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Common Ion Effect

Le Châtelier's Principle states that if an equilibrium gets out of balance, the reaction will shift to restore the balance. If a common ion is added to a weak acid or weak base equilibrium, then the equilibrium will shift towards the reactants, in this case the weak acid or base.

Introduction

The common-ion effect is used to describe the effect on an equilibrium involving a substance that adds an ion that is a part of the equilibrium.

Example 1

Consider the lead(II) ion concentration in this saturated solution of PbCl2. The balanced reaction is

\[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \]

Assuming the concentration of dissolved lead(II) chloride is \(s\), then:

\[[Pb^{2+}] = s \]

\[[Cl^- ] = 2s\]

Put these values into the solubility product expression, and do the sum.

So the concentration of lead(II) ions in the solution is 1.62 x 10-2 M. What happens if you add some sodium chloride to this saturated solution? Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them; this is the origin of the term "common ion effect".

Look at the original equilibrium expression again:

\[ PbCl_2 \; (s) \rightleftharpoons Pb^{2+} \; (aq) + 2Cl^- \; (aq) \]

What would happen to that equilibrium if you added extra chloride ions? According to Le  Châtelier, the position of equilibrium will shift to counter what you have just done. In this case, it would tend to remove the chloride ions by making extra solid lead(II) chloride.

 


Note:  Actually, of course, the concentration of lead(II) ions in the solution is so small to start with, that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride.


The lead(II) chloride will become even less soluble - and, of course, the concentration of lead(II) ions in the solution will decrease. Something similar happens whenever you have a sparingly soluble substance. It will be less soluble in a solution which contains any ion which it has in common. This is the common ion effect.

A Simple Example

Suppose you tried to dissolve some lead(II) chloride in some \(0.100 M\) sodium chloride solution instead of in water. What would the concentration of the lead(II) ions be this time? As before, let's call the concentration of the lead(II) ions \(s\).

\[[Pb^{2+}] = s\]

Now the sum gets different. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 M coming from the sodium chloride solution.

In calculations like this, you can always assume that the concentration of the common ion is entirely due to the other solution. This makes the math a lot easier. In fact if you don't make this assumption, the math of this can become impossible to do at this level.

So we assume:

\[[Cl^- ] = 0.100 M\]

The rest of the sum looks like this:

\begin{equation} \begin{split} K_{sp}& = [Pb^{2+}][Cl^-]^2 \\ & = s \times (0.100)^2 \\ 1.7 \times 10^{-5} & = s \times 0.00100 \end{split} \end{equation}

therefore

\begin{equation} \begin{split} s & = \dfrac{1.7 \times 10^{-5}}{0.0100} \\ & = 1.7 \times 10^{-3} \, \text{M} \end{split} \end{equation}

Finally, compare that value with the simple saturated solution we started with:

Original solution:

\[Pb^{2+}] = 0.0162 \, M\]

Solution in \(0.100 \, M\, NaCl\) solution:

\[ [Pb^{2+}] = 0.0017 \, M \]

The concentration of the lead(II) ions has fallen by a factor of about \(10\). If you tried the same sum with more concentrated solutions of sodium chloride, the solubility would fall still further. Try it yourself with chloride ion concentrations of \(0.5\) and \(1.0 M\).

Common Ion Effect on Weak Acids and Bases

 Adding a common ion prevents the weak acid or weak base from ionizing as much as it would without the added common ion. The common ion effect suppresses the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.

Example 2

The common ion effect of H3O+ on the ionization of acetic acid

common ion 1.jpg

The common ion effect suppresses the ionization of a weak base by adding more of an ion that is a product of this equilibrium.

Example 3

Consider the common ion effect of OH- on the ionization of ammonia

common ion 2.jpg

Therefore, adding the common ion of hydroxide, shifts the reaction towards the left to decrease the stress (Le Châtelier's Principle), forming more reactants. This decreases the reaction quotient, for the reaction is being pushed towards the left to reach equilibrium. The equilibrium constant, Kb=1.8*10-5, does not change. The reaction is put out of balance, or equilibrium.

QA= [NH4+][OH-]/[NH3]

At first, when more hydroxide is added, the quotient is greater than the equilibrium constant. The reaction then shifts right, causing the denominator to increase, decreasing the reaction quotient and pulling towards equilibrium and causing Q to decrease towards K.

Common Ion Effect on Solubility

Adding a common ion decreases solubility, as the reaction will shift towards the left to relieve the stress of the excess product. Adding a common ion to a dissociation reaction causes the equilibrium to shift left, towards the reactants, causing precipitation.

Example 4

Consider the reaction:

\[ PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)\]

What happens to the solubility of PbCl2(s) when you add 0.1 M NaCl? 

SOLUTION

Ksp=1.7*10-5

Qsp= 1.8*10-5

Identify the common ion: Cl-

Notice: Qsp > Ksp The addition of NaCl has caused the reaction to shift out of equilibrium since there are more ions dissociated. Typically, the solve for the molarities of the following reactions you would do the following assume the solubility of PbCl2 to be equivalent to the concentration of Pb2+ produced because they are in a 1:1 ratio. 

Since Ksp for the reaction is 1.7*10-5, the overall reaction would be (s)(2s)2= 1.7*10-5. Solving for s, you would get, s= 1.62*10-2 M. The coefficient on Cl- is 2, so it is assumed that twice as much Cl- is produced as Pb2+, hence the '2s.' You use the solubility equilibrium constant to solve for the molarities of the ions at equilibrium.

Now let's go back to the common ion being added to the reaction at equilibrium. The molarity of Cl- added would be 0.1 M because Na+ and Cl- are in a 1:1 ration in the ionic salt, NaCl. Therefore, the OVERALL molarity of Cl- would be 2s + 0.1, with 2s being the contribution of the chloride ion from the dissociation of lead chloride. 

 
Qsp = [Pb2+][Cl-]2

1.8*10-5 = (s)(2s + 0.1)2

s= [Pb2+]= 1.8*10-3 M , 2s= [Cl-] = approximately 0.1 M

Notice that the molarity of Pb2+ is lower when NaCl is added. The equilibrium constant stays the same because of the increased concentration of the chloride ion. To simplify the reaction, it can be assumed that [Cl-] is approximately 0.1M since the formation of the chloride ion from the dissociation of lead chloride is so small. The reaction quotient for PbCl2 is greater than the equilibrium constant because of the added Cl-. This therefore shift the reaction left towards equilibrium, causing precipitation and lowering the current solubility of the reaction. Overall, the solubility of the reaction decreases with the added sodium chloride.

References

  1. Harwood, William S., F. G. Herring, Jeffry D. Madura, and Ralph H. Petrucci. General Chemistry Principles and Modern Applications. 9th ed. New Jersey: Prentice Hall, 2007. 670-a32.
  2. "Common Ion Effect." Common Ion Effect. Web. May 2012. <http://www.cartage.org.lb/en/themes/...nIonEffect.htm>.
  3. "Solubility Equilibrium." ThinkQuest. Oracle Foundation. Web. May 2012. <http://library.thinkquest.org/C00666...olubility.html>.

Contributors

  • Emmellin Tung, Mahtab Danai (UCD)
  • Jim Clark (ChemGuide)

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