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Mastering the Common Ion Effect in Chemistry
Dive into the fascinating world of the common ion effect and its impact on solubility and equilibrium. Our expert-led video lessons will guide you through this crucial chemistry concept, from theory to real-world applications.

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Now Playing:The common ion effect – Example 0a
Intros
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  1. What is the common ion effect?
  2. What is the common ion effect?
    Changing the solubility equilibrium.
  3. What is the common ion effect?
    How do we decrease solubility of a compound?
Examples
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  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
    Solubility and ion concentration
    Jump to:NotesConceptFAQsPrerequisites
    Notes

    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.
    Concept

    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!

    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.

    Prerequisites

    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.