AB₂ salt (PbI₂), where one formula unit releases one Pb²⁺ and two I^-:

$K sp = s(2s) 2 = 4s 3$

Always write the dissolution equation first: the stoichiometry sets the exponents and the relation between ion concentrations and s.

The reaction quotient Q and precipitation

Compare the quotient of actual ion concentrations with K_sp:

Q < K_sp: unsaturated — more solid can dissolve.
Q = K_sp: saturated — equilibrium.
Q > K_sp: supersaturated — precipitation occurs until Q falls to K_sp.

The classic exam problem: when two solutions are mixed, recompute ion concentrations after dilution to the total volume, then compare Q with K_sp.

The common-ion effect

If the solution already contains one of the salt's ions (e.g., Cl⁻ from NaCl), Le Châtelier's principle pushes the equilibrium toward the solid, and solubility drops — often by orders of magnitude. This is why precipitates are washed with a common-ion solution to keep them from dissolving, and why AgCl is far less soluble in seawater than in distilled water.

Selective precipitation

When a reagent forms insoluble salts with several ions, the salt whose Q exceeds its K_sp at the lowest reagent concentration precipitates first. Adding Ag⁺ slowly to a mixture of 0.010 M Br⁻ and 0.010 M Cl⁻ precipitates AgBr first: AgBr (K_sp = 5.0 × 10^-13) needs only [Ag⁺] = 5.0 × 10^-11 M, whereas AgCl (K_sp = 1.8 × 10^-10) needs 1.8 × 10^-8 M — 360 times more.

How It Works / Step-by-Step Process

Computing molar solubility from K_sp

Write the balanced dissolution equation and the K_sp expression.
Let s = molar solubility; express each ion concentration in terms of s.
Substitute into K_sp and solve for s.
Optionally convert to g/L by multiplying s by the salt's molar mass.

Predicting precipitation when solutions are mixed

Calculate the final concentration of each ion after mixing (dilution: M₁V₁ = M₂V₂ for each ion).
Write the K_sp expression for the possible precipitate and substitute the mixed concentrations to get Q.
Compare Q with K_sp: Q > K_sp means a precipitate forms.

Common Confusions

Do Not Confuse	With	The Difference
K_sp	Solubility	K_sp is an equilibrium constant (units vary with stoichiometry); solubility is a concentration. Related but not interchangeable
K_sp of AgCl and K_sp of PbI₂	Directly comparable numbers	Only salts with the same stoichiometry (same expression form) can be compared directly
Q	K_sp	Q uses actual current concentrations; K_sp is the equilibrium value. The comparison predicts precipitation
"Insoluble" salt	Zero solubility	Every salt dissolves a little; K_sp tells you how little
Concentrations before mixing	Concentrations after mixing	Mixing dilutes both ions; always recompute for the total volume
Common-ion effect	A different equilibrium	It is the same dissolution equilibrium shifted by Le Châtelier when a product ion is added
Solubility of AB₂	Solubility of AB with the same K_sp	The 2:1 stoichiometry inserts a factor of 4: K_sp = 4s³, not s²

Eli explains

The same idea, in plain words

Explain it like I’m 10

Even "solid rock" salts lose a few pieces into the water — like sugar dissolving, but barely. The K_sp is like a rule that says how many pieces can float free before the solid says "no more." If you already have lots of one kind of piece floating in the water, the solid dissolves even less — that's the common-ion effect — and when two solids could form, the one that needs the fewest pieces forms first.

Worked example

Example 1: Molar solubility of silver chloride, in mol/L and g/L

For AgCl, K_sp = 1.8 × 10^-10. Let s = molar solubility. Each formula unit gives one Ag⁺ and one Cl⁻:

$K sp = [Ag +][Cl -] = s \cdot s = s 2$

$s = \sqrt K sp = \sqrt 1.8 \times 10 -10 = 1.3 \times 10 -5 mol/L$

Convert to grams per liter using the molar mass of AgCl (107.87 + 35.45 = 143.32 g/mol):

$1.3 \times 10 -5 mol L \times 143.32 g mol = 1.9 \times 10 -3 g/L$

So "insoluble" AgCl dissolves about 1.9 mg per liter — tiny, but measurable, and enough to matter below.

Example 2: Molar solubility of lead(II) iodide (2:1 stoichiometry)

For PbI₂, K_sp = 7.1 × 10^-9. The dissolution:

$PbI 2 (s) ⇌ Pb 2+ (aq) + 2 I - (aq)$

With s = molar solubility, [Pb²⁺] = s and [I^-] = 2s:

$K sp = [Pb 2+][I -] 2 = s(2s) 2 = 4s 3$

$s = 3 \sqrt K sp 4 = 3 \sqrt 7.1 \times 10 -9 4 = 3 \sqrt 1.775 \times 10 -9 = 1.2 \times 10 -3 mol/L$

The trap: forgetting the factor of 2 on I⁻ (using s³ = K_sp) gives the wrong answer. The 2:1 stoichiometry makes PbI₂ about 90 times more soluble than AgCl in molar terms, despite its larger K_sp.

Example 3: The common-ion effect on AgCl

How much AgCl dissolves in 0.010 M NaCl? The solution already contains [Cl^-] = 0.010 M before any AgCl dissolves. With s = molar solubility, [Ag⁺] = s and [Cl^-] = 0.010 + s:

$K sp = s(0.010 + s) \approx s(0.010)$

because s is tiny compared with 0.010 M. Solve:

$s = 1.8 \times 10 -10 0.010 = 1.8 \times 10 -8 mol/L$

That is roughly 700 times less soluble than in pure water (1.3 × 10^-5 mol/L). The common ion suppresses dissolution almost entirely.

Example 4: Will a precipitate form when two solutions are mixed?

Mix 20.0 mL of 1.0 × 10^-4 M AgNO₃ with 20.0 mL of 1.0 × 10^-4 M NaCl. Could AgCl precipitate?

Compute the concentration of each ion after mixing (total volume = 40.0 mL):

$[Ag +] = [Cl -] = 20.0 mL \times 1.0 \times 10 -4 M 40.0 mL = 5.0 \times 10 -5 M$

Form the quotient and compare with K_sp:

$Q = [Ag +][Cl -] = (5.0 \times 10 -5)(5.0 \times 10 -5) = 2.5 \times 10 -9$

Since Q = 2.5 × 10^-9 > K_sp = 1.8 × 10^-10, the solution is supersaturated and AgCl will precipitate until Q falls to K_sp. Using the pre-mixing concentrations would overestimate Q by a factor of four.

Key takeaways

K_sp is an equilibrium constant for a solid dissolving; the pure solid is excluded from the expression.
Molar solubility: AB salt s = K_sp; AB₂ salt s = ³K_sp/4 — write the dissolution equation first.
Q < K_sp dissolves; Q > K_sp precipitates; Q = K_sp is saturated.
Compare K_sp values only between salts of the same stoichiometry.
Common-ion effect: adding an ion already in the salt's formula lowers its solubility.
After mixing solutions, always use concentrations in the total volume.
The salt needing the smallest precipitating-ion concentration to exceed K_sp precipitates first.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

Write the K_sp expression for CaF₂ dissolving into Ca²⁺ and F^-.
Show answer
CaF₂(s) ⇌ Ca²⁺ + 2F^-; K_sp = [Ca²⁺][F^-]².
The K_sp of AgCl is 1.8 × 10^-10. What is its molar solubility in pure water?
Show answer
s = K_sp = 1.8 × 10^-10 = 1.3 × 10^-5 mol/L.
A solution has [Ag⁺] = [Cl^-] = 1.0 × 10^-4 M. Will AgCl precipitate?
Show answer
Q = (1.0 × 10^-4)² = 1.0 × 10^-8 > 1.8 × 10^-10, so yes — supersaturated, AgCl precipitates.
Why does AgCl dissolve less in seawater than in distilled water?
Show answer
Seawater contains high Cl⁻, a common ion. By Le Châtelier's principle, the extra Cl⁻ pushes the equilibrium toward solid AgCl, cutting solubility to about 1.8 × 10^-8 mol/L in 0.010 M Cl⁻.
A salt with AB stoichiometry has K_sp = 1.0 × 10^-12. If [A²⁺] = [B^2-] = 1.0 × 10^-6 M, is the solution saturated, unsaturated, or supersaturated?
Show answer
Q = (1.0 × 10^-6)² = 1.0 × 10^-12 = K_sp — exactly saturated.
In a mixture of 0.010 M Cl⁻ and 0.010 M I⁻, which precipitates first — AgCl (K_sp = 1.8 × 10^-10) or AgI (K_sp = 8.5 × 10^-17)? Why?
Show answer
AgI: it needs only [Ag⁺] = 8.5 × 10^-17/0.010 = 8.5 × 10^-15 M, far less than the 1.8 × 10^-8 M needed for AgCl. The smaller the K_sp, the earlier the salt precipitates as Ag⁺ is added.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

solubility product (Ksp): Equilibrium constant for a slightly soluble salt dissolving, equal to the product of ion concentrations raised to their coefficients
molar solubility (s): Moles of salt that dissolve per liter of saturated solution
saturated solution: A solution in equilibrium with undissolved solid (Q = K_sp)
reaction quotient (Q): The same product expression as K_sp, but for actual current concentrations
common-ion effect: The drop in solubility caused by an ion already in solution being part of the salt
selective precipitation: Separating ions by precipitating them one at a time using differences in K_sp

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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