8.2: The Common-Ion Effect

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Learning Objectives

Recognize common ions from various salts.
Calculate solubility of solutions involving common ions.
Apply the common-ion effect to the pH effect on the solubility of weakly acidic or basic salts.

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. The concentrations of ions of dissolved salts are described by their solubility products (K_sp). If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, when \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common to both salts.

\[\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}\]

\[\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}\]

When \(\ce{KCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{KCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[K+]}\). Instead, the ions will have the following relationship:

\[\mathrm{[Na^+] + [K^+] = [Cl^-]} \label{1}\]

Consideration of charge balance or mass balance or both leads to the same conclusion. The following examples show how the concentration of the common ion is calculated.

Example \(\PageIndex{1}\)

What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)?

SOLUTION

Each of these salts is considered fully soluble, so they will dissociate in solution according to the following equilibria:

\[\mathrm{NaCl \rightleftharpoons Na^+ + Cl^-}\]

\[\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + 2Cl^-}\]

\[\mathrm{HCl \rightleftharpoons H^+ + Cl^-}\]

Due to the conservation of ions, we have

\[\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M} \nonumber \]

but for the common ion

\(\begin{alignat}{3}
\nonumber &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\\
\nonumber & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\\
\nonumber & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\\
\nonumber & &&= && &&\mathrm{\:0.40\: M}
\nonumber \end{alignat}\)

Exercise \(\PageIndex{1}\)

John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution?

Solution

\[\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M} \nonumber\]

The Common Ion Effect and Solubility

The solubility product expression tells us that the equilibrium concentrations of the cation and the anion are inversely related. That is, as the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases—and vice versa—so that K_sp is constant. Consequently, the solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. Adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le Châtelier’s principle. Le Châtelier's Principle states that if an equilibrium becomes unbalanced, the reaction will shift to restore the balance. As a result, the solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.

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

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

Defining \(s\) as the concentration of dissolved lead(II) chloride, then:

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

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

These values can be substituted into the solubility product expression, which can be solved for \(s\):

\[\begin{eqnarray} K_{sp} &=& [Pb^{2+}] [Cl^-]^2 \\ &=& s \times (2s)^2 \\ 1.7 \times 10^{-5} &=& 4s^3 \\ s^3 &=& \frac{1.7 \times 10^{-5}}{4} \\ &=& 4.25 \times 10^{-6} \\ s &=& \sqrt[3]{4.25 \times 10^{-6}} \\ &=& 1.62 \times 10^{-2}\ mol\ dm^{-3} \end{eqnarray} \]The concentration of lead(II) ions in the solution is 1.62 x 10-2 M.

Consider what happens if sodium chloride is added 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:

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

What happens to that equilibrium if extra chloride ions are added? According to Le Châtelier, the position of equilibrium will shift to counter the change, in this case, by removing the chloride ions by making extra solid lead(II) chloride.

Of course, the concentration of lead(II) ions in the solution is so small that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride. The lead(II) chloride becomes even less soluble, and the concentration of lead(II) ions in the solution decreases. This type of response occurs with any sparingly soluble substance: it is less soluble in a solution which contains any ion which it has in common. This is the common ion effect.

If an attempt is made to dissolve some lead(II) chloride in some 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions this time? As before, define s to be the concentration of the lead(II) ions.

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

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, it can be assumed that the concentration of the common ion is entirely due to the other solution. This simplifies the calculation.

So we assume:

\[[Cl^- ] = 0.100\; M \label{3} \nonumber\]

The rest of the mathematics 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} \nonumber \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} \label{4} \nonumber\end{equation}

Finally, compare that value with the simple saturated solution:

Original solution:

\[[Pb^{2+}] = 0.0162 \, M \label{5} \nonumber\]

Solution in 0.100 M NaCl solution:

\[ [Pb^{2+}] = 0.0017 \, M \label{6} \nonumber \]

The concentration of the lead(II) ions has decreased by a factor of about 10. If more concentrated solutions of sodium chloride are used, the solubility decreases further.

The common ion effect usually decreases the solubility of a sparingly soluble salt.

Example \(\PageIndex{2}\)

Calculate the solubility of calcium phosphate [Ca₃(PO₄)₂] in 0.20 M CaCl₂.

Given: concentration of CaCl₂ solution

Asked for: solubility of Ca₃(PO₄)₂ in CaCl₂ solution

Strategy:

Write the balanced equilibrium equation for the dissolution of Ca₃(PO₄)₂. Tabulate the concentrations of all species produced in solution.
Substitute the appropriate values into the expression for the solubility product and calculate the solubility of Ca₃(PO₄)₂.

Solution:

A The balanced equilibrium equation is given in the following table. If we let x equal the solubility of Ca₃(PO₄)₂ in moles per liter, then the change in [Ca²⁺] is once again +3x, and the change in [PO₄³⁻] is +2x. We can insert these values into the ICE table.

\[Ca_3(PO_4)_{2(s)} \rightleftharpoons 3Ca^{2+}_{(aq)} + 2PO^{3−}_{4(aq)}\]

	Ca3(PO4)2	[Ca²⁺]	[PO₄³⁻]
initial	pure solid	0.20	0
change	—	+3x	+2x
final	pure solid	0.20 + 3x	2x

B The K_sp expression is as follows:

K_sp = [Ca²⁺]³[PO₄³⁻]² = (0.20 + 3x)³(2x)²= 2.07×10⁻³³

Because Ca₃(PO₄)₂ is a sparingly soluble salt, we can reasonably expect that x << 0.20. Thus (0.20 + 3x) M is approximately 0.20 M, which simplifies the K_sp expression as follows:

\[\begin{align}K_{\textrm{sp}}=(0.20)^3(2x)^2&=2.07\times10^{-33}
\\x^2&=6.5\times10^{-32}
\\x&=2.5\times10^{-16}\textrm{ M}\end{align}\]

This value is the solubility of Ca₃(PO₄)₂ in 0.20 M CaCl₂ at 25°C. It is approximately nine orders of magnitude less than its solubility in pure water, as we would expect based on Le Châtelier’s principle. With one exception, this example is identical to Example \(\PageIndex{2}\)—here the initial [Ca²⁺] was 0.20 M rather than 0.

Exercise \(\PageIndex{2}\)

Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The solubility of silver carbonate in pure water is 8.45 × 10⁻¹² at 25°C.

Answer

2.9 × 10⁻⁶ M (versus 1.3 × 10⁻⁴ M in pure water)

Solubility and pH

In the case of a weak acid/base equilibrium, changing the pH of a solution by adding H⁺ or OH^- ions is also an example of the common-ion effect. This can also affect the solubility of ionic salts in which the cation or anion is either acidic or basic. In this case, changing the pH can increase or decrease the solubility because one of the ions will react with H⁺ or OH^- ions.

For example, consider the dissociation of calcium carbonate, the main component of limestone and marble:

\[CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2−}_{3(aq)}\]

Carbonates are not neutral salts, but rather are weak bases because of the equilibrium between carbonate and bicarbonate:

\[CO^{2−}_{3(aq)} + H_2O_{(l)} \rightleftharpoons OH^{-}_{(aq)} + HCO^{-}_{3(aq)}\]

Thus, the concentration of carbonate can be influenced by the pH. Decreasing the pH of the solution (making it more acidic) will cause carbonate to be converted to bicarbonate, shifting the above equilibrium to the right, or alternatively, driving the following equilibrium forward:

\[CO^{2−}_{3(aq)} + H^+_{(aq)} \rightleftharpoons HCO^{-}_{3(aq)}\]

When \([CO^{2−}_3]\) decreases, this then shifts the dissociation equilibrium to the right as well according to Le Chatelier's Principle, thus increasing the solubility. This effect is seen in nature by the formation of underground caverns, which are carved out of limestone rock over many years by groundwater that is slightly acidic due to dissolved CO₂ (Figure \(\PageIndex{1}\)). Slight differences in the solubility of \(CaCO_{3(s)}\) as the groundwater drips into the open space of the cave cause limestone to be deposited at the top and bottom of the cavern as stalactites and stalagmites.

Figure \(\PageIndex{1}\). Limestone caverns are formed by the action of acidic groundwater on calcium carbonate rock. Image By Juloml - Own work, CC BY-SA 4.0, https://commons.wikimedia.org/w/inde...?curid=9647226

This phenomenon holds true for any ionic compound containing a weakly basic anion - increasing the acidity (decreasing the pH) will increase the solubility of the salt by reacting with the weak base. The opposite would be the case for an ionic compound containing a weakly acidic cation, such as ammonium salts; in that case, decreasing the acidity (increasing the pH) would increase their solubility by deprotonating the cation.

Decreasing the pH increases the solubility of salts containing a weakly basic anion.

Example \(\PageIndex{3}\)

What effect will adding 0.1 M HCl to a solution of each of the following salts have on their solubility?

Calcium fluoride, CaF₂

Sodium carbonate, Na₂CO₃

Potassium bromide, KBr

Solution

Consider whether any of the ions in each salt are acidic or basic.

CaF₂: \(Ca^{2+}\) is a neutral ion. \(F^-\) is weakly basic (the conjugate base of the weak acid, \(HF\)). Adding HCl will increase the solubility of \(CaF_2\) by removing \(F^-\) by the reaction \[H^+ + F^- \rightleftharpoons HF \]

Na₂CO₃: \(Na^+\) is a neutral ion. \(CO^{2-}_3\) is weakly basic (the conjugate base of the weak acid, \(HCO^-_3\)). Adding HCl will increase the solubility of \(Na_2CO_3\) by removing \(CO^{2-}_3\) by the reaction \[H^+ + CO^{2-}_3\rightleftharpoons HCO^-_3\]

KBr: Both \(K^+\) and \(Br^-\) are neutral ions (\(Br^-\) is the conjugate base of the strong acid, \(HBr\)). Adding HCl to a solution of KBr will have no effect on its solubility.

Exercise \(\PageIndex{3}\)

For which of the following salts will the solubility NOT be affected by pH?

Magnesium iodide, MgI₂

Potassium phosphate, K₃PO₄

Lead (II) sulfide, PbS

Answer: MgI₂(both neutral ions)

Summary

The common-ion effect is an application of Le Chatelier's Principle to solubility equilibria. The solubility of a slightly soluble salt is decreased when a common ion (in the form of another, more soluble, salt) is added. For salts that contain an acidic or basic ion, pH can also affect solubility. Decreasing pH increases the solubility of weakly basic salts by reaction of the basic anion with H⁺.

References

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.

Contributors

Emmellin Tung, Mahtab Danai (UCD)
Jim Clark (ChemGuide)
Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)
Anna Christianson, Bellarmine University