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Common Ion Effect

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Common Ion Effect

Quite a charge

Lithium carbonate is an essential component of lithium batteries, which tend to be longer-lasting than regular alkaline batteries. The material is obtained from lithium ores by adding CO 2 under high pressure to form the more soluble LiHCO 3 . The mixture is then depressurized to remove the carbon dioxide and the lithium carbonate precipitates out of solution.

Common Ion Effect

In a saturated solution of calcium sulfate, an equilibrium exists between the solid calcium sulfate and its ions in solution.

\text{CaSO}_4(s) \rightleftarrows \text{Ca}^{2+}(aq)+\text{SO}_4^{2-}(aq) \qquad K_{sp}=2.4 \times 10^{-5}

Suppose that some calcium nitrate were added to this saturated solution. Immediately, the concentration of the calcium ion in the solution would increase. As a result, the ion product of the [Ca 2+ ] times the [SO 4 2− ] would increase and now be greater than the K_{sp} . According to LeChâtelier’s principle, the equilibrium above would shift to the left in order to relieve the stress of the added calcium ion. Additional calcium sulfate would precipitate out of the solution until the ion product once again becomes equal to the K_{sp} . Note that in the new equilibrium the concentrations of the calcium ion and the sulfate ion would no longer be equal to each other. The calcium ion concentration would be larger than the sulfate ion concentration.

This situation describes the common ion effect. A common ion is an ion that is in common to both salts in a solution. In the above example, the common ion is Ca 2+ . The common ion effect is a decrease in the solubility of an ionic compound as a result of the addition of a common ion. Adding calcium ion to the saturated solution of calcium sulfate causes additional CaSO 4 to precipitate from the solution, lowering its solubility. The addition of a solution containing sulfate ion, such as potassium sulfate, would result in the same common ion effect.

Sample Problem: The Common Ion Effect

What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH has been added?

Step 1: List the known quantities and plan the problem .


  • K_{sp}= 3.0 \times 10^{-16} (from table in the concept Conversion of Ksp to Solubility )
  • moles of added NaOH = 0.040 mol
  • volumes of solution = 1.00 L


  • [Zn 2+ ] = ? M

Express the concentrations of the two ions relative to the variable s . The concentration of the zinc ion will be equal to s , while the concentration of the hydroxide ion will be equal to 0.040 + 2s .

Step 2: Solve .

The  K_{sp} expression can be written in terms of the variable s .


Because the value of the  K_{sp} is so small, we can make the assumption that the value of  s will be very small compared to 0.040. This simplifies the mathematics involved in solving for s .

K_{sp}&=(s)(0.040)^2=0.0016 \ s=3.0 \times 10^{-16} \\s&=\frac{K_{sp}}{\left [\text{OH}^- \right ]^2}=\frac{3.0 \times 10^{-16}}{0.0016}=1.9 \times 10^{-13} \ \text{M}

The concentration of the zinc ion is equal to  s and so [Zn 2+ ] = 1.9 × 10 -13  M.

Step 3: Think about your result .

The relatively high concentration of the common ion, OH , results in a very low concentration of zinc ion. The molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in water.


  • The common ion and common ion effect are described.
  • Calculations involving the common ion effect are described.


Work the problems at the link below:



  1. How is Le Châtelier’s principle involved in the common-ion effect?
  2. Could the common ion effect ever increase the solubility of a compound?
  3. In the sample problem, what would the effect be of adding Zn(NO 3 ) 2 ?

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