A solution of Ca(OH) 2 at equilibrium contains = 0.0108 M, = 0.0216 M, and some solid salt. The example below, involving the dissolution of a slightly soluble salt, illustrates this important concept.Įxample 6.1.3 – Common Ion Added to a System at EquilibriumĬalcium hydroxide is a sparingly soluble salt that exists in equilibrium in aqueous solution with its ions. Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
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In general, we can conclude that adding a common ion prevents a weak acid (or similarly, a weak base) from ionizing as much as it would without the added common ion. According to Le Chatelier’s Principle, increased means that the weak acid equilibrium will shift towards the reactants. Thus, the presence of the sodium salt increases the concentration of conjugate base. The A – ion is the common ion to both chemical equations, but note their differences: the first is an equilibrium but the second goes to completion. If we were to add a second substance, for example, the sodium salt of the conjugate base (NaA) to this solution, then this soluble salt would increase the concentration of the A – ion in solution: Consider the following chemical equation, representing the ionization of a generic weak acid: Le Châtelier’s Principle states that if an equilibrium becomes unbalanced, the reaction will shift in order to restore the balance. But what happens if we involve substances that only partially ionize in solution? What effect would a common ion have on an equilibrium? In the above examples, since all of the compounds undergo 100% ionization, we could use simple stoichiometry to find the concentrations of the common ion. John poured 10.0 mL of 0.10 M NaCl, 10.0 mL of 0.10 M KOH, and 5.0 mL of 0.20 M HCl solutions together and then he made the total volume to be 100.0 mL. Therefore, the total amount of chloride ion in solution is: Therefore, due to the conservation of ions, we have:Įach compound produces chloride ions however, we note that calcium chloride produces 2 mol of Cl – ions for every mol of CaCl 2 dissolved.
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Using stoichiometry, we see that each mol of dissolved compound produces a mol of cations. What are, ,, and in a solution containing 0.10 M each of NaCl, CaCl 2, and HCl?Īll three species ionize completely in water:ĬaCl 2 ( s) → Ca 2+ ( aq ) + 2Cl – ( aq ) You may recognize the above equivalence as an application of the concepts of charge balance and mass balance.Įxample 6.1.1 – Concentrations of Common Ions Since both salts ionize completely in solution, we can use stoichiometry to express the concentrations of the dissolved ions in solution in the following equation: In a system containing NaCl and KCl, the Cl – ions are common ions. When NaCl and KCl, both very soluble salts, are dissolved in the same solution, the Cl − ions are common to both salts. Contributions from all salts must be included in the calculation of concentration of the common ion.
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If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. If several salts are present in a system, they all ionize in the solution. The common ion effect is used to describe the effect on an existing equilibrium by the addition of a second substance that contains an ion common to the equilibrium.