AP Chemistry 7.5: Solubility Equilibria, Ksp, and Precipitation

Some ionic compounds are considered insoluble, but this does not mean absolutely none of the solid dissolves.

• Instead, a small amount may dissolve until an equilibrium is established between the solid and its dissolved ions.
• For example:
• AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
• This is a solubility equilibrium.
• The equilibrium constant for this type of process is called the solubility-product constant, written:
• Ksp
• Khan Academy's current Unit 7 lesson covers calculating solubility from Ksp, predicting precipitation with Q vs. Ksp, the common-ion effect, and the effect of pH on solubility.

For:

• AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
• the solid is omitted.
• Therefore:
• Ksp = [Ag⁺][Cl⁻]

For:

• CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq)
• the expression is:
• Ksp = [Ca²⁺][F⁻]²
• Again, the coefficient becomes the exponent.

Molar solubility, often represented by s, is the number of moles of solute that dissolve per liter of solution to form a saturated solution.

• Suppose:
• AgCl(s) ⇌ Ag⁺ + Cl⁻
• If AgCl has molar solubility s:
• [Ag⁺] = s
• [Cl⁻] = s
• Therefore:
• Ksp = s²
• So:
• s = √Ksp

Consider:

• CaF₂(s) ⇌ Ca²⁺ + 2F⁻
• Suppose molar solubility = s.
• Then:
• [Ca²⁺] = s
• [F⁻] = 2s
• Therefore:
• Ksp = [Ca²⁺][F⁻]²
• Ksp = (s)(2s)²
• Ksp = 4s³

If Ksp is known, solve:

• s = ∛(Ksp/4)
• This is why you cannot assume Ksp equals molar solubility.
• The relationship depends on stoichiometry.

A larger Ksp often corresponds to greater solubility when comparing compounds with the same dissolution stoichiometry.

• However, directly comparing Ksp values between compounds with different ion ratios can be misleading.
• For example:
• MX:
• Ksp = s²
• while:
• MX₂:
• Ksp = 4s³
• Therefore you may need to calculate actual molar solubilities instead of simply comparing Ksp values.

A saturated solution is in equilibrium with undissolved solute.

• At saturation:
• Qsp = Ksp
• An unsaturated solution contains less dissolved solute than the equilibrium maximum.
• Then:
• Qsp < Ksp
• An oversaturated/supersaturated condition has:
• Qsp > Ksp
• and precipitation is thermodynamically favored until Qsp returns to Ksp.

Use the ion reaction quotient Qsp.

• It has the same expression as Ksp, but uses the current ion concentrations.
• Then compare.

The solution is unsaturated.

• No precipitate is expected from the equilibrium calculation.
• More solid could dissolve if available.

The system is saturated and at equilibrium.

There are too many dissolved ions.

• A precipitate forms until:
• Qsp = Ksp
• Khan Academy uses exactly this comparison in its current precipitation material.

This is a major exam trap.

• Suppose you mix two solutions containing ions that could form a precipitate.

Before calculating Qsp, determine the new ion concentrations after mixing.

If volumes are additive:

• new concentration = moles ion / total volume
• For a solution being diluted by mixing:
• Mnew = Mold Vold / Vtotal
• Then substitute those new concentrations into Qsp.

Do not use the original concentrations before mixing.

Suppose:

• AgCl(s) ⇌ Ag⁺ + Cl⁻
• Now add NaCl.
• NaCl provides additional:
• Cl⁻
• The chloride ion is already part of the AgCl dissolution equilibrium, so it is called a common ion.
• Adding Cl⁻ makes the equilibrium shift left:
• Ag⁺ + Cl⁻ → AgCl(s)
• Therefore the molar solubility of AgCl decreases.
• This is the common-ion effect.
• The key relationship is:
• adding a common ion generally decreases the solubility of a sparingly soluble ionic compound

For:

• AgCl(s) ⇌ Ag⁺ + Cl⁻
• suppose the solution already contains 0.100 M Cl⁻ from another soluble salt.
• If AgCl dissolves by s:
• [Ag⁺] = s
• [Cl⁻] = 0.100 + s
• Then:
• Ksp = s(0.100+s)
• If s is very small relative to 0.100:
• 0.100+s ≈ 0.100
• so:
• Ksp ≈ s(0.100)
• This shows mathematically why adding a common ion makes s smaller.

pH can affect the solubility of ionic compounds when one of the ions reacts with H⁺ or OH⁻.

• Consider a salt containing a basic anion.
• If H⁺ reacts with that anion, lowering the concentration of the free anion, the dissolution equilibrium can shift right to replace it.
• Therefore adding acid can increase the solubility of certain salts containing basic anions.
• For example, carbonates often become more soluble in acidic solution because:
• CO₃²⁻
• can react with H⁺.
• Removing CO₃²⁻ from the dissolution equilibrium pulls more solid into solution.
• Khan Academy's current Unit 7 notes that salts containing basic anions can become more soluble as pH decreases, while salts with anions of negligible basicity are much less affected by pH changes.

Consider:

• CaCO₃(s) ⇌ Ca²⁺ + CO₃²⁻
• Adding H⁺ consumes carbonate through acid-base chemistry.
• This decreases [CO₃²⁻].
• The dissolution equilibrium responds by shifting:
• right
• So more CaCO₃ dissolves.

Solubility equilibria use exactly the same logic as the rest of Unit 7:

• write an equilibrium expression;
• omit the pure solid;
• compare Q and K;
• use Le Châtelier's principle;
• use ICE-table reasoning;
• apply stoichiometric coefficients correctly.
• Ksp is not a separate type of chemistry. It is an application of the same equilibrium principles.

Do not include the solid in the Ksp expression.

Do not automatically set every ion concentration equal to s. Use the dissolution coefficients.

Do not compare Ksp values directly to determine solubility when the compounds have different dissolution stoichiometries.

When solutions are mixed, account for dilution before calculating Qsp.

• Remember:

Qsp > Ksp → precipitate forms

For precipitation:

• Qsp < Ksp → no precipitate
• Qsp = Ksp → saturated equilibrium
• Qsp > Ksp → precipitate forms

Consider these dissolution reactions.

MX(s) ⇌ M⁺ + X⁻

• If molar solubility = s:
• Ksp = s²

MX₂(s) ⇌ M²⁺ + 2X⁻

• Then:
• [M²⁺] = s
• [X⁻] = 2s
• so:
• Ksp = 4s³

M₂X₃(s) ⇌ 2M³⁺ + 3X²⁻

• Then:
• [M³⁺] = 2s
• [X²⁻] = 3s
• so:
• Ksp = (2s)²(3s)³
• Ksp = 108s⁵
• The important lesson is:
• use the balanced dissolution equation before relating Ksp to molar solubility.

Earlier you learned how to predict and write precipitation reactions.

• Unit 7 explains precipitation using equilibrium.

For:

• AgCl(s) ⇌ Ag⁺ + Cl⁻
• if:
• Qsp > Ksp
• the dissolved-ion concentrations are too high to remain at equilibrium.
• Therefore AgCl precipitates until:
• Qsp = Ksp
• So precipitation reactions are really another application of equilibrium.