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Solubility & Ksp Lab

Select a sparingly soluble salt, calculate its molar solubility from Ksp, explore the common ion effect, and predict whether mixing two solutions will cause precipitation by comparing the ion product Q to Ksp.

Guided Experiment: Molar Solubility from Ksp

How does the stoichiometry of a salt's dissolution reaction affect its molar solubility? Do salts with similar Ksp values always have similar molar solubilities?

Write your hypothesis in the Lab Report panel, then click Next.

Solution Visualization

++++++++++++++−−−−−−−−−−−−−−AgCl in WaterAg⁺Cl⁻Solid AgClSaturated

Controls

Results

Dissolution Equilibrium
AgCl → Ag⁺ + Cl⁻
Ksp=[Ag+]⋅[Cl−]K_{sp} = [\text{Ag}^{+}] \cdot [\text{Cl}^{−}]
Molar Solubility Derivation
Ksp=s⋅s=s2K_{sp} = s \cdot s = s^{2}
s=1.33e−5 Ms = 1.33e-5 \text{ M}
Ksp
1.77e-10
Molar Solubility
1.33e-5 M
Salt Type
AB
[Ag+]
1.33e-5 M
[Cl−]
1.33e-5 M
Ion Product (Q) vs Ksp
Q
1.77e-10
Ksp
1.77e-10
Result
Q = Ksp
Solution is at equilibrium (saturated)

Data Table

(0 rows)
#TrialSaltKspMolar Solubility(M)Common Ion(M)[Cation](M)[Anion](M)QQ vs Ksp
0 / 500
0 / 500
0 / 500

Reference Guide

Solubility Product (Ksp)

For a sparingly soluble salt AₘBₙ that dissolves as AₘBₙ → mA⁺ + nB⁻, the solubility product expression is written as the product of ion concentrations raised to their stoichiometric coefficients.

Ksp=[An+]m⋅[Bm−]nK_{sp} = [A^{n+}]^m \cdot [B^{m-}]^n

Ksp is a constant at a given temperature. A small Ksp means the salt is very sparingly soluble. Ksp does not change when common ions are added.

Molar Solubility

Molar solubility (s) is the number of moles of salt that dissolve per liter of solution. The relationship between s and Ksp depends on the stoichiometry of the salt.

AB: Ksp=s2⇒s=Ksp\text{AB: } K_{sp} = s^2 \quad\Rightarrow\quad s = \sqrt{K_{sp}}
AB2: Ksp=4s3⇒s=Ksp43\text{AB}_2\text{: } K_{sp} = 4s^3 \quad\Rightarrow\quad s = \sqrt[3]{\frac{K_{sp}}{4}}

Salts with different stoichiometry cannot be compared by Ksp alone. A salt with a larger Ksp can have a smaller molar solubility if its dissolution produces more ions.

Common Ion Effect

Adding a soluble salt that shares an ion with the sparingly soluble salt shifts the equilibrium, reducing the molar solubility. This is an application of Le Chatelier's principle.

For example, adding NaCl to a saturated AgCl solution increases [Cl⁻], which shifts the equilibrium to the left, decreasing [Ag⁺] and the amount of AgCl that can dissolve.

Ksp=s⋅(s+c)≈s⋅c(c≫s)K_{sp} = s \cdot (s + c) \approx s \cdot c \quad (c \gg s)

When the common ion concentration c is much larger than s, the molar solubility is approximately Ksp / c for AB-type salts.

Precipitation Prediction (Q vs Ksp)

The ion product Q has the same form as Ksp but uses the actual (non-equilibrium) ion concentrations in solution.

Q=[An+]m⋅[Bm−]nQ = [A^{n+}]^m \cdot [B^{m-}]^n

Q < Ksp means the solution is unsaturated. More salt can dissolve. No precipitate forms.

Q = Ksp means the solution is exactly saturated. The system is at equilibrium.

Q > Ksp means the solution is supersaturated. A precipitate will form until Q decreases to Ksp.