Solubility and Solubility Product

Solubility refers to the maximum amount of a solute that can dissolve in a given quantity of solvent at a specific temperature and pressure to form a saturated solution. It is usually expressed in units like grams per litre (g/L) or moles per litre (mol/L).

The solubility product is an equilibrium constant that applies to sparingly soluble ionic compounds. It is the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients.

1.0Introduction

1. Solubility as an Equilibrium: The solubility of ionic solids can be treated as an equilibrium process, where the solid is in dynamic balance with its dissociated ions in solution.
                                                         Solid↔Ions in Solution
2. Complete Dissociation: When an ionic solid dissolves in water, it dissociates completely into its constituent ions.
3. Quantitative Definition: Solubility refers to the amount of solid (expressed in moles or grams per litre) that dissolves in a solvent to form a saturated solution, resulting in the production of ions. This value can be calculated for specific solids.
4. Qualitative Definition: Solubility can also be described qualitatively. There is no precise boundary separating "soluble" and "insoluble" substances. The terms "soluble," "slightly soluble," and "insoluble" are relative and context-dependent.

2.0Solubility Product (K_sp)

The solubility product (K_sp) is a special equilibrium constant that describes how slightly soluble ionic compounds dissolve in water.

For example, barium sulfate is nearly insoluble, but a tiny amount does dissolve. When mixed with water, some barium and sulfate ions separate from the solid and enter the solution. Over time, some dissolved ions return to the solid, creating a balance.

At equilibrium, most of the compound remains as a solid, with only a tiny amount dissolved. This balance can be described using K_sp, which depends on the concentration of ions in the solution.

The solubility product (K_sp​) relates the concentrations of ions in a saturated solution of a sparingly soluble compound to its solubility (S).

3.0Common Ion Effect on Solubility

The solubility of substances always decreases in the presence of a common ion. According to Le-Chatelier's principle on increasing common ion concentration equilibrium shifts in a backward direction until the equilibrium is re-established, so, the solubility of substances decreases but Ksp remains the same because it is an equilibrium constant which depends only on temperature.

4.0Condition of Precipitation

The ionic product (IP) of an electrolyte is defined in the same way as Ksp. The only difference is that ionic product (IP) expression contains the initial concentration of ions or the concentration at any time (Solution is under-saturated or over-saturated) whereas the expression of Ksp contains only saturated condition (equilibrium) concentration.

• Soluble: A solid is considered soluble if at least 0.1 mol L⁻¹ dissolves.

IP < K_sp: The solution is unsaturated and precipitation will not occur.

• Slightly Soluble: Solubility falls between 0.001 mol L⁻.

IP = K_sp: The solution is saturated and solubility equilibrium exists.

• Insoluble: A solid is considered insoluble if less than 0.001 mol L⁻¹ dissolves.

IP > K_sp: The solution is supersaturated and hence precipitation of the compound will occur

5.0Relationship Between Solubility and Solubility Product ($K_{sp}$)

The solubility product (K_sp​) relates the concentrations of ions in a saturated solution of a sparingly soluble compound to its solubility (S).

General Dissociation Equation:

A_xB_y(s) ↔ xA^y+(aq) + yB^x−(aq)

At equilibrium:

• The concentration of A^y+is  xS
• The concentration of B^x− is yS

Expression for K_sp​:

K_sp=[A^y+]^x⋅[B^x−]^y

Substitute the concentrations:

K_sp=(xS)^x⋅(yS)^y

Simplify:

K_sp=S^x+y⋅x^x⋅y^y

Final Relationship:

K_sp=[S]^x+y⋅[x]^x⋅[y]^y

Here:

• S: Solubility of the compound in mol/L
• x and y: Stoichiometric coefficients of A and B in the dissociation equation.

Example: For a solution of AgCl in water, the equilibrium can be expressed as:

                     AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

• At equilibrium, all three components—solid AgCl, Ag⁺ ions, and Cl⁻ ions—must be present to confirm that the system has reached equilibrium.
• The equilibrium constant for this reaction is denoted as K, and in this case, it is called the Solubility Product Constant, symbolized as K_sp.
• The Ksp(AgCl) represents the solubility product constant for AgCl when it dissolves (or attempts to dissolve) according to the equation above.
• Similar to other equilibrium constants (K_eq), the value of K_sp depends on the temperature. Typically, more solids dissolve at higher temperatures.

6.0Solved Example 

To calculate the solubility of AgCl (AgCl(s)) in mol/L, we use its solubility product constant (K_sp).

Given:

Ksp(AgCl)=1.8×10⁻¹⁰

Step 1: Write the dissociation equation

$AgCl(s)\leftrightarrow A{g}^{+}(aq)+C{l}^{-}(aq)$

Step 2: Express K_sp​

$K_{sp}=[A{g}^{+}][C{l}^{-}]$

Let the solubility of AgCl be s. In solution:

• The concentration of Ag⁺ is s.
• The concentration of Cl⁻ is also s.

Substitute into the K_sp expression:

K_sp = s*s=s²

Step 3: Solve for s

s²=1.8 ×10^-10 

$s=\sqrt{1.8\times 10^{-10}}$

The solubility of AgCl is 1.34×10⁻⁵ mol/L.

Solved Example:

The solubility of magnesium hydroxide (Mg(OH)₂ at 298 K is 1.71×10−4 mol dm⁻³. Calculate the solubility product (K_sp​).

Solution:
The dissolution equilibrium for Mg(OH)₂​ is:

$Mg(OH{)}_2(s)\leftrightarrow M{g}^{2+}(aq)+2O{H}^{-}(aq)$

For every mole of Mg(OH)₂ that dissolves:

• 1 mole of Mg²⁺ ions is produced.
• 2 moles of OH⁻ ions are produced.

Concentration of ions:

• The concentration of Mg²⁺ ions is equal to the solubility of

$[M{g}^{2+}]=1.71\times 10^{-4}mold{m}^{-3}$

• The concentration of OH⁻ ions is twice that of Mg²⁺:

$[O{H}^{-}]=2\times 1.71\times 10^{-4}=3.42\times 10^{-4}mold{m}^{-3}$

Solubility product expression:

$K_{sp}=[M{g}^{2+}]\times [O{H}^{-}{]}^2$

Substitute the values:

$K_{sp}=(1.71\times 10^{-4})\times (3.42\times 10^{-4}{)}^2$

Simplify the calculation:

First Calculate $(3.42\times 10^{-4}{)}^2:$

$(3.42\times 10^{-4}{)}^2=11.7\times 10^{-8}$

Next, multiply by $1.71\times 10^{-4}$:

$K_{sp}=1.71\times 11.7\times 10^{-12}=20.0\times 10^{-12}$

Final result:    

$K_{sp}=2.00\times 10^{-11}$