Solubility Product and Precipitation

Introduction

A solubility rule suggests which solid might form, but Ksp gives a quantitative equilibrium threshold. Precipitation depends on the product of the free ion activities at the present concentrations. Compare that current product Qsp with Ksp at the same temperature: Qsp below Ksp means undersaturation, equality means saturation, and Qsp above Ksp means precipitation is thermodynamically favoured.

Core explanation

Start with a balanced dissolution equation. For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp = a(Ag⁺)a(Cl⁻), where a denotes activity. In a dilute teaching approximation, use [Ag⁺][Cl⁻]. The solid's activity is effectively one and does not appear in the expression. For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, Ksp = a(Ca²⁺)a(F⁻)². The exponent two comes from the stoichiometric coefficient, not from the fluoride ion's charge.

Qsp has the same algebraic form but uses the current free-ion activities before equilibrium is established. If Qsp > Ksp, a solid is favoured and free-ion levels decrease as precipitation proceeds. If Qsp < Ksp and solid is present, some solid may dissolve. When solid and solution reach equilibrium, Qsp = Ksp. A value greater than Ksp predicts direction, not the final mass of precipitate or how fast it visibly appears.

After mixing two solutions, dilution must be calculated first. Suppose 10.0 mL of 0.010 M AgNO₃ is mixed with 90.0 mL of 0.010 M NaCl and volumes are additive. The just-mixed concentrations are [Ag⁺] = 0.0010 M and [Cl⁻] = 0.0090 M before precipitation, so Qsp = 9.0×10⁻⁶. This greatly exceeds AgCl's small Ksp under ordinary conditions, so precipitation is expected. Multiplying 0.010 by 0.010 instead would use stock concentrations and overstate Qsp.

An analytical concentration need not equal free-ion concentration. If ammonia complexes Ag⁺, some dissolved silver is [Ag(NH₃)₂]⁺; simple AgCl Ksp still uses free Ag⁺. If acid protonates CO₃²⁻, total dissolved inorganic carbon includes HCO₃⁻ and CO₂ , but CaCO₃ Ksp uses free CO₃²⁻. In a simple exam problem, absence of complexants and pH effects may be assumed; in an advanced one they must be included explicitly.

Ksp describes equilibrium at a given temperature. It can change with temperature, whereas merely adding a common ion at fixed temperature changes equilibrium concentrations without changing thermodynamic Ksp. Activity coefficients make concentration-based apparent products sensitive to ionic strength, so a rigorous comparison uses activities or conditional constants appropriate to the medium.

Selective precipitation uses different onset thresholds. For AgCl in a solution with known chloride, the free silver threshold is [Ag⁺]onset ≈ Ksp/[Cl⁻]. For M(OH)₂, the hydroxide threshold is √(Ksp/[M²⁺]). The relative thresholds show which solid can begin forming first as a reagent is added. A complete separation also requires asking how much first ion remains when the second reaches its threshold.

Finally, thermodynamic favourability and visible timing differ. A supersaturated liquid can remain clear temporarily if nucleation is slow. Conversely, a cloudy suspension may contain a different solid than the one hypothesized. Pair Qsp reasoning with observation and confirmatory evidence.

Step-by-step reasoning

1. Write the correct solid formula and balanced dissolution. 2. Form Ksp and Qsp with stoichiometric exponents. 3. Calculate free-ion concentrations after mixing or speciation. 4. Compare Qsp with Ksp at the same temperature. 5. If final amount is needed, add material balances and solve equilibrium.

Visual explanation

Draw a number line with Qsp < Ksp, Qsp = Ksp and Qsp > Ksp regions. Place arrows toward dissolution on the left and precipitation on the right, both ending at equality.

Real-world analogy

A room has a capacity limit for pairs of two kinds of items. Qsp describes the current pairing pressure, while Ksp is the equilibrium limit; exceeding it drives items into a separate storage form, the solid.

Real-world example

Silver halide precipitation is a familiar analytical separation. Whether a visible AgCl solid appears depends on the free Ag⁺ and Cl⁻ levels after reagent addition, not simply on their names.

Why?

Why does CaF₂ use [F⁻]² in Ksp? One formula unit releases two fluoride ions, so the law-of-mass-action expression contains the fluoride activity raised to the second power.

Common misconception

“Qsp > Ksp tells the precipitate mass.” It only predicts the thermodynamic direction; final amounts require conservation and the new equilibrium ion concentrations.

Worked example

A hypothetical MX salt has Ksp = 2.0×10⁻⁸. Mixing gives free [M⁺] = 1.0×10⁻³ M and [X⁻] = 5.0×10⁻⁴ M. Qsp = 5.0×10⁻⁷, twenty-five times Ksp. A precipitate is favoured. The final free-ion concentrations cannot be obtained by dividing both initial values by 25, because precipitation consumes one M⁺ and one X⁻ per formula unit and their initial amounts differ.

Quick check

1. Which value predicts precipitation direction: the total dissolved metal concentration or Qsp from free ions? Answer: Qsp from the free-ion activities or their justified concentration approximations, compared with Ksp at the same temperature.

Exam focus

Dilute after mixing, use free ions and preserve the dissolution stoichiometric powers. State direction separately from final mass.

Advanced insight

OpenStax develops Qsp and Ksp comparison at https://openstax.org/books/chemistry-2e/pages/15-1-precipitation-and-dissolution. A supersaturated state can be metastable, so kinetics may delay an equilibrium-predicted solid.

Summary

Ksp is the equilibrium free-ion product for a specified solid; Qsp is the current product. Comparing them predicts dissolution or precipitation direction. Concentrations after mixing, complexation and acid-base speciation determine the actual Qsp, while final precipitate amount requires additional balances.

Practice questions

1. Write Ksp for Al(OH)₃(s) ⇌ Al³⁺ + 3OH⁻. Answer: Ksp = a(Al³⁺)a(OH⁻)³, approximated by [Al³⁺][OH⁻]³ in dilute solution. 2. What does Qsp < Ksp indicate for a proposed precipitate? Answer: The solution is undersaturated with respect to that solid. 3. Does adding a common ion change thermodynamic Ksp at fixed temperature? Answer: No. It changes the equilibrium free-ion concentrations. 4. Why is stock concentration not generally used directly after mixing? Answer: The ions are diluted into the final combined volume before Qsp is evaluated.