Kidney stones form when the ionic product of calcium and oxalate ions in urine exceeds Ksp of calcium oxalate — and the common ion effect explains why some dietary choices that seem healthy can actually increase the risk of stone formation.
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A doctor tells a patient with kidney stones: "Reduce your oxalate intake." The patient reasons: "But if I drink lots of water and eat a high-calcium diet, the calcium should bind to any oxalate in my gut before it reaches my kidneys." The doctor responds: "Actually, increasing calcium intake can sometimes make kidney stones worse, not better."
Who do you think is correct, and why? What chemistry principle is the doctor applying? Write your reasoning before reading on.
Wrong: Chemical equations can be balanced by changing subscripts in formulas.
Right: Chemical equations must be balanced by changing coefficients only. Subscripts in chemical formulas define the identity of the compound — changing them creates a different substance. If you cannot balance an equation with whole-number coefficients, check that your formulas are correct.
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.
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
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.
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.
| Dissolution type | Temperature increases | Ksp change | Solubility change | Example |
|---|---|---|---|---|
| Endothermic (LE > HE) most ionic compounds |
Shifts RIGHT (endothermic direction) | Ksp increases | Increases | KNO₃, NaCl, most salts |
| Exothermic (HE > LE) | Shifts LEFT | Ksp decreases | Decreases | Ce₂(SO₄)₃ (retrograde) |
This is the same LCP temperature rule applied to dissolution equilibria: increase T shifts equilibrium in the endothermic direction.
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.
25.0 mL of 4.0 × 10⁻³ mol/L Pb(NO₃)₂ is mixed with 75.0 mL of 2.0 × 10⁻⁴ mol/L Na₂SO₄. Ksp(PbSO₄) = 1.6 × 10⁻⁸. (a) Calculate [Pb²⁺] and [SO₄²⁻] after mixing. (b) Calculate Qsp. (c) Predict whether a precipitate forms.
V_total = 25.0 + 75.0 = 100.0 mL
[Pb²⁺]new = (4.0 × 10⁻³)(25.0/100.0) = (4.0 × 10⁻³)(0.250) = 1.0 × 10⁻³ mol/L
[SO₄²⁻]new = (2.0 × 10⁻⁴)(75.0/100.0) = (2.0 × 10⁻⁴)(0.750) = 1.5 × 10⁻⁴ mol/L
$Q_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] = (1.0 \times 10^{-3})(1.5 \times 10^{-4}) = 1.5 \times 10^{-7}$
Qsp = 1.5 × 10⁻⁷ > Ksp = 1.6 × 10⁻⁸. Qsp > Ksp → solution is supersaturated with respect to PbSO₄ → white PbSO₄ precipitate forms. Ions deposit as solid until [Pb²⁺][SO₄²⁻] decreases to 1.6 × 10⁻⁸.
Calculate the molar solubility of CaF₂ (Ksp = 3.9 × 10⁻¹¹) in (a) pure water and (b) 0.10 mol/L NaF solution. Compare and explain using Le Chatelier's Principle.
CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). [Ca²⁺] = s; [F⁻] = 2s.
$K_{sp} = 4s^3 = 3.9 \times 10^{-11}$
$s^3 = 9.75 \times 10^{-12}$ → $s = \sqrt[3]{9.75 \times 10^{-12}} = \mathbf{2.14 \times 10^{-4}}$ mol/L
NaF fully dissolves → [F⁻] = 0.10 mol/L initially. CaF₂ dissolves by s mol/L.
[Ca²⁺] = s; [F⁻] = 0.10 + 2s ≈ 0.10 mol/L (s << 0.10 — verify after solving)
$K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 = s(0.10)^2 = s(0.010) = 3.9 \times 10^{-11}$
$s = 3.9 \times 10^{-11}/0.010 = \mathbf{3.9 \times 10^{-9}}$ mol/L
Verify: 2s = 7.8 × 10⁻⁹ << 0.10 ✓ (assumption valid)
s in pure water = 2.14 × 10⁻⁴ mol/L; s in 0.10 mol/L NaF = 3.9 × 10⁻⁹ mol/L.
CaF₂ is approximately 55,000× less soluble in 0.10 mol/L NaF than in pure water.
LCP: NaF adds F⁻ (a product of CaF₂ dissolution). Adding product → Qsp increases above Ksp → LCP shifts dissolution equilibrium LEFT → more CaF₂ precipitates → solubility decreases dramatically. Ksp unchanged.
100.0 mL of 1.0 × 10⁻⁵ mol/L CaCl₂ is mixed with 100.0 mL of 1.0 × 10⁻⁵ mol/L Na₂SO₄. Ksp(CaSO₄) = 4.9 × 10⁻⁵. Will CaSO₄ precipitate? Show all working including the dilution step.
V_total = 100.0 + 100.0 = 200.0 mL
[Ca²⁺]new = (1.0 × 10⁻⁵)(100.0/200.0) = 5.0 × 10⁻⁶ mol/L
[SO₄²⁻]new = (1.0 × 10⁻⁵)(100.0/200.0) = 5.0 × 10⁻⁶ mol/L
$Q_{sp} = (5.0 \times 10^{-6})(5.0 \times 10^{-6}) = 2.5 \times 10^{-11}$
Qsp = 2.5 × 10⁻¹¹ << Ksp = 4.9 × 10⁻⁵. Qsp < Ksp → solution is highly unsaturated → no precipitate forms. The solution can dissolve much more CaSO₄ before reaching saturation.
Comparing Qsp to Ksp tells you whether a precipitate will form: if Qsp > Ksp, precipitation occurs; if Qsp < Ksp, the solution is unsaturated; if Qsp = Ksp, it is exactly saturated. The common-ion effect decreases the solubility of a salt when one of its ions is already present in solution (shifts the dissolution equilibrium left). This explains why drinking large volumes of water with a high-calcium diet can still promote kidney stones if oxalate is present — the ion product may exceed Ksp.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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?
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