Qsp, Precipitation and the Common Ion Effect

Qsp is to Ksp exactly what Q is to Keq, the same expression evaluated at non-equilibrium concentrations, used to predict which direction the system must shift to reach equilibrium.

Qsp uses current (non-equilibrium) ion concentrations. When the system is at equilibrium, Qsp = Ksp. At any other moment, Qsp ≠ Ksp.

This is identical to the Q vs Keq framework from L12 applied to dissolution equilibria. The only new elements are the name (Qsp vs Q) and the dilution step required when two solutions are mixed.

Qsp uses current (non-equilibrium) ion concentrations in the same expression as Ksp. Decision rule: Qsp > Ksp → supersaturated → precipitate forms (shift left); Qsp = Ksp → exactly saturated, at equilibrium; Qsp < Ksp → unsaturated → no precipitate, more can dissolve. Only Qsp > Ksp causes precipitation, the most common error is reversing this rule.

Draw the three-box Qsp decision diagram into your notes before the check below.

We just saw the Qsp decision rule: Qsp > Ksp → precipitate; Qsp < Ksp → unsaturated. That raises a question: when two solutions are mixed together, what do you use for the ion concentrations in the Qsp calculation, the original concentrations or something else? This card answers it → by showing that mixing always dilutes every ion, and demonstrating the dilution formula that must be applied before calculating Qsp.

The single most commonly dropped step in Qsp calculations is accounting for the dilution that occurs when two solutions are mixed, failing to dilute the concentrations before computing Qsp is the number one source of wrong answers in IQ4.

When two solutions of volumes V₁ and V₂ are mixed:

$$c_{new} = c_{original} \times \frac{V_{original}}{V_{total}}$$

This dilution step must be applied to every ion before Qsp is calculated.

Example: 50.0 mL of 2.0 × 10⁻³ mol/L AgNO₃ mixed with 50.0 mL of 2.0 × 10⁻³ mol/L NaCl. Ksp(AgCl) = 1.8 × 10⁻¹⁰.

V_total = 50.0 + 50.0 = 100.0 mL

[Ag⁺]_new = (2.0 × 10⁻³)(50.0/100.0) = 1.0 × 10⁻³ mol/L

[Cl⁻]_new = (2.0 × 10⁻³)(50.0/100.0) = 1.0 × 10⁻³ mol/L

Qsp = [Ag⁺][Cl⁻] = (1.0 × 10⁻³)(1.0 × 10⁻³) = 1.0 × 10⁻⁶

Qsp = 1.0 × 10⁻⁶ > Ksp = 1.8 × 10⁻¹⁰ → precipitate forms

Mixing dilution step (non-negotiable): c_new = c_original × V_original / V_total. Apply to EVERY ion before calculating Qsp. Example: 50 mL of 2.0×10⁻³ mol/L AgNO₃ + 50 mL of 2.0×10⁻³ mol/L NaCl → V_total = 100 mL → [Ag⁺]_new = [Cl⁻]_new = 1.0×10⁻³ mol/L → Qsp = 1.0×10⁻⁶ > Ksp(AgCl) = 1.8×10⁻¹⁰ → precipitate forms.

Write the dilution formula and the 50+50 mL example into your notes before the check below.

We just saw that mixing solutions requires a dilution step before calculating Qsp, original concentrations cannot be used. That raises a question: what happens to the solubility of a sparingly soluble salt when you deliberately add a solution containing one of its constituent ions, how does Qsp explain the decrease in solubility? This card answers it → by working through the AgCl + NaCl common ion effect calculation showing a 7,500× reduction in solubility.

If you add a salt that shares an ion with a sparingly soluble compound already at equilibrium, Le Chatelier's Principle shifts the dissolution equilibrium left, the compound becomes less soluble in the presence of its own ions.

Mechanism: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), Ksp = 1.8 × 10⁻¹⁰.

Molar solubility in pure water: s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol/L.

Add NaCl to 0.10 mol/L: [Cl⁻] immediately increases to ≈ 0.10 mol/L.

New Qsp = [Ag⁺][Cl⁻] = (1.34 × 10⁻⁵)(0.10) = 1.34 × 10⁻⁶ >> Ksp.

Qsp > Ksp → equilibrium shifts LEFT → AgCl precipitates → [Ag⁺] decreases.

New equilibrium [Ag⁺]: [Ag⁺] = Ksp/[Cl⁻] = 1.8 × 10⁻¹⁰/0.10 = 1.8 × 10⁻⁹ mol/L

Solubility decreases from 1.34 × 10⁻⁵ mol/L to 1.8 × 10⁻⁹ mol/L, approximately 7,500× less soluble.

Common ion effect: adding a common ion to a saturated equilibrium → Qsp increases > Ksp → shifts LEFT → compound precipitates → solubility decreases. Ksp is UNCHANGED (only temperature changes Ksp). Calculation: new s = Ksp / [common ion concentration] when common ion dominates. AgCl in 0.10 mol/L NaCl: [Ag⁺] = 1.8×10⁻¹⁰/0.10 = 1.8×10⁻⁹ mol/L, 7,500× less soluble than in pure water.

Write the common ion effect mechanism and the AgCl calculation into your notes before the check below.

We just saw that adding a common ion decreases solubility by raising Qsp above Ksp, while Ksp itself stays fixed. That raises a question: what actually changes Ksp, and how does temperature affect it for different types of ionic compounds? This card answers it → by connecting dissolution thermodynamics (ΔH sign) to LCP temperature shifts and the resulting Ksp and solubility changes.

Since dissolution is an equilibrium process, changing temperature shifts the equilibrium, and the direction depends on whether dissolution is endothermic or exothermic, exactly as for any other equilibrium.

This is the same LCP temperature rule applied to dissolution equilibria: increase T shifts equilibrium in the endothermic direction.

Temperature effect on Ksp (4 steps): (1) determine ΔH sign for dissolution; (2) apply LCP, increase T shifts toward endothermic direction; (3) state Ksp change; (4) state solubility change. Endothermic dissolution (most salts): increase T → Ksp increases → more soluble. Exothermic dissolution (retrograde, e.g. Ce₂(SO₄)₃): increase T → Ksp decreases → less soluble. Only temperature changes Ksp.

Write the four-step temperature-Ksp procedure and the two types (endo vs exo) into your notes before the check below.

We just saw that temperature changes Ksp by shifting the dissolution equilibrium, endo dissolution becomes more soluble at higher T, exo less so. That raises a question: how do Qsp, common ion effect, and dissolution thermodynamics combine in a real physiological context where chemistry determines human health outcomes? This card answers it → by tracing how calcium oxalate Qsp exceeding Ksp in urine causes kidney stones, and why the relationship between dietary calcium and stone risk is more complex than it first appears.

A kidney stone forms when Qsp exceeds Ksp in urine, so the same rule you just used on a beaker of mixed solutions applies to a far messier system. The much-discussed link between calcium intake and stone risk is a good test of how carefully you state a chemical argument.

Calcium oxalate stone formation: $\text{CaC}_2\text{O}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq)$

Ksp(CaC₂O₄) = 2.3 × 10⁻⁹ at 37°C. If [Ca²⁺] × [C₂O₄²⁻] > 2.3 × 10⁻⁹ → Qsp > Ksp → stones form.

The patient's logic (partly correct): Dietary calcium in the GI tract binds dietary oxalate → forms insoluble CaC₂O₄ excreted in faeces → less oxalate absorbed → lower urinary [C₂O₄²⁻] → Qsp decreases → lower stone risk. This IS chemically sound for moderate calcium intake from food.

The doctor's concern (common ion effect): If calcium absorption is high AND oxalate absorption is not fully blocked, urinary [Ca²⁺] increases → even a modest [C₂O₄²⁻] may give Qsp = [Ca²⁺] × [C₂O₄²⁻] > Ksp. The increase in [Ca²⁺] can outweigh the decrease in [C₂O₄²⁻].

What the chemistry actually establishes: Qsp = [Ca²⁺] × [C₂O₄²⁻], so a stone can form only when that product exceeds Ksp. Both concentrations matter, and a change that raises one while lowering the other has no chemically predetermined outcome. That is as far as equilibrium reasoning takes you.

What is a medical finding, not a chemical deduction: observational studies have reported that calcium taken as food is associated with lower stone risk, plausibly because calcium binds oxalate in the gut so less oxalate is absorbed, while high-dose calcium supplements have been associated with higher risk. Those findings come from clinical and epidemiological research and depend on dose, timing with meals, hydration and the individual. They are not consequences of the Ksp expression, and an exam answer should not present them as if they were. Cite them as context, and keep the chemistry claim to the Qsp mechanism.