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The common ion effect

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Intros
Lessons
  1. What is the common ion effect?
  2. Changing the solubility equilibrium.
  3. How do we decrease solubility of a compound?
  4. How do we increase solubility of a compound?
  5. Changing solubility: Worked example (MgSO4).
  6. pH and solubility.
  7. (AP) Entropy of dissolution.
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Examples
Lessons
  1. Recall the use of the common-ion effect to change the solubility of a saturated solution.
    A student has a saturated solution of AgCl, which is in equilibrium with its ions.

    AgCl (s)  ⇌ \, \rightleftharpoons \, Ag+ (aq) + Cl- (aq)

    What will adding the following salts to the saturated solution do to the solubility equilibrium?
    Explain your answer.
    1. Pb(NO3)2
    2. Na2S
    3. AgNO3
    4. KCl
      1. Apply the common ion effect to suggest ways of changing the solubility of a saturated solution.
        A student has a saturated solution of lead chloride, with the precipitate in equilibrium with its aqueous ions.

        PbCl2  ⇌ \, \rightleftharpoons \, Pb2+ (aq) + 2Cl- (aq)

        What compounds could be added to have the following effects on the PbCl2 solution?
        1. Increase solubility.
        2. Decrease solubility.
      Topic Notes
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      Introduction to the Common Ion Effect

      Welcome to our exploration of the common ion effect, a fascinating concept in chemistry that plays a crucial role in understanding solubility and equilibrium. This phenomenon occurs when a compound is added to a solution containing one of its ions, affecting the solubility equilibrium. Our introduction video will guide you through this concept, making it easier to grasp its significance in chemical reactions. As your math tutor, I'm excited to help you understand how the common ion effect influences the solubility of compounds and shifts chemical reactions. You'll discover its applications in various fields, from environmental science to pharmaceutical research. By the end of this lesson, you'll have a solid foundation in how the common ion effect impacts solubility and equilibrium constants. So, let's dive in and unravel the mysteries of this important chemical principle together!

      Understanding the Common Ion Effect

      Introduction to the Common Ion Effect

      The common ion effect is a phenomenon in chemistry that plays a crucial role in understanding solubility equilibrium. This effect occurs when an ion that is common to two solutes is added to a solution, affecting the solubility of one or both solutes. To comprehend this concept fully, we'll explore it using the example of calcium hydroxide, as demonstrated in the video.

      Calcium Hydroxide: A Common Ion Effect Example

      Calcium hydroxide (Ca(OH)) is a sparingly soluble compound that dissociates in water to form calcium ions (Ca²) and hydroxide ions (OH). In a saturated solution of calcium hydroxide, there exists an equilibrium between the solid Ca(OH) and its ions in solution. The solubility product constant (Ksp) governs this equilibrium.

      When we add a common ion, such as calcium ions from calcium chloride (CaCl) or hydroxide ions from sodium hydroxide (NaOH), to this saturated solution, we observe the common ion effect in action. The addition of these common ions shifts the equilibrium, affecting the solubility of calcium hydroxide.

      The Impact on Solubility Equilibrium

      In a saturated solution of calcium hydroxide, adding extra calcium ions or hydroxide ions disturbs the existing equilibrium. According to Le Chatelier's principle, which we'll discuss in more detail shortly, the system will respond to this disturbance by shifting the equilibrium to counteract the change.

      For instance, if we add calcium ions to the saturated Ca(OH) solution, the equilibrium shifts towards the formation of more solid Ca(OH). This shift reduces the solubility of calcium hydroxide in the solution. Similarly, adding hydroxide ions would have the same effect, pushing the equilibrium towards the solid phase and decreasing solubility.

      Le Chatelier's Principle and the Common Ion Effect

      Le Chatelier's principle is fundamental to understanding the common ion effect. This principle states that when a system at equilibrium is subjected to a change, the system will adjust to partially counteract that change and establish a new equilibrium. In the context of the common ion effect, Le Chatelier's principle explains why adding a common ion reduces the solubility of a sparingly soluble salt.

      When we add a common ion to a saturated solution, we increase the concentration of that ion in the solution. According to Le Chatelier's principle, the system will respond by shifting the equilibrium in a direction that reduces the concentration of the added ion. In our calcium hydroxide example, this means the equilibrium shifts towards the formation of more solid Ca(OH), effectively reducing its solubility.

      Quantitative Analysis of the Common Ion Effect

      To understand the common ion effect quantitatively, we can use the solubility product constant (Ksp). For calcium hydroxide, the Ksp expression is:

      Ksp = [Ca²][OH]²

      When we add a common ion, the concentration of that ion increases, but the Ksp value remains constant at a given temperature. To maintain the Ksp, the concentration of the other ion must decrease, resulting in a reduction of the overall solubility of the compound.

      Applications of the Common Ion Effect

      The common ion effect has numerous practical applications in chemistry and related fields. It's used in analytical chemistry for selective precipitation of ions, in environmental science for water treatment processes, and in the pharmaceutical industry for controlling drug solubility and absorption. Understanding this effect is crucial for predicting and controlling the behavior of ions in solution across various scientific and industrial applications.

      Conclusion: The Significance of the Common Ion Effect

      The common ion effect, exemplified by the behavior of calcium hydroxide in solution, is a powerful demonstration of how chemical systems respond to changes in their environment. By applying Le Chatelier's principle to solubility equilibria, we can

      Decreasing Solubility Using the Common Ion Effect

      The common ion effect is a powerful tool for decreasing solubility in chemical solutions. This phenomenon occurs when a soluble salt with an ion in common with a sparingly soluble salt is added to a solution, resulting in a decrease in the solubility of the sparingly soluble salt. To understand this concept better, let's explore the magnesium sulfate example from the video and delve into the process, impact on equilibrium, and real-world applications.

      In the case of magnesium sulfate (MgSO4), a moderately soluble salt, we can decrease its solubility by adding a soluble salt that shares a common ion. The dissociation equation for magnesium sulfate in water is:

      MgSO4 (s) Mg2+ (aq) + SO42- (aq)

      To decrease the solubility of magnesium sulfate, we can add a soluble salt containing either Mg2+ or SO42- ions. For instance, if we add sodium sulfate (Na2SO4), a highly soluble salt, to the solution, it will introduce additional sulfate ions. The dissociation equation for sodium sulfate is:

      Na2SO4 (s) 2Na+ (aq) + SO42- (aq)

      The introduction of extra sulfate ions from sodium sulfate shifts the equilibrium of the magnesium sulfate dissociation to the left, according to Le Chatelier's principle. This principle states that when a system at equilibrium is disturbed, it will adjust to counteract the change. In this case, the increase in sulfate ion concentration causes the system to favor the formation of solid magnesium sulfate, effectively decreasing its solubility.

      The common ion effect on solubility can be quantified using the solubility product constant (Ksp). For magnesium sulfate, the Ksp expression is:

      Ksp = [Mg2+][SO42-]

      When additional sulfate ions are introduced, the concentration of SO42- increases, causing a decrease in the concentration of Mg2+ to maintain the constant Ksp value. This results in the precipitation of magnesium sulfate from the solution.

      The common ion effect has numerous practical applications and real-world examples. In water treatment, it is used to remove hardness-causing ions like calcium and magnesium. By adding sodium carbonate (Na2CO3) to hard water, the common carbonate ion causes the precipitation of calcium carbonate (CaCO3) and magnesium carbonate (MgCO3), effectively softening the water.

      In the field of dentistry, the common ion effect is utilized to prevent tooth decay. Fluoride-containing toothpaste introduces fluoride ions, which can replace hydroxide ions in tooth enamel. This substitution forms fluorapatite, a compound more resistant to acid attacks than the original hydroxyapatite, thus reducing tooth decay.

      The common ion effect also plays a crucial role in environmental science. In natural water systems, the solubility of calcium carbonate (limestone) is influenced by the presence of carbon dioxide. As CO2 levels in the atmosphere increase, more carbonic acid forms in water bodies, leading to an increase in carbonate ions. This common ion effect can impact the formation and dissolution of limestone formations in caves and marine environments.

      In the pharmaceutical industry, the common ion effect is employed in drug formulation and delivery. By manipulating the solubility of certain drugs through the addition of common ions, pharmacists can control the rate of drug release in the body, enhancing the effectiveness of medications.

      The mining industry also benefits from the common ion effect. In the extraction of certain metals, such as silver, the addition of common ions can help precipitate the desired metal from solution, improving the efficiency of the extraction process.

      Understanding and applying the common ion effect is crucial in various scientific and industrial processes. By manipulating the solubility of compounds through the introduction of common ions, we can control chemical equilibria, precipitate desired substances, and optimize numerous practical applications across different fields. This phenomenon demonstrates the intricate balance of chemical equilibria and the importance of the solubility product constant in predicting and controlling solubility.

      Increasing Solubility and Manipulating Equilibrium

      Understanding how to increase solubility by manipulating equilibrium is crucial in various chemical processes. This concept is fundamental in fields ranging from pharmaceuticals to environmental science. Let's explore the methods to enhance solubility and the role of equilibrium manipulation in this process.

      Methods to Increase Solubility

      One of the primary ways to increase solubility is by manipulating the chemical equilibrium of a solution. This can be achieved through several methods:

      • Common Ion Effect: Adding a common ion can decrease solubility due to the shift in equilibrium.
      • pH Adjustment: Changing the pH of a solution can significantly affect the solubility of certain compounds.
      • Temperature Manipulation: Generally, increasing temperature enhances solubility for most solids in liquids.
      • Pressure Changes: For gases, increasing pressure typically increases solubility in liquids.

      Example: Sodium Phosphate and Magnesium Sulfate

      A classic example of manipulating equilibrium to increase solubility is the addition of sodium phosphate to a saturated magnesium sulfate solution. In this scenario:

      1. The saturated magnesium sulfate solution is at equilibrium.
      2. Adding sodium phosphate introduces phosphate ions (PO4^3-).
      3. These phosphate ions react with magnesium ions (Mg^2+) to form magnesium phosphate.
      4. As magnesium ions are removed from the solution, more magnesium sulfate dissolves to restore equilibrium.
      5. This process continues, effectively increasing the solubility of magnesium sulfate.

      The Role of Spectator Ions

      Spectator ions play a crucial role in understanding solubility and equilibrium. These are ions that do not participate directly in a reaction but are present in the solution. Key points about spectator ions include:

      • They do not affect the chemical reaction but contribute to the overall ionic strength of the solution.
      • In the context of solubility, spectator ions can influence the solubility product constant (Ksp).
      • Understanding which ions are spectators helps in predicting reaction outcomes and solubility changes.

      Using Solubility Tables Effectively

      Solubility tables are invaluable tools for predicting and understanding the behavior of various compounds in solution. Here's how to use them effectively:

      1. Identify the Compound: Locate your compound of interest in the table.
      2. Check Solubility Classification: Tables often use terms like "soluble," "slightly soluble," or "insoluble."
      3. Note Exceptions: Pay attention to any exceptions or special conditions listed.
      4. Consider Temperature: Some tables provide solubility data at different temperatures.
      5. Use Ksp Values: For more precise calculations, use the solubility product constant (Ksp) if provided.

      By effectively using solubility tables, you can:

      • Predict precipitation reactions
      • Estimate the solubility of compounds under various conditions
      • Design experiments to manipulate solubility
      • Understand the behavior of ions in solution

      Practical Applications

      Understanding and manipulating solubility has numerous practical applications:

      • Pharmaceutical Industry: Enhancing drug solubility for better absorption

      pH Effects on Solubility and the Common Ion Effect

      The solubility of compounds in aqueous solutions is a crucial concept in chemistry, and understanding how pH changes can affect solubility is essential for many chemical processes. This relationship between pH and solubility becomes even more intricate when considering the common ion effect. Let's explore these concepts in detail, using specific examples to illustrate the principles at work.

      One of the most illustrative examples of pH sensitivity in solubility is calcium hydroxide, Ca(OH). This compound, also known as slaked lime, demonstrates a remarkable sensitivity to pH changes. In its natural state, calcium hydroxide has limited solubility in water. However, when the pH of the solution is lowered (becoming more acidic), the solubility of calcium hydroxide increases dramatically.

      This increase in solubility occurs because the hydroxide ions (OH) from the calcium hydroxide react with the added hydrogen ions (H) from the acid. This reaction forms water molecules, effectively removing hydroxide ions from the solution. As a result, more calcium hydroxide dissolves to replace the hydroxide ions, maintaining the equilibrium described by its solubility product constant (Ksp).

      The equation for this process can be written as:

      Ca(OH) + 2H Ca² + 2HO

      This pH sensitivity is a characteristic of many hydroxides and other bases. It's important to note that the reverse is also true: increasing the pH (making the solution more basic) can decrease the solubility of calcium hydroxide.

      In contrast, some compounds show little to no sensitivity to pH changes. Calcium chloride (CaCl) is an excellent example of this. The solubility of calcium chloride remains relatively constant across a wide range of pH values. This is because neither calcium ions (Ca²) nor chloride ions (Cl) participate in acid-base reactions under normal conditions. The dissociation of calcium chloride in water can be represented as:

      CaCl Ca² + 2Cl

      This dissociation is not significantly affected by the presence of additional H or OH ions in the solution, making calcium chloride's solubility largely independent of pH.

      The role of strong acids and solubility in these processes is crucial to understanding pH effects on solubility. Strong acids, such as hydrochloric acid (HCl), completely dissociate in water, providing a high concentration of H ions. These H ions can dramatically affect the solubility of pH-sensitive compounds like calcium hydroxide.

      Weak acids, on the other hand, only partially dissociate in water. Acetic acid (CHCOOH), for example, exists in an equilibrium with its ions:

      CHCOOH CHCOO + H

      The partial dissociation of weak acids and solubility means they have a less dramatic effect on pH and, consequently, on the solubility of pH-sensitive compounds. However, weak acids can still significantly influence solubility, especially in concentrated solutions or when dealing with compounds that are particularly sensitive to pH changes.

      The common ion effect further complicates the relationship between pH and solubility. This effect occurs when a compound is dissolved in a solution that already contains one of its ions. For example, if calcium chloride is added to a solution already containing chloride ions (from another source), the solubility of calcium chloride will decrease due to the common ion effect.

      In the context of pH and solubility, the common ion effect can be observed when adding a salt that shares an ion with the dissolving compound. For instance, adding sodium hydroxide (NaOH) to a saturated solution of calcium hydroxide will decrease the solubility of calcium hydroxide due to the additional hydroxide ions introduced.

      Understanding these interactions between pH, solubility, and the common ion effect is crucial in various fields, including environmental science, wastewater treatment, and pharmaceuticals.

      Thermodynamics of Dissolving Processes

      The process of dissolving substances in solvents is fundamentally governed by thermodynamic principles, with enthalpy and entropy playing crucial roles. Understanding these factors is essential for comprehending the behavior of solutions and predicting solubility. This section delves into the thermodynamic aspects of dissolving processes, exploring the interplay between enthalpy, entropy, and Gibbs free energy.

      Enthalpy in Dissolving Processes

      Enthalpy, denoted as H, represents the heat content of a system. In the context of dissolving processes, enthalpy changes (ΔH) indicate whether heat is absorbed or released during dissolution. The enthalpy of solution is the sum of three main components: the energy required to break solute-solute interactions, the energy needed to create spaces in the solvent for solute particles, and the energy released when new solute-solvent interactions form. The net result determines whether the process is endothermic (heat-absorbing) or exothermic (heat-releasing).

      Entropy and Its Role in Dissolution

      Entropy, symbolized by S, is a measure of the disorder or randomness in a system. In dissolving processes, entropy changes (ΔS) are typically positive, as the dissolution of a solid into a liquid generally increases the overall disorder of the system. This increase in entropy is a driving force for many dissolving processes, even when the enthalpy change is unfavorable.

      The Gibbs Free Energy Equation

      The Gibbs free energy (G) is a thermodynamic potential that combines the effects of enthalpy and entropy, providing a comprehensive measure of the spontaneity of a process. The equation for Gibbs free energy change is:

      ΔG = ΔH - TΔS

      Where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the absolute temperature, and ΔS is the change in entropy. A negative ΔG indicates a spontaneous process, while a positive ΔG indicates a non-spontaneous process. This equation is fundamental in predicting solubility and understanding the balance between enthalpy and entropy in dissolving processes.

      Sodium Chloride: An Example of Entropy-Driven Dissolution

      The dissolution of sodium chloride (NaCl) in water serves as an excellent example of an entropy-driven process. When NaCl dissolves, the highly ordered crystal structure breaks apart, and the ions become hydrated by water molecules. Although breaking the ionic bonds requires energy (positive ΔH), the significant increase in disorder (positive ΔS) as the ions disperse throughout the solution outweighs the unfavorable enthalpy change. The result is a negative ΔG, making the process spontaneous at room temperature.

      Endothermic vs. Exothermic Dissolving Processes

      Dissolving processes can be classified as endothermic or exothermic based on their enthalpy changes:

      Endothermic Dissolving Processes

      In endothermic dissolving processes, heat is absorbed from the surroundings, resulting in a positive ΔH. Examples include the dissolution of ammonium chloride or potassium nitrate in water. These processes often result in a cooling effect on the surrounding solution. Despite the unfavorable enthalpy change, many endothermic dissolving processes occur spontaneously due to a significant increase in entropy.

      Exothermic Dissolving Processes

      Exothermic dissolving processes release heat to the surroundings, characterized by a negative ΔH. The dissolution of sulfuric acid in water or the mixing of ethanol and water are examples of exothermic processes. These reactions often result in an increase in the temperature of the solution. Exothermic dissolving processes are generally more likely to

      Conclusion: Mastering the Common Ion Effect

      The common ion effect is a crucial concept in chemistry, significantly impacting solubility and equilibrium. This phenomenon occurs when a common ion is added to a solution, shifting the equilibrium and reducing solubility. Understanding its relationship with pH and thermodynamics is essential for predicting chemical behaviors. The common ion effect influences various processes, from buffer solutions to precipitation reactions. It's closely tied to Le Chatelier's principle and plays a vital role in maintaining pH balance in biological systems. Mastering this concept enables chemists to manipulate solubility in practical applications, such as water treatment and pharmaceutical formulations. By grasping the common ion effect, students can better comprehend complex chemical systems and their dynamics. We encourage you to apply this knowledge to real-world problems and continue exploring related topics like Ksp and buffer solutions. Remember, a solid understanding of the common ion effect is fundamental for success in advanced chemistry studies and research.

      What is the common ion effect?

      What is the common ion effect? Changing the solubility equilibrium.

      Step 1: Introduction to the Common Ion Effect

      In this lesson, we will explore the common ion effect. The primary objective is to understand how the common ion effect can be used to decrease the solubility of a saturated solution. A saturated solution is one where the maximum amount of a substance has dissolved, and no more can dissolve at a given temperature, reaching an equilibrium point.

      Step 2: Understanding Solubility Equilibrium

      Solubility equilibrium refers to the balance between the dissolved ions in a solution and the undissolved solid phase. For example, consider a salt like calcium chloride (CaCl2). In a saturated solution, there is a dynamic equilibrium between the dissolved ions (Ca2+ and Cl-) and the undissolved solid. This equilibrium can be represented as:

      CaCl2 (s) Ca2+ (aq) + 2Cl- (aq)

      At equilibrium, the rate of dissolution (forward reaction) is equal to the rate of crystallization (backward reaction).

      Step 3: The Role of Common Ions

      The common ion effect involves adding a substance that contains an ion already present in the equilibrium. For instance, if we add magnesium chloride (MgCl2) to the saturated solution of CaCl2, the chloride ion (Cl-) is common to both salts. The addition of MgCl2 increases the concentration of Cl- ions in the solution, which disturbs the equilibrium.

      Step 4: Shifting the Equilibrium

      According to Le Chatelier's principle, the system will adjust to counteract the disturbance caused by the added common ion. In this case, the increased concentration of Cl- ions will shift the equilibrium to the left, favoring the formation of the undissolved solid (CaCl2). This results in a decrease in the solubility of CaCl2 in the solution.

      Step 5: Practical Applications

      The common ion effect is a useful tool in various applications, such as controlling the solubility of salts in chemical reactions, purifying compounds, and even in biological systems where ion concentrations need to be regulated. By understanding and manipulating the common ion effect, chemists can achieve desired outcomes without changing the temperature of the system.

      Step 6: Conclusion

      In summary, the common ion effect is a phenomenon where the addition of a common ion to a saturated solution affects the solubility equilibrium. By increasing the concentration of a common ion, the solubility of the original substance decreases, leading to the formation of more undissolved solid. This principle is widely used in chemistry to manipulate solubility and achieve specific results in various processes.

      FAQs

      Here are some frequently asked questions about the common ion effect:

      1. What happens in the common ion effect?

      The common ion effect occurs when a soluble compound is added to a solution containing one of its ions, resulting in a decrease in the solubility of a sparingly soluble salt with that same ion. This effect shifts the equilibrium towards the formation of the solid, reducing overall solubility.

      2. Does the common ion effect increase Ksp?

      No, the common ion effect does not increase the solubility product constant (Ksp). The Ksp remains constant at a given temperature. However, the common ion effect decreases the solubility of the sparingly soluble salt while maintaining the same Ksp value.

      3. What is a real-world example of the common ion effect?

      A common real-world example is the use of fluoride in toothpaste. The presence of fluoride ions in toothpaste creates a common ion effect with the hydroxyapatite in tooth enamel, reducing its solubility and helping to prevent tooth decay.

      4. How does the common ion effect relate to Le Chatelier's principle?

      The common ion effect is an application of Le Chatelier's principle. When a common ion is added to a solution, the equilibrium shifts to counteract this change by favoring the formation of the less soluble compound, thus reducing its solubility in accordance with Le Chatelier's principle.

      5. What is the difference between the solubility product principle and the common ion effect?

      The solubility product principle describes the equilibrium between a sparingly soluble salt and its ions in a saturated solution. The common ion effect, on the other hand, is a phenomenon that occurs when additional common ions are introduced, affecting this equilibrium and reducing the salt's solubility while maintaining the same solubility product constant.

      Prerequisite Topics for Understanding the Common Ion Effect

      To fully grasp the concept of the common ion effect in chemistry, it's crucial to have a solid foundation in related topics. Two key prerequisite subjects that play a significant role in understanding this phenomenon are the solubility product constant and buffer solutions.

      The common ion effect is a principle that describes how the solubility of a slightly soluble ionic compound is affected by the presence of a common ion in solution. This concept is intimately connected to the solubility product constant, which quantifies the extent to which a compound dissolves in water. Understanding the solubility product is essential because it provides the mathematical framework for predicting how the common ion effect will influence solubility.

      When studying the common ion effect, you'll often encounter scenarios involving equilibrium shifts. This is where knowledge of buffer solutions becomes invaluable. Buffer solutions resist changes in pH when small amounts of acid or base are added, and they operate on principles similar to those governing the common ion effect. Both concepts involve the manipulation of equilibrium to maintain certain conditions in a solution.

      The relationship between the common ion effect and solubility product constant is particularly important. When a common ion is introduced to a solution, it shifts the equilibrium according to Le Chatelier's principle, which is a fundamental concept you'll encounter when studying both topics. This shift typically results in decreased solubility of the ionic compound, a phenomenon that can be quantitatively analyzed using the solubility product constant.

      Similarly, understanding buffer solutions provides insight into how systems resist changes in concentration, which is analogous to how the common ion effect influences solubility. In both cases, the presence of certain ions in solution affects the behavior of the system as a whole. The principles of equilibrium that govern buffer action are closely related to those that explain the common ion effect.

      By mastering these prerequisite topics, you'll be better equipped to understand the nuances of the common ion effect. You'll be able to predict how adding a common ion will affect solubility, calculate the magnitude of these effects, and understand their practical applications in various chemical processes. Moreover, this knowledge forms a crucial foundation for more advanced topics in chemistry, such as precipitation reactions, selective precipitation, and complex ion formation.

      In conclusion, a thorough understanding of the solubility product constant and buffer solutions is essential for mastering the common ion effect. These topics provide the conceptual and mathematical tools necessary to analyze and predict the behavior of ionic compounds in solution, making them indispensable prerequisites for any student of chemistry aiming to comprehend this important phenomenon.

      In this lesson, we will learn:

      • How to use the common ion effect to decrease the solubility of a saturated solution.
      • How to use a solubility table to suggest ways to increase the solubility of a saturated solution.
      • How pH and thermodynamics influence the solubility of salts.

      Notes:

      • Whenever we talk about a compound with low solubility or a saturated solution, we should always write the equilibrium that has been created. For any salt MmXx.

        MmXx (s)  ⇌ \, \rightleftharpoons \, Mx+(aq) + Xm-(aq)

        This can be broken up into the individual forward and reverse reactions. The forward reaction is the dissolving process that changes the substance from solid to aqueous state:

        MmXx (s)  \, →  \, Mx+(aq) +  \, Xm-(aq)

        The reverse reaction is the crystallization process that changes the substance from aqueous back to the solid state, as a precipitate:

        Mx+ (aq) +  \, Xm-(aq)  \, →  \, Mm Xx (s)


        This is useful for when we want to reduce or increase the solubility of some compounds. Even though Ksp cannot change, there are ways to change solubility in a given solution without changing the temperature!

      • The common ion effect is a way to change the solubility of a compound by adding a soluble salt that has an ion in common with the compound you are trying to change the solubility of.
        Like any process at equilibrium, the common ion effect is governed by Le Chatelier’s principle. This is important in predicting how the solubility will change.
        • For example, the salt calcium hydroxide, Ca(OH)2, when saturated has the equilibrium:

          • Ca(OH)2 (s)  ⇌ \, \rightleftharpoons \, Ca2+ (aq) + 2OH- (aq)

          To decrease the solubility of Ca(OH)2, we need the equilibrium to shift to the left and favor the precipitate. To do this, we add a compound that will dissolve to produce more Ca2+ or OH- \,ions in solution. Applying Le Chatelier’s principle, increasing [Ca2+] or [OH-] will cause the system to shift away from them, which is towards more Ca(OH)2 (s).
          Using a solubility table and our Predicting the solubility of salts lesson recall that compounds with a nitrate (NO3-) anion are highly soluble in water, If we added some calcium nitrate (Ca(NO3)2) to the solution, the following happens:

          Ca(NO3)2 (s)  \, →  \, Ca2+ (aq) + 2NO3- (aq)

          As it is highly soluble, this is not an equilibrium, it is a straightforward dissolving process. But the extra Ca2+ (aq) will now disturb the Ca(OH)2 equilibrium:

          Ca(OH)2 (s)  ⇌ \, \rightleftharpoons \, Ca2+ (aq) + 2OH- (aq)

          To maintain the Ksp concentration of Ca(OH)2 at equilibrium, the equilibrium must shift to the left (favoring the crystallization reaction) and makes more Ca(OH)2 (s). When this happens, we will have decreased the solubility of Ca(OH)2 as more of it is in precipitate form now.

        • To increase the solubility of Ca(OH)2, we need to do the opposite; the equilibrium must shift to the right and favor the dissolving reaction. To do this, we need to add a compound that will reduce the amount of Ca2+ or OH- ions in solution by precipitating one of the ions out of solution.
          Using a solubility table, we can see that compounds containing sodium ions (Na+) will dissolve in water, and that calcium carbonate, CaCO3, has low solubility. If we added sodium carbonate, Na2CO3, we would begin to precipitate CaCO3 (s) while reducing the Ca2+ (aq) concentration. The Ca(OH)2 equilibrium will respond by shifting to the right to produce more Ca2+ ions. Remember that Na+ like NO3-  \, is a spectator ion and will not form a precipitate!
          The same could be done with the OH- ions dissolved; adding Pb(NO3)2 to the solution would cause Pb(OH)2 (s) to precipitate and the equilibrium will shift to the right. This would produce more OH- ions to rebalance those that were lost when Pb(OH)2 (s) started forming.

      • As the above hopefully shows, the common ion effect is governed by Le Chatelier’s principle.
        Le Chatelier’s principle also applies to acid-base equilibria so the solubility of many salts is pH sensitive; if one of the aqueous ions is a conjugate pair with a weak acid or base, a change in pH will affect its concentration and the solubility equilibrium of the salt.
        • For example, Ca(OH)2 has the solubility equilibrium:

          • Ca(OH)2 (s)  ⇌  \, \rightleftharpoons \, Ca2+(aq) + 2OH-(aq)

          Where Ksp = 5.02 * 10-6 = [Ca2+(aq)][OH-(aq)]2

          Adding aqueous acid (H+) to a saturated solution of Ca(OH)2 to lower the pH will form water, using the hydroxide ions from Ca(OH)2 that dissolved.

          H+ (aq) + OH- (aq)  ⇌  \, \rightleftharpoons \, H2O (l)

          This removal of OH- (aq) as water forms means the solubility equilibrium is disturbed. There is less OH (aq), so according to Le Chatelier’s principle the equilibrium will shift towards the aqueous products to restore the original OH- that was lost. To do this, more Ca(OH)2 (s) must dissolve.

          You can combine the two equations above which shows the process in one step.

          Ca(OH)2 (s) + 2H+ (aq) + 2OH- (aq)  ⇌  \, \rightleftharpoons \, Ca2+ (aq) + 2OH- (aq) + 2H2O (l)

          With the equations combine you can see the process clearly: making the solution more acidic will increase the solubility of Ca(OH)2 by forming water.
        • If you have a solubility equilibrium with an aqueous ion that is the conjugate base of a strong acid (chlorides, bromides, sulfates, nitrates), pH has no effect on solubility.
          This is because adding aqueous acid or base will not lead to the strong acid/base being formed; strong acids and bases have 100% dissociation.
          For example, calcium chloride, CaCl2, has the following solubility equilibrium:

          • CaCl2 (s)  ⇌  \, \rightleftharpoons \, Ca2+ (aq) + 2Cl- (aq)

          Where Ksp = [Ca2+ (aq)][Cl-]2

          Adding H+ (aq) to a saturated solution of CaCl2 containing Cl- ions will not cause HCl to form because as a strong acid, it stays (virtually) 100% dissociated. So the Cl- (aq) will remain aqueous chloride ions, and the solubility equilibrium is undisturbed.

      • (AP CHEMISTRY ONLY)
        When a substance dissolves, its solubility is influenced by several thermodynamic factors. In Entropy and Gibbs free energy, we looked at the thermodynamics behind a chemical change – is it feasible or not? Recall that the Gibbs free energy equation has two thermodynamic factors involved:

        • ΔG\Delta G = ΔH\Delta H - TΔST \Delta S

        • ΔH\Delta H is the enthalpy change and ΔS\Delta S is the entropy change. A process is feasible provided that ΔG\Delta G is negative overall.

        There are many examples where a species dissolves (ΔG\Delta G is negative) but the temperature of the solution decreases (ΔH\Delta H is positive).
        The explanation for this is that a large enough positive entropy change offsets a positive enthalpy change to make an endothermic process feasible. When a substance dissolves, make a note of the chemical changes occurring.
        With sodium chloride dissolving in water, the following three chemical changes are occurring:
        • The NaCl lattice breaks up, so the Na+ Cl- ionic bonds are overcome.
        • Some solvent-solvent interactions are broken up, as ions will occupy some space that solvent molecules previously did. In a water solvent, this will be some hydrogen bonds being disrupted and reorganized.
        • As the ionic lattice is broken up, the resulting aqueous Na+ and Cl- ions will make many new polar interactions with the water solvent.

        A positive enthalpy change tells us that overall, more energy was needed to break the existing bonds/interactions than was released by forming new ion-solvent interactions.
        Does this surprise you? Dissolving an ionic substance breaks strong ionic bonds!
        Still, the large entropy increase of a lattice breaking into aqueous ions offset this.
        Every ionic species will have a different balance of ionic bonds broken, solvent-ion interactions formed and an associated entropy change. This is what gives rise to the variety in solubility we find in ionic substances.

      • In short, the common ion effect and increasing solubility works by these principles:
        • When a solution is saturated and the Ksp equilibrium is established, changing the ion concentrations will change the equilibrium position which, here, is the compound’s solubility.
        • To decrease the solubility of a saturated solution of a compound, add a soluble salt with an ion in common to it. Use a solubility table to find a salt and remember spectator ions make soluble salts!
        • To increase the solubility, adding soluble salts with an ion that will form compounds of low solubility with one of the aqueous ions at equilibrium. For example, to a saturated AgCl solution, adding soluble Pb(NO3)2 which will precipitate PbCl2 with the Cl- ions in solution.