Topic B. Factors Affecting Solubility

Site: UPLB e-Learning Based Instruction
Course: CHEM 32 - C (2nd sem 20-21)
Book: Topic B. Factors Affecting Solubility
Printed by: Guest user
Date: Saturday, 16 November 2024, 12:56 AM

Description

     Le Châtelier's principle governs the response of a system in equilibrium when a stress is introduced. Some “stressors” that affect solubility equilibria are presence of common ion, pH and presence of a complexing agent.

Subtopic B.1. Common-Ion Effect

     Le Châtelier's principle governs the response of a system in equilibrium when a stress is introduced. Some “stressors” that affect solubility equilibria are presence of common ionpH and presence of a complexing agent.


     The common-ion effect results when an ionic compound is mixed in a solution that contains any of its ions that is already at equilibrium.


Review Reading: 

The Common-Ion Effect” by Tung, E. et al. (2020)

This online resource discusses how common ion influences the solubility of an ion. Special emphasis on Example 17.1.3, which compares solubilities in water and in a solution with the common ion. Answer the self-test in this material to assess your understanding of the topic. Start reviewing by clicking the title

Subtopic B.2. Effect of Temperature on Solubility of Solids

     Le Châtelier's principle governs the response of a system in equilibrium when a stress is introduced. Some “stressors” that affect solubility equilibria are presence of common ionpH and presence of a complexing agent.


     Effect of temperature in solubility values has been determined for solid solutes in liquid solvents. These effects differ depending on whether the reaction is endothermic or exothermic. Using Le Châtelier's principle, the effects of temperature in both scenarios can be determined.

1.   First, consider an endothermic reaction (ΔHsolvation > 0): Increasing the temperature results in a stress on the reactants’ side from the additional heat. Le Châtelier’s principle predicts that the system shifts toward the product side in order to alleviate this stress. By shifting towards the product side, more of the solid is dissociated when equilibrium is again established, resulting in increased solubility.

2.   Second, consider an exothermic reaction (ΔHsolvation<0): Increasing the temperature results in a stress on the products’ side from the additional heat. Le Châtelier’s principle predicts that the system shifts toward the reactant side in order to alleviate this stress. By shifting towards the reactant's side, less of the solid is dissociated when equilibrium is again established, resulting in decreased solubility.

 

(Text excerpt adapted from Solubility and Factors Affecting Solubility by Tran and Liu, 2020)

Subtopic B.3. Effect of pH

     Le Châtelier's principle governs the response of a system in equilibrium when a stress is introduced. Some “stressors” that affect solubility equilibria are presence of common ionpH and presence of a complexing agent.


     When any of the ions in the salt has acidic or basic properties, solubility of precipitates will depend on pH. When the anion is a weak base B- , the solubility of salt increases as pH is decreased (Case 1); when the cation is a weak acid AH+ , the solubility of salt increases as pH is increased (Case 2); when both ions are pH-dependent, effect of pH will depend on equilibrium constant values (Case 3).


Incorporating the effect of pH in solubility calculations may be done in two ways: using the systematic approach or using distribution (α).

 

1.      Using the systematic approach

     To demonstrate this approach, consider Example 11-7 below (page 260-262 of Skoog et al., 2013).

Calculate the molar solubility of calcium oxalate in a solution that has been buffered so that its pH is constant and equal to 4.00.

To solve for the molar solubility using the systematic approach, the step by step process is as follows:

STEP 1: Write Pertinent Equilibria


STEP 2: Define the Unknown


STEP 3: Write All the Equilibrium-Constant Expressions


STEP 4: Mass- and Charge-Balance Expressions


STEP 5: Count the Number of Independent Equations and Unknowns. There are four independent equations (Ksp, Ka1, Ka2, MBE) and four unknowns ([Ca2+], [C2O42-], [HC2O4-], [H2C2O4]). Therefore, an exact solution can be obtained.

STEP 6: Make Approximations. It is already easy to solve the system of equations exactly, so we no longer need to make approximations.

STEP 7: Solve.



2.    Using distribution α  

      With the same example given above (Example 11-7), let’s solve for the molar solubility using α.


Quantitative Separation of Metal Sulfides

     By controlling pH or [H3O+], selective separation of metal ions using saturated H2S can be performed. Precipitates containing sulfide ion have very low Ksp values (10-10 to 10-90). In order to say that precipitation has completed, the amount of solute remaining in the solution should be very small (99.9% or more) leaving less than 0.1% (1/1000) of the ion in solution.


Required Reading: 

The Effects of pH on Solubility

Read on section, Selective Precipitation Using pH. This portion of the online resource shows that cadmium (II) and zinc (II) are separated using selective precipitation with sulfide. Click the title start reading the resource. 


Supplementary Reading: 

“Chapter 11C. Separation of ions by control of the concentration of the precipitating agent” (page 268-272 of Skoog et al., 2013)

Note in Example 11-11 how cadmium (II) and thallium (I) were separated in the solution.

Subtopic B.4. Presence of Complexing Agent

     Le Châtelier's principle governs the response of a system in equilibrium when a stress is introduced. Some “stressors” that affect solubility equilibria are presence of common ionpH and presence of a complexing agent.


     Molar solubility of precipitates significantly increases in the presence of complexing agents that form complexes with either ions. Their presence shifts the equilibrium forward due to a decrease in free ions.


     Solubility calculation in the presence of complexing agents may be done using metal distribution (α):

Let’s take AgSCN in a solution with 0.0150 M free ammonia. To simplify this problem, neglect all acid/base properties of Ag+ and SCNand only account for the presence of NH3 as the complexing agent.

AgSCN: Ksp=1.1×10-12Ammonia complexes: Kf1=2.04×103 Kf2=8.13×103


Solubility of AgSCN is 1.0x10-6 M  in water (that is, without the auxiliary complexing agent). Observe that the solubility increased in the presence of ammonia.