Solubility Equilibria of Sparingly Soluble Salts: Definitions, Applications, Examples. & Problems

The salt that dissolves slightly in water is called a sparingly soluble salt. Even though they are only slightly dissolved in water, they establish an equilibrium between the undissolved solid and ions in solution. This is known as solubility equilibrium. For example: CaF₂ (Calcium fluoride), AgCl (Silver chloride), BaSO₄ (Barium sulfate).

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What happens when sparingly soluble salt is added to water?
Consider a general sparingly soluble salt:
$\mathrm{AB}\left(s\right)\rightleftharpoons A^{+}\left(\mathrm{aq}\right)+B^{-}\left(\mathrm{aq}\right)$

1. Free ions are gained when a small amount of salt is dissolved in water.
2. As the ions increase, they recombine to form a solid salt.
3. A dynamic equilibrium is established where the rate of dissolution = rate of precipitation.

An equilibrium constant used to describe the extent to which a sparingly soluble salt dissolved in water is called the solubility product constant. It is denoted by Ksp. The solubility product constant is applied to the salts that do not completely dissolve, instead establishing a dynamic equilibrium between the solid salt and its ions in solution.

Example: 

Sparingly soluble salt: $\mathrm{AB}\left(s\right)\rightleftharpoons A^{+}\left(\mathrm{aq}\right)+B^{-}\left(\mathrm{aq}\right)$

Ksp = $K_{\mathrm{sp}}=\left[A^{+}\right]\left[B^{-}\right]$

Application of Solubility Product Constant (Ksp)

• Common ion effect
• Predicting precipitation
• Determining solubility

Key Points:

• Ksp does not use the concentration of the undissolved solid
• At a given temperature solubility product constant
• Higher Ksp, greater solubility
• Lower Ksp, lesser solution 

When ions are already present in the solution from another source, they decrease the solubility of the ionic salt; this phenomenon is called the common ion effect. 

Related Topics: NCERT Solution | Class 11 Chemistry NCERT Solutions

Example:   

Sparingly soluble salt silver chloride (AgCl):

$\mathrm{AgCl}\left(s\right)\rightleftharpoons A{g}^{+}\left(\mathrm{aq}\right)+C{l}^{-}\left(\mathrm{aq}\right)$

1. If NaCl is added to the solution, it will provide more Cl⁻ ions, which are common to AgCl.
2. As per Le Chatelier's Principle, the equilibrium shifts to the left.

Here we have various application of solubility equilibria of sparingly soluble salts.

1. Prediction of precipitation
2. Qualitative Inorganic Analysis
3. Water Purification
4. Formulation of Medicines
5. Construction and scale formation
6. Soil Chemistry and Agriculture
7. Geology and Mineral Formation

Also Check: NCERT Solution for Class 11 Chemistry Equilibrium 

Solubility equilibria involve:

• Solubility Product: $K_{sp}$ quantifies ion concentrations at equilibrium.
• Common Ion Effect: Added ions reduce solubility.
• Precipitation Occurs when $Q>K_{sp}$ .
• Stoichiometry: Complex salts (e.g., $\mathrm{A}\mathrm{l}(\mathrm{O}\mathrm{H}{)}_3$ ) require careful $K_{sp}$ expressions.

Problem 1: Calculate the solubility of AgCl ( $K_{sp}=1.8\times {10}^{-10}$ ) in pure water.

$\begin{array}{c}\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{A}\mathrm{g}}^{+}+{\mathrm{C}\mathrm{l}}^{-},\mathrm{}{K}_{sp}=s^2=1.8\times {10}^{-10}\\ s=\sqrt[]{1.8\times {10}^{-10}}\approx 1.34\times {10}^{-5}\mathrm{M}\end{array}$

Problem 2: Find the solubility of AgCl in 0.01 M NaCl .

$\begin{array}{c}\left[{\mathrm{C}\mathrm{l}}^{-}\right]\approx 0.01\mathrm{M},\mathrm{}{K}_{sp}=\left[{\mathrm{A}\mathrm{g}}^{+}\right]\left[{\mathrm{C}\mathrm{l}}^{-}\right]=s\cdot 0.01=1.8\times {10}^{-10}\\ s=\frac{1.8\times {10}^{-10}}{0.01}\approx 1.8\times {10}^{-8}\mathrm{M}\end{array}$

Problem 3: Does AgCl precipitate when 10 mL of $0.02\mathrm{M}{\mathrm{A}\mathrm{g}\mathrm{N}\mathrm{O}}_3$ mixes with 10 mL of 0.02 M NaCl ? ( $K_{sp}=1.8\times {10}^{-10}$ )

Final volume $=20\mathrm{m}\mathrm{L}$ . Concentrations: $\left[{\mathrm{A}\mathrm{g}}^{+}\right]=\left[{\mathrm{C}\mathrm{l}}^{-}\right]=\frac{0.02\times 10}{20}=0.01\mathrm{M}$ .

$Q=\left[{\mathrm{A}\mathrm{g}}^{+}\right]\left[{\mathrm{C}\mathrm{l}}^{-}\right]=(0.01{)}^2={10}^{-4}$

Since $Q>K_{sp},\mathrm{A}\mathrm{g}\mathrm{C}\mathrm{l}$ precipitates.
Problem 4: For ${\mathrm{C}\mathrm{a}}_3{\left({\mathrm{P}\mathrm{O}}_4\right)}_2,K_{sp}=2.07\times {10}^{-33}$ . Calculate solubility in pure water.

$\begin{array}{c}{\mathrm{C}\mathrm{a}}_3{\left({\mathrm{P}\mathrm{O}}_4\right)}_2\left(\mathrm{s}\right)\rightleftharpoons 3{\mathrm{C}\mathrm{a}}^{2+}+2{\mathrm{P}\mathrm{O}}_4^{3-},\mathrm{}{K}_{sp}=\left(3s{)}^3\right(2s{)}^2=108s^5\\ 108s^5=2.07\times {10}^{-33},\mathrm{}{s}^5\approx 1.92\times {10}^{-35},\mathrm{}s\approx 1.27\times {10}^{-7}\mathrm{M}\end{array}$