Solubility Equilibria of Sparingly Soluble Salts
💧 Salt Solubility Basics
Salts dissolve differently in water! Their solubility depends on:
- 🔋 Lattice enthalpy (ion-ion attraction in solid)
- 💦 Solvation enthalpy (ion-water attraction, always releases energy!)
For dissolving: Solvation enthalpy > Lattice enthalpy. Polar solvents (like water) help this!
📊 Solubility Categories
| Category | Solubility | Examples |
|---|---|---|
| Soluble | > 0.1 M | Calcium chloride (hygroscopic!) |
| Slightly soluble | 0.01 M – 0.1 M | – |
| Sparingly soluble | < 0.01 M | Lithium fluoride (“insoluble”) |
✨ Solubility Product Constant (Ksp)
For a salt: \( \ce{M_xX_y(s) <=> xM^{p+}(aq) + yX^{q-}(aq)} \)
Ksp = \( [\ce{M^{p+}}]^x [\ce{X^{q-}}]^y \) (Concentrations at equilibrium)
Example for \(\ce{BaSO4}\):
\( \ce{BaSO4(s) <=> Ba^{2+}(aq) + SO4^{2-}(aq)} \)
Ksp = \( [\ce{Ba^{2+}}][\ce{SO4^{2-}}] = 1.1 \times 10^{-10} \) at 298K
📐 Calculating Solubility (S) from Ksp
Case 1: 1:1 salts like BaSO4
If \( \ce{S} \) = solubility, then \( [\ce{Ba^{2+}}] = \ce{S} \), \( [\ce{SO4^{2-}}] = \ce{S} \)
\( \ce{K_{sp} = S^2} \) → \( \ce{S} = \sqrt{\ce{K_{sp}}} = \sqrt{1.1 \times 10^{-10}} = 1.05 \times 10^{-5} \text{ mol/L} \)
Case 2: Salts with multiple ions like A2X3
\( \ce{A2X3 -> 2A^{3+} + 3X^{2-}} \)
If \( \ce{S} \) = solubility, then \( [\ce{A^{3+}}] = 2\ce{S} \), \( [\ce{X^{2-}}] = 3\ce{S} \)
\( \ce{K_{sp} = (2S)^2 (3S)^3 = 108S^5} \)
🔍 Comparing Solubility of Salts
Problem: Which is more soluble? \(\ce{Ni(OH)2}\) (Ksp = \( 2.0 \times 10^{-15} \)) or \(\ce{AgCN}\) (Ksp = \( 6 \times 10^{-17} \))?
Solution:
For \(\ce{AgCN}\): \( \ce{S_1^2 = 6 \times 10^{-17}} \) → \( \ce{S_1 = 7.8 \times 10^{-9} \text{ M}} \)
For \(\ce{Ni(OH)2}\): \( \ce{[Ni^{2+}][OH^-]^2 = S_2 \cdot (2S_2)^2 = 4S_2^3 = 2.0 \times 10^{-15}} \) → \( \ce{S_2 = 0.58 \times 10^{-4} \text{ M}} \)
Result: \(\ce{S_2 > S_1}\) → \(\ce{Ni(OH)2}\) is more soluble! ✅
⚠️ Common Ion Effect
Adding a common ion decreases solubility (Le Chatelier’s principle).
Example: Solubility of \(\ce{Ni(OH)2}\) in 0.10 M NaOH:
\( \ce{K_{sp} = [Ni^{2+}][OH^-]^2 = S \cdot (0.10 + 2S)^2 = 2.0 \times 10^{-15}} \)
Since Ksp is small, \( \ce{0.10 + 2S ≈ 0.10} \), so:
\( \ce{S \cdot (0.10)^2 = 2.0 \times 10^{-15}} \) → \( \ce{S} = 2.0 \times 10^{-13} \text{ M} \)
Compare: In pure water, solubility was \( 5.8 \times 10^{-5} \text{ M}\) – way higher! 😲
📈 pH Effect on Solubility
Salts of weak acids (e.g., phosphates, carbonates) dissolve better in acidic pH!
Why? H+ reacts with the anion (X–), reducing its concentration:
\( \ce{H+ + X- <=> HX} \) → More solid dissolves to maintain Ksp!
For salt MX: \( \ce{S} = \sqrt{\ce{K_{sp} \left(1 + \frac{[H^+]}{\ce{K_a}}\right)} \)
Lower pH (↑ [H+]) → ↑ solubility!
🔑 NEET Must-Knows
- Calculate solubility (S) from Ksp for any salt type (e.g., BaSO4, A2X3).
- Compare solubility of salts using Ksp (like AgCN vs. Ni(OH)2).
- Common ion effect: Predict solubility in solutions with common ions (e.g., Ni(OH)2 in NaOH).
- pH effect: Explain why salts like CaCO3 dissolve in acid 🍋.
- Write Ksp expressions from dissociation equations.
