Grade 11

Grade 11Balance


Solubility equilibrium of sparingly soluble salts


Introduction to solubility equilibrium

Sparingly soluble salts are those that dissolve in water to a minimal extent. Understanding their solubility is important in many fields, including chemistry, environmental science, and engineering. Solubility equilibrium helps us determine how these salts behave in aqueous systems and calculate their solubility product.

Basic definitions and concepts

Before we discuss solubility equilibrium, let us discuss some fundamental concepts.

Solubility

Solubility is the maximum amount of solute that can dissolve in a solvent at a given temperature, resulting in a saturated solution. For less soluble salts, this amount is much less, often expressed as moles per liter (mol/L).

Saturated solution

A saturated solution contains the maximum concentration of solute that can dissolve at a particular temperature. Any more solute added will not dissolve and result in the formation of additional solid.

Solubility product constant (K sp)

The solubility product constant, K sp, is an equilibrium constant that applies to the solubility of sparingly soluble salts. It shows the extent to which the compound dissociates in water.

Expression for K sp

To split a common salt AB into its ions:

AB(s) ⇋ A + (aq) + B - (aq)

K sp expression is given as:

K sp = [A + ][B - ]

Visual examples of solubility equilibrium

Consider the solubility equilibrium of silver chloride (AgCl):

AgCl(s) ⇋ Ag + (aq) + Cl - (aq)
AgCl2 undissolved Ag+ Cl-

Here, K sp = [Ag + ][Cl - ]

General calculations related to K sp

Let us see how the solubility product constant is calculated for different situations.

Determining solubility from K sp

For a salt AB where K sp = s^2 :

s = √(K sp )

For example, if K sp is given for AgCl, then consider AgCl ⇋ Ag + (aq) + Cl - (aq) :

K sp = s^2

Calculating ion concentration

If the solubility s is given then:

[A + ] = [B - ] = s

Illustrative examples

Consider calcium sulfate (CaSO₄):

CaSO₄(s) ⇋ Ca 2+ (aq) + SO₄ 2- (aq)

Given K sp = 2.4 × 10 -5, solve for s:

s = √(2.4 × 10 -5 ) = 4.9 × 10 -3 mol/L

Factors affecting solubility

To apply these concepts in practical scenarios it is necessary to understand what factors affect solubility.

Common-ion effect

When a common ion is already present in a solution, the solubility of the salt decreases.

For example, in a solution containing NaCl, the solubility of AgCl is reduced due to the Cl - ion common to both salts.

pH of the solution

Solubility can be increased or decreased depending on the pH of the solution. In some cases, the addition of an acid or base affects the equilibrium.

Illustrative example using pH

For lead carbonate (PbCO₃), the addition of acid shifts the equilibrium:

PbCO₃(s) ⇋ Pb 2+ (aq) + CO₃ 2- (aq)

The H + from the acid reacts with CO₃ 2-, increasing the solubility.

Le Chatelier's principle and solubility

Le Chatelier's principle states that if an external condition changes, causing a shift in equilibrium, the system will adjust to counteract this change.

Applications to solubility

When the concentration of an ion increases, the equilibrium shifts, causing more of the solid to form, which decreases solubility, and vice versa.

Real-world applications

The solubility equilibrium of sparingly soluble salts has many practical applications.

Water treatment

Understanding solubility helps in removing unwanted ions from water by precipitating them as insoluble salts.

Medicines

Sparingly soluble compounds are used to control drug delivery, where solubility determines the release rate.

Practice problems

Practicing calculations related to solubility equilibrium will deepen your understanding. Here are some problems you can solve:

Problem 1

K sp = 1.1 × 10 -10. Calculate the solubility of BaSO₄ in mol/L.

Problem 2

If the solubility of PbI₂ is 1.3 × 10 -3 mol/L, find K sp.

Problem 3

Find the new solubility of AgCl in the presence of 0.1 M NaCl solution.

Conclusion

Understanding the solubility equilibrium of sparingly soluble salts is essential for predicting how different salts will behave under different conditions. Along with practical applications, recognizing the factors that affect solubility provides deep insight into chemical reactions in natural and industrial processes.


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