Calculate Ksp, predict precipitation, and determine molar solubility
Solubility Product (Ksp): For MmAn ⇌ mM+ + nA-
Ksp = [M+]m[A-]n
Use scientific notation (e.g., 1.8e-10)
The solubility product constant (Ksp) describes the equilibrium between a solid ionic compound and its dissolved ions in a saturated solution.
General Form:
MmAn(s) ⇌ mM+(aq) + nA-(aq)
Ksp = [M+]m[A-]n
Qsp > Ksp
Supersaturated → Precipitate forms
Qsp = Ksp
Saturated → At equilibrium
Qsp < Ksp
Unsaturated → No precipitation
Problem:
Will AgCl precipitate if 100 mL of 0.001 M AgNO₃ is mixed with 100 mL of 0.001 M NaCl? (Ksp of AgCl = 1.8×10⁻¹⁰)
Solution:
1. After mixing, volume = 200 mL
2. [Ag⁺] = (0.001 M)(100 mL)/(200 mL) = 5×10⁻⁴ M
3. [Cl⁻] = (0.001 M)(100 mL)/(200 mL) = 5×10⁻⁴ M
4. Qsp = [Ag⁺][Cl⁻] = (5×10⁻⁴)(5×10⁻⁴)
5. Qsp = 2.5×10⁻⁷
Qsp (2.5×10⁻⁷) > Ksp (1.8×10⁻¹⁰)
✓ Yes, AgCl will precipitate
Problem:
Calculate the molar solubility of PbI₂ in pure water. (Ksp = 7.1×10⁻⁹)
Solution:
PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)
If s = molar solubility:
[Pb²⁺] = s, [I⁻] = 2s
Ksp = [Pb²⁺][I⁻]² = (s)(2s)² = 4s³
7.1×10⁻⁹ = 4s³
s³ = 1.775×10⁻⁹
s = 1.21×10⁻³ M
Molar solubility = 1.21×10⁻³ M
Or 1.21 millimoles per liter
| Compound | Formula | Ksp | Solubility |
|---|---|---|---|
| Silver chloride | AgCl | 1.8×10⁻¹⁰ | Very low |
| Barium sulfate | BaSO₄ | 1.1×10⁻¹⁰ | Very low |
| Calcium carbonate | CaCO₃ | 3.4×10⁻⁹ | Low |
| Lead(II) iodide | PbI₂ | 7.1×10⁻⁹ | Low |
| Silver chromate | Ag₂CrO₄ | 1.1×10⁻¹² | Very low |
| Iron(III) hydroxide | Fe(OH)₃ | 2.8×10⁻³⁹ | Extremely low |
| Calcium phosphate | Ca₃(PO₄)₂ | 2.1×10⁻³³ | Extremely low |
Adding a common ion (an ion already present in the equilibrium) decreases the solubility of a sparingly soluble salt.
Example: AgCl is less soluble in 0.1 M NaCl than in pure water because of the added Cl⁻ ions.
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Adding Cl⁻ shifts equilibrium left, reducing [Ag⁺] and thus the solubility of AgCl.
This principle is used in qualitative analysis to selectively precipitate ions.