⚗️ Visual Chemistry
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Solubility Equilibrium & Ksp

20min Part 2 / Ch4 / Lesson 4
Prerequisites: 2-4-2

Objectives

  • Understand the meaning of the solubility product Ksp
  • Use Ksp to predict whether a precipitate will form
  • Explain the common-ion effect

Solubility Product Ksp

A sparingly soluble salt AmBn\mathrm{A_mB_n} dissolves slightly in water to reach equilibrium:

AmBn(s)⇌mAn+(aq)+nBm−(aq)\mathrm{A_mB_n(s) \rightleftharpoons mA^{n+}(aq) + nB^{m-}(aq)}

Solubility product: Ksp=[An+]m[Bm−]nK_{sp} = [\mathrm{A^{n+}}]^m [\mathrm{B^{m-}}]^n

A smaller KspK_{sp} means a less soluble salt.

SaltDissolution equilibriumKspK_{sp}
AgClAg⁺ + Cl⁻1.8×10−101.8 \times 10^{-10}
BaSO₄Ba²⁺ + SO₄²⁻1.1×10−101.1 \times 10^{-10}
PbI₂Pb²⁺ + 2I⁻9.8×10−99.8 \times 10^{-9}
CaCO₃Ca²⁺ + CO₃²⁻3.4×10−93.4 \times 10^{-9}

Representative Ksp values (25°C)

Predicting Precipitation

Compare the ion-product QQ with KspK_{sp}:

  • Q>KspQ > K_{sp} → precipitate forms
  • Q=KspQ = K_{sp} → saturated (at equilibrium)
  • Q<KspQ < K_{sp} → no precipitate (undersaturated)
Will AgCl precipitate when equal volumes of 0.010 mol/L AgNO₃ and 0.0010 mol/L NaCl are mixed?

Equal volumes → each ion concentration is halved:

[Ag+]=0.0050 mol/L[\mathrm{Ag^+}] = 0.0050\,\mathrm{mol/L}, [Cl−]=0.00050 mol/L[\mathrm{Cl^-}] = 0.00050\,\mathrm{mol/L}

Q=[Ag+][Cl−]=0.0050×0.00050=2.5×10−6Q = [\mathrm{Ag^+}][\mathrm{Cl^-}] = 0.0050 \times 0.00050 = 2.5 \times 10^{-6}

Q=2.5×10−6≫Ksp=1.8×10−10Q = 2.5 \times 10^{-6} \gg K_{sp} = 1.8 \times 10^{-10}

Since Q>KspQ > K_{sp}, AgCl precipitates.

Common-Ion Effect

Adding NaCl to a saturated AgCl solution increases [Cl⁻]:

[Ag+][Cl−]=Ksp[\mathrm{Ag^+}][\mathrm{Cl^-}] = K_{sp}

[Cl⁻] rises → [Ag⁺] must decrease → more AgCl precipitates.

This decrease in solubility caused by adding an ion already present in the equilibrium is the common-ion effect.

Check Your Understanding

Q1 A salt with a smaller Ksp is...

Q2 When Q > Ksp, what happens?

Q3 Adding NaCl to saturated AgCl solution causes AgCl solubility to...

Exercises

Q1. Given Ksp(PbI2)=9.8×10−9K_{sp}(\mathrm{PbI_2}) = 9.8 \times 10^{-9}, find [Pb2+][\mathrm{Pb^{2+}}] in a saturated PbI₂ solution.

Solution

PbI2⇌Pb2++2I−\mathrm{PbI_2 \rightleftharpoons Pb^{2+} + 2I^-}

Let [Pb2+]=s[\mathrm{Pb^{2+}}] = s, then [I−]=2s[\mathrm{I^-}] = 2s

Ksp=s⋅(2s)2=4s3K_{sp} = s \cdot (2s)^2 = 4s^3

s=Ksp43=9.8×10−943=2.45×10−93s = \sqrt[3]{\dfrac{K_{sp}}{4}} = \sqrt[3]{\dfrac{9.8 \times 10^{-9}}{4}} = \sqrt[3]{2.45 \times 10^{-9}}

s=1.35×10−3 mol/Ls = 1.35 \times 10^{-3}\,\mathrm{mol/L}