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.

Kidney stones are the biological consequence of Qsp exceeding Ksp in urine — and the controversy over high-calcium diets and stone risk is a direct application of the common ion effect.

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⁻].

Research finding: Moderate calcium (1000–1200 mg/day from food) reduces stone risk by binding dietary oxalate in the gut. Excessive supplemental calcium can increase risk by raising urinary [Ca²⁺] without sufficient corresponding oxalate reduction. The Qsp = [Ca²⁺] × [C₂O₄²⁻] product — both concentrations — determines stone risk.

Q1. 50.0 mL of 1.0 × 10⁻⁴ mol/L AgNO₃ is mixed with 50.0 mL of 1.0 × 10⁻⁴ mol/L NaCl. Ksp(AgCl) = 1.8 × 10⁻¹⁰. Will a precipitate form?

Q2. NaCl is dissolved in a saturated AgCl solution at equilibrium with excess AgCl solid. Which correctly describes the effect?

Q3. The molar solubility of BaSO₄ in pure water is 1.05 × 10⁻⁵ mol/L. In a 0.010 mol/L Na₂SO₄ solution, which prediction is correct?

SAQ 1 (5 marks): A chemist mixes 30.0 mL of 5.0 × 10⁻³ mol/L Ba(NO₃)₂ with 70.0 mL of 3.0 × 10⁻³ mol/L Na₂SO₄. Ksp(BaSO₄) = 1.1 × 10⁻¹⁰. (a) Calculate [Ba²⁺] and [SO₄²⁻] after mixing. (b) Calculate Qsp. (c) Predict whether a precipitate forms and justify using the Qsp vs Ksp comparison.

SAQ 2 (4 marks): The molar solubility of AgCl in pure water is 1.34 × 10⁻⁵ mol/L. A chemist adds excess solid NaCl to a saturated AgCl solution until [Cl⁻] = 0.050 mol/L. (a) Calculate the new [Ag⁺] in the presence of the common ion. (b) Describe the shift in equilibrium using Le Chatelier's Principle. (c) State whether Ksp changes and explain why or why not.