Equilibrium · TOPIC LESSON
Separate free-ion concentrations, total dissolved material and the amount of solid present.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
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No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
A sparingly soluble ionic solid exchanges ions with its surrounding solution. At saturation with solid present, dissolution and precipitation occur at equal macroscopic rates, giving constant equilibrium activities. The solid has not become chemically inactive; the opposing processes balance. Adding more of the same pure solid changes the amount of solid available, not its activity in the equilibrium expression.
An unsaturated solution can dissolve additional solid under the specified conditions. A supersaturated solution has an ion-activity product above the equilibrium value but may persist temporarily because nucleation and growth have kinetic barriers. Thermodynamics predicts the favored change, while the appearance and speed of visible precipitation require kinetics as well.
Molar solubility s is moles of formula units dissolved per litre of saturated solution under stated conditions. It is not automatically the concentration of every ion. For MX₂ dissolving to M²⁺+2X⁻ with no other reactions, [M²⁺]=s but [X⁻]=2s. Mass solubility requires multiplication by the solid’s molar mass.
Temperature, solvent, common ions, acidity, complexing ligands and activity effects can change solubility. Ksp belongs to the particular dissolution reaction at a temperature; the solubility is a composition-dependent result. Distinguishing those two ideas prevents the mistake of treating a change in s as proof that the thermodynamic constant itself changed.
02 / UNDERSTAND & APPLY
For M_pX_q(s) ⇌ pM(aq)+qX(aq), Ksp=a(M)^p a(X)^q, with the actual ion charges included in the balanced equation. The pure solid has unit activity in the usual standard-state model. In dilute ideal calculations, activities are approximated by concentration ratios and the familiar bracket expression is used.
The coefficients become powers, while ion charges do not independently become powers. For CaF₂(s) ⇌ Ca²⁺+2F⁻, Ksp=[Ca²⁺][F⁻]². For Ag₂CO₃(s) ⇌ 2Ag⁺+CO₃²⁻, Ksp=[Ag⁺]²[CO₃²⁻]. The charge on carbonate does not make its concentration squared; its stoichiometric coefficient is one.
In pure water with no significant competing chemistry, use [M]=ps and [X]=qs, giving Ksp=p^p q^q s^(p+q). For a 1:1 salt this reduces to s². For a 1:2 or 2:1 salt it gives 4s³. For a 2:3 salt it gives 108s⁵. Derive the expression instead of applying the square-root formula to every salt.
This simple relation assumes dissolution is the only important source and sink of the ions. Hydrolysis, protonation, complexation or an already present common ion changes the mass balances. The correct Ksp expression still holds, but the free-ion concentrations are no longer just stoichiometric multiples of one s.
Ksp = [M]^p[X]^q = p^p q^q s^(p+q)
Ideal concentration model in pure water, with no coupled reactions or initial common ions.
ORIGINAL WORKED EXAMPLE
A hypothetical MX₂ solid has Ksp=3.20×10⁻¹¹ and dissolves only as M²⁺+2X⁻ in dilute water. Find s.
Answer s=2.00×10⁻⁴ M; [X⁻]=4.00×10⁻⁴ M.
03 / UNDERSTAND & APPLY
For salts with the same dissolution stoichiometry in the same medium and with no important side reactions, a larger Ksp gives a larger molar solubility. Across different stoichiometries, numerical Ksp values cannot be ranked directly as if they were all s². One may involve s³ or s⁵ and different coefficient factors.
For example, a hypothetical 1:1 salt with Ksp=1.00×10⁻¹⁰ has s=1.00×10⁻⁵ M, whereas a 1:2 salt with the same numerical Ksp has s=(Ksp/4)^(1/3)≈2.92×10⁻⁴ M. Equal Ksp numbers therefore do not imply equal dissolved formula-unit concentrations. The comparison is model-specific, not a universal ranking of “insolubility.”
Also distinguish molar solubility from grams per litre. A heavier formula unit can give a greater mass solubility at the same molar solubility. If the question asks which sample dissolves more grams, convert after solving the equilibrium. “More soluble” is ambiguous without the unit basis and conditions.
Activities become important at larger ionic strength. A concentration-based conditional product can vary as activity coefficients change even when the thermodynamic Ksp at fixed temperature is constant. The introductory concentration approximation is most reliable when its assumptions are stated and the ionic environment is not extreme.
ORIGINAL WORKED EXAMPLE
A salt has molar solubility 2.00×10⁻⁴ mol L⁻¹ and molar mass 150 g mol⁻¹. How much dissolves in 250 mL of saturated solution under the same conditions?
Answer 7.50 mg.
04 / UNDERSTAND & APPLY
For MX(s) ⇌ M⁺+X⁻ in a solution initially containing X⁻ at concentration C, dissolution gives [M⁺]=s and [X⁻]=C+s if no other reactions occur. The equilibrium equation is Ksp=s(C+s). If s is much smaller than C, approximate s≈Ksp/C, then check that the neglected contribution is indeed small.
For MX₂ in an added X⁻ solution, Ksp=s(C+2s)², not s(C+s)². If the initial common ion is the metal instead, the equation is (C+s)(2s)². Which ion was added and its coefficient both matter. A memorized common-ion formula with no mass balance can silently put the wrong concentration into the wrong power.
The common-ion effect usually reduces the simple salt’s dissolution by increasing one product activity. However, the same reagent may also create complexes or change pH, which can increase total dissolved material. Chloride, for example, can act as a common ion for one equilibrium and a ligand in another. When multiple roles are relevant, all coupled equilibria must be considered.
Adding a common ion does not guarantee a precipitate if the solution was initially far below saturation. Calculate the current ion product after dilution and mixing. The same chemical equilibrium explains both reduced saturated solubility and precipitation onset, but they are different initial-state questions.
ORIGINAL WORKED EXAMPLE
A hypothetical 1:1 salt has Ksp=4.00×10⁻¹². Find its solubility in 0.0200 M X⁻, assuming no complexation or acid–base reaction.
Answer s≈2.00×10⁻¹⁰ M.
05 / UNDERSTAND & APPLY
The ion product Qsp has the same form as Ksp but uses the current free-ion activities. If Qsp<Ksp, no precipitation is thermodynamically required and more solid could dissolve. If Qsp=Ksp, the solution is at the saturation condition. If Qsp>Ksp, precipitation is favored until the dissolved activities fall toward the equilibrium relation, subject to kinetics and available material.
When solutions are mixed, calculate moles of each relevant ion and divide by the total volume before forming Qsp, assuming volumes are additive if the problem permits. Using the unmixed concentrations can overestimate the product substantially. Spectator ions may matter for ionic strength but are not inserted into the dissolution quotient merely because they are present.
Once precipitation begins, stoichiometry determines how much of each ion is consumed. For a large supersaturation, a limiting-reagent subtraction can estimate the remaining excess ion, followed by Ksp to find the tiny residual counterion. Near the threshold, the amount precipitated may be small enough that an exact mass balance is preferable.
A valid precipitate calculation cannot remove all free ions exactly while solid and solution remain at equilibrium: Ksp fixes a nonzero residual product for finite Ksp. “Essentially complete” means negligible at the problem’s precision, not mathematically zero. Check the sign of the predicted precipitated amount; a negative result indicates that precipitation was incorrectly assumed.
ORIGINAL WORKED EXAMPLE
Mix 50.0 mL of 2.00×10⁻⁴ M M⁺ with 50.0 mL of 4.00×10⁻⁴ M X⁻. For MX, Ksp=1.00×10⁻⁹. Predict precipitation and estimate final free M⁺.
Answer Precipitation occurs; final [M⁺]≈9.16×10⁻⁶ M.
06 / UNDERSTAND & APPLY
A salt containing a basic anion may become more soluble in acid because protonation lowers the free-anion activity. For a carbonate, CO₃²⁻ can become HCO₃⁻ and then dissolved carbon dioxide/carbonic-acid species. The dissolution quotient responds to free CO₃²⁻, not total dissolved inorganic carbon, so consuming that free ion can drive further dissolution.
Anions that are negligibly basic in the relevant medium do not gain the same strong acid-solubility effect by protonation. Likewise, adding acid may introduce a common ion or alter other chemistry. State which coupled equilibrium is being invoked instead of treating “acid dissolves every salt” as a rule.
For M(OH)₂(s) ⇌ M²⁺+2OH⁻ at a maintained pH, Ksp=[M²⁺][OH⁻]². At 25 °C in the ideal aqueous approximation, [OH⁻]=10^(pH−14) M. The metal solubility from this expression can be calculated if hydroxide activity is externally controlled and complexation or hydrolysis does not change the model.
“Maintained pH” is important. Dissolving a hydroxide consumes acid and changes solution composition. A finite unbuffered acid supply may not hold pH constant, so solve the proton and charge balances rather than treating the initial pH as fixed forever. Amphoteric hydroxides can also dissolve in excess base by hydroxo-complex formation, reversing a naive common-ion-only prediction.
ORIGINAL WORKED EXAMPLE
A hypothetical M(OH)₂ has Ksp=1.00×10⁻¹². At 25 °C a buffer maintains pH=10.00. Ignoring other metal reactions, find free [M²⁺] at saturation.
Answer Free [M²⁺]=1.00×10⁻⁴ M.
07 / UNDERSTAND & APPLY
A ligand can bind dissolved metal ions into a complex. For M⁺+2L⇌ML₂⁺, an overall formation constant is β₂=[ML₂⁺]/([M⁺][L]²) in the ideal approximation. Removing free M⁺ into the complex lowers the ion product of a salt MX and can allow more solid to dissolve.
Ksp still uses free M⁺, not total metal [M]_T=[M⁺]+[ML₂⁺]+other metal species. If only that one complex matters and free ligand is maintained, [M]_T=[M⁺](1+β₂[L]²). The enhancement factor can be large even though the free metal concentration remains constrained by Ksp.
Adding the dissolution and complex-formation reactions gives MX(s)+2L⇌ML₂⁺+X⁻, with net equilibrium constant Kspβ₂. This multiplication is an equilibrium-cycle result. It does not mean β₂ alone determines solubility: anion balance, ligand consumption and other species still have to be included.
If ligand is not in large excess, distinguish free [L] from total added ligand. Complexation consumes ligand, and protonation can reduce the fraction available to bind metal. Treating total ammonia or another base ligand as free ligand in an acidic solution can drastically overestimate complex formation and salt dissolution.
ORIGINAL WORKED EXAMPLE
At saturation, a 1:1 salt has free [M⁺]=2.00×10⁻⁸ M. A maintained free ligand concentration of 0.100 M forms only ML₂⁺ with β₂=1.00×10⁸ M⁻² in the concentration convention. Find total dissolved metal.
Answer Total dissolved metal≈0.0200 M despite free metal only 2.00×10⁻⁸ M.
08 / UNDERSTAND & APPLY
To determine which salt precipitates first as a reagent is added, calculate the free precipitating-ion concentration needed to reach each Ksp. For a 1:1 MX salt, [X⁻]_threshold=Ksp/[M⁺]. Different initial metal concentrations can change the order, so ranking Ksp alone may be insufficient even for matching stoichiometry.
When one precipitate is already present, its free metal concentration follows Ksp/[X⁻] as the reagent increases. Evaluate that concentration at the onset threshold of the second precipitate to determine how much of the first ion remains. This is the basis of an ideal selective-separation calculation.
“Selective” needs a quantitative criterion. If a problem requires 99.9% removal before the second ion begins to precipitate, compare the residual first-ion concentration with 0.001 of its initial concentration. Merely showing different onset points does not establish near-complete separation.
These calculations assume controlled volume or appropriately corrected dilution, no significant complexation, suitable equilibration and the specified free-ion chemistry. Real analytical precipitation can involve coprecipitation, adsorption and kinetic effects not captured by the simple Ksp model. State when the calculation is an ideal equilibrium separation estimate.
ORIGINAL WORKED EXAMPLE
M⁺ and N⁺ each begin at 0.0100 M. Their 1:1 X⁻ salts have Ksp values 1.00×10⁻¹² and 1.00×10⁻⁸ respectively. Ignore dilution and side reactions. Which precipitates first, and how much M⁺ remains when NX just starts?
Answer MX first; [M⁺]=1.00×10⁻⁶ M at NX onset, corresponding to 99.99% removal.
09 / UNDERSTAND & APPLY
These original problems use dilute concentration approximations and exclude side reactions unless stated. A hypothetical salt name or constant is an instructional model, not a tabulated-data claim. Write the dissolution equation and free-ion mass balances before applying a root.
Check whether solid remains, whether the solution was diluted, whether an ion is protonated or complexed, and whether an approximation was validated. A correct Ksp expression with an incorrect free-ion concentration is still an incorrect equilibrium calculation.
ORIGINAL PRACTICE · 1
Ksp=(2s)²s=4s³.
The cation concentration is 2s and the anion concentration is s. Charges balance but do not replace stoichiometric powers.
ORIGINAL PRACTICE · 2
3.00×10⁻⁵ M.
Both ions have concentration s, so s²=Ksp and s=√Ksp.
ORIGINAL PRACTICE · 3
s≈3.00×10⁻⁸ M.
s≈Ksp/C=9.00×10⁻¹⁰/0.0300. Since s/C=10⁻⁶, neglecting added s in the common-anion concentration is justified.
ORIGINAL PRACTICE · 4
2.50×10⁻⁷.
Each concentration halves to 5.00×10⁻⁴ M after mixing. Their product is 2.50×10⁻⁷, not 10⁻⁶.
ORIGINAL PRACTICE · 5
No.
The solution is unsaturated with respect to that solid in the stated model. If solid is available, dissolution rather than net precipitation is favored.
ORIGINAL PRACTICE · 6
It falls by a factor of four.
Ksp=[M²⁺][OH⁻]² stays fixed at fixed temperature, so doubling hydroxide requires one-quarter the free metal concentration in this isolated equilibrium model.
ORIGINAL PRACTICE · 7
The ligand converts free metal into dissolved complexes.
Ksp constrains free-ion activity; total dissolved metal includes the complexed forms as well. These are different concentrations.
ORIGINAL PRACTICE · 8
1.00×10⁻⁹ M.
At onset Qsp=Ksp, so [X⁻]=Ksp/[M⁺]=2.00×10⁻¹¹/0.0200, before substantial precipitation changes the metal concentration.
KEEP THE BIG IDEAS
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