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What is the common ion effect?

Short Answer

Expert verified
The common ion effect is the decrease in solubility of an ionic compound when another compound with a common ion is added to the solution due to a shift in equilibrium.

Step by step solution

01

Understanding the concept

The common ion effect refers to the phenomenon where the addition of an ion common to a solute in a solution affects the solubility of the solute. This is seen when the solubility of an ionic compound decreases upon the addition of another ionic compound that shares a common ion.
02

Applying Le Chatelier's Principle

To understand the common ion effect, apply Le Chatelier's Principle, which states that if an equilibrium system is disturbed, the system will shift its equilibrium position so as to counteract the disturbance. In the case of the common ion effect, the addition of an ion that is already present in the solution will shift the equilibrium to decrease the solubility of the solute.
03

Observing the effect on solubility

When a common ion is added, the ion product of the solute can exceed the solubility product constant (Ksp), resulting in precipitation if the solution was originally saturated, or simply a reduction in the amount of solute that can dissolve if the solution was unsaturated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Imagine a seesaw perfectly balanced in a playground. If a weight is added to one side, the seesaw will tip to accomodate the added weight. Similarly, in the world of chemistry, Le Chatelier's Principle describes a system's response to maintain equilibrium after a change in conditions.

When a chemical reaction at equilibrium experiences a change, such as a change in concentration, temperature, or pressure, the equilibrium shifts to counteract the change. Considering the common ion effect, when an ion is added to a solution where its salt is already present, the reaction dips towards the left to reduce the disturbance caused by the excess ion, thus reducing the solubility of the salt.

Understanding this principle is akin to finding a path of least resistance. The system naturally adjusts, striving to maintain a delicate balance—much like a tightrope walker correcting their stance when a gust of wind threatens their poise.
Solubility Product Constant (Ksp)
The solubility product constant (Ksp) serves as an indicator of how much of a compound can dissolve in a solution before reaching the saturation point. It's a unique value for each ionic compound at a given temperature and represents the equilibrium between the solid substance and its ions in solution.

Ksp is essentially a mathematical expression of solubility for ionic compounds. For example, if we have a salt represented by the formula AB, which disassociates into its ions A+ and B-, its Ksp would be given by the expression \( Ksp = [A+][B-] \). When a common ion is added, the surplus of either A+ or B- ions influences this product; if the product exceeds the Ksp value, additional solid precipitates out, since the solution capacity to dissolve more salt is compromised.

Grasping the Ksp concept aids in predicting solubility, anticipating precipitate formation, and navigating the complexities of ionic reactions in aqueous solutions.
Equilibrium in Chemistry
Chemical equilibrium is the state of balance where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time. It doesn't imply that the reactants and products are equal in concentration but that their ratios are stable.

Envisioning a busy street where cars are moving at such a rate that the number of cars entering is equal to those leaving. Similarly, in a chemical reaction at equilibrium, the conversion of reactants to products is matched by the reverse process. The common ion effect is a perfect study of equilibrium in practice. When a common ion is introduced to a balanced system, it momentarily shifts the equilibrium. However, reactions work tirelessly to re-establish their state of balance, impacting solubilities and concentrations in the process.

Mastering chemical equilibrium is crucial for chemists to manipulate reactions for desired outcomes, whether it is to maximize the yield of product or to understand reaction dynamics.
Ionic Compounds Solubility
The solubility of ionic compounds is an intriguing affair influenced by various factors such as temperature, the dielectric constant of the solvent, and the presence of other ions in solution. Solubility is a measure of how well a substance can dissolve in a solvent to form a homogeneous mixture.

In the context of common ions, the solubility of an ionic compound in water can dramatically decrease with the introduction of a common ion. This process, underpinned by the common ion effect and Le Chatelier's Principle, shows how a balanced system reacts in an attempt to maintain equilibrium. Gaining insights into the solubility of ionic compounds empowers us to make informed decisions in fields ranging from medicine formulation to environmental chemistry.

Overall, understanding the fundamentals of how ionic compounds dissolve, how equilibrium is established, and how external factors affect solubility is fundamental for students and professionals in the scientific community.

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Most popular questions from this chapter

Describe the solubility of CaF \(_{2}\) in each solution compared to its solubility in water. a. in a 0.10 \(\mathrm{M} \mathrm{NaCl}\) solution b. in a 0.10 \(\mathrm{M}\) NaF solution c. in a 0.10 \(\mathrm{M}\) HCl solution

Without doing any calculations, determine if \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}, \mathrm{pH}>\mathrm{p} K_{\mathrm{a}},\) or \(\mathrm{pH}<\mathrm{p} K_{\mathrm{a}} .\) Assume that \(\mathrm{HA}\) is a weak monoprotic acid. a. 0.10 mol \(\mathrm{HA}\) and 0.050 \(\mathrm{mol}\) of \(\mathrm{A}^{-}\) in 1.0 \(\mathrm{L}\) of solution b. 0.10 \(\mathrm{mol} \mathrm{HA}\) and 0.150 \(\mathrm{mol}\) of \(\mathrm{A}^{-}\) in 1.0 \(\mathrm{L}\) of solution c. 0.10 mol \(\mathrm{HA}\) and 0.050 \(\mathrm{mol}\) of \(\mathrm{OH}^{-}\) in 1.0 \(\mathrm{L}\) of solution d. 0.10 \(\mathrm{mol} \mathrm{HA}\) and 0.075 \(\mathrm{mol}\) of \(\mathrm{OH}^{-}\) in 1.0 \(\mathrm{L}\) of solution

Consider the titration of a 25.0 \(\mathrm{mL}\) sample of 0.115 \(\mathrm{M} \mathrm{RbOH}\) with 0.100 \(\mathrm{M} \mathrm{HCl} .\) Determine each quantity. a. the initial pH b. the volume of added acid required to reach the equivalence point c. the pH at 5.0 \(\mathrm{mL}\) of added acid d. the pH at the equivalence point e. the pH after adding 5.0 \(\mathrm{mL}\) of acid beyond the equivalence point

Calculate the molar solubility of MX \(\left(K_{\mathrm{sp}}=1.27 \times 10^{-36}\right)\) in each liquid or solution. a. pure water b. 0.25 \(\mathrm{M} \mathrm{MCl}_{2}\) c. 0.20 \(\mathrm{M} \mathrm{Na}_{2} \mathrm{X}\)

What is an indicator? How can an indicator signal the equivalence point of a titration?

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