AP Chemistry: Common Ion Effect

Why does adding table salt to a solution make it harder for other salts to dissolve? Why do kidney stones form more readily in certain conditions? The answer lies in the common ion effect, a powerful application of equilibrium principles that explains how the solubility of a salt is suppressed when another source of one of its ions is introduced. Mastering this concept is essential for predicting solubility in complex solutions, designing selective precipitation experiments in the lab, and understanding crucial biological and engineering processes, from water treatment to pathological calcification.

Foundational Principle: Le Chatelier’s Stress on Solubility

At its core, the common ion effect is a direct consequence of Le Chatelier's principle. This principle states that when a stress is applied to a system at equilibrium, the system shifts to counteract that stress. Consider a saturated solution of a sparingly soluble salt like silver chloride, AgCl. The dynamic dissolution equilibrium is represented as: $$AgCl(s)\rightleftharpoons A{g}^{+}(aq)+C{l}^{-}(aq)$$

In a solution containing only dissolved AgCl, the concentrations of $A{g}^{+}$ and $C{l}^{-}$ are equal. Now, imagine we add a soluble chloride salt like NaCl to this saturated solution. This addition introduces a large amount of $C{l}^{-}$ ions, which are a product of the AgCl dissolution equilibrium.

According to Le Chatelier, the system will shift to reduce the increased concentration of the product ion, $C{l}^{-}$. It does this by shifting the equilibrium position to the left, toward the solid AgCl. This shift reduces the concentration of the other product ion, $A{g}^{+}$, and causes more solid AgCl to precipitate from solution. The net result is that the solubility of AgCl—the amount that dissolves to form $A{g}^{+}$ and $C{l}^{-}$—is decreased in the presence of the common ion $C{l}^{-}$ from NaCl. The same effect would occur if we added a source of $A{g}^{+}$ ions, such as $AgN{O}_3$.

The Quantitative Backbone: The Solubility Product Constant ($K_{sp}$)

To move from a qualitative understanding to quantitative prediction, we use the solubility product constant ($K_{sp}$). For a generic salt $A_x{B}_y$ that dissociates as $A_x{B}_y(s)\rightleftharpoons x{A}^{y+}(aq)+y{B}^{x-}(aq)$, the $K_{sp}$ expression is: $$K_{sp}=[A^{y+}{]}^x[B^{x-}{]}^y$$

$K_{sp}$ is a constant at a given temperature. For AgCl ($AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$), $K_{sp}=[A{g}^{+}][C{l}^{-}]=1.8\times 10^{-10}$ at 25°C. This constant relationship is the key to all solubility calculations. Crucially, $K_{sp}$ is an equilibrium constant; it is only dependent on temperature. The actual individual ion concentrations can change, but their product at equilibrium must always equal $K_{sp}$.

Calculating Molar Solubility with a Common Ion

Molar solubility is defined as the number of moles of a solid that dissolve per liter of solution to form a saturated solution. Calculating it with a common ion present is a classic AP Chemistry problem. The strategy is to account for the initial concentration of the common ion from the soluble salt before any of the sparingly soluble salt dissolves.

Worked Example: Calculate the molar solubility of lead(II) iodide, $Pb{I}_2$, in a 0.10 M solution of sodium iodide, NaI. $K_{sp}$ for $Pb{I}_2=8.5\times 10^{-9}$.

Step 1: Write the equilibrium and $K_{sp}$ expression. $$Pb{I}_2(s)\rightleftharpoons P{b}^{2+}(aq)+2I^{-}(aq)$$ $$K_{sp}=[P{b}^{2+}][I^{-}{]}^2=8.5\times 10^{-9}$$

Step 2: Define variables and account for the common ion. Let $s$ = the molar solubility of $Pb{I}_2$ (mol/L).

• From the dissolution of $Pb{I}_2$: For every 1 mol of $Pb{I}_2$ that dissolves, 1 mol of $P{b}^{2+}$ and 2 mol of $I^{-}$ are produced.
• From the NaI solution: NaI is fully soluble, providing an initial $[I^{-}]=0.10M$.

Therefore, at equilibrium: $[P{b}^{2+}]=s$ $[I^{-}]=0.10+2s$ (The total iodide comes from the NaI and the dissolved $Pb{I}_2$).

Step 3: Apply the $K_{sp}$ expression and solve for $s$. $$K_{sp}=(s)(0.10+2s{)}^2=8.5\times 10^{-9}$$ Because $K_{sp}$ is very small, $2s$ will be negligible compared to 0.10 M. This is the common ion approximation: $0.10+2s\approx 0.10$. The equation simplifies to: $$(s)(0.10{)}^2=8.5\times 10^{-9}$$ $$s(0.01)=8.5\times 10^{-9}$$ $$s=8.5\times 10^{-7}M$$

Step 4: Verify the approximation. $2s=1.7\times 10^{-6}$, which is indeed much less than 0.10, so the approximation is valid. The molar solubility of $Pb{I}_2$ in 0.10 M NaI is $8.5\times 10^{-7}M$. Compare this to its solubility in pure water, which you can calculate as $s=\sqrt[3]{K_{sp}/4}\approx 1.3\times 10^{-3}M$. The presence of the common ion has decreased the solubility by over a thousandfold.

Connection to Selective Precipitation

The dramatic reduction in solubility caused by a common ion is harnessed in selective precipitation, a technique used to separate a mixture of metal ions in solution. By carefully controlling the concentration of a precipitating anion, you can precipitate one ion while leaving others in solution.

For example, suppose a solution contains $0.010MA{g}^{+}$ and $0.010MP{b}^{2+}$. You want to separate them by adding NaCl. The relevant $K_{sp}$ values are $AgCl=1.8\times 10^{-10}$ and $PbC{l}_2=1.6\times 10^{-5}$.

The ion that precipitates first is the one whose $K_{sp}$ is exceeded first as you add $C{l}^{-}$. You calculate the minimum $[C{l}^{-}]$ needed to begin precipitating each cation.

• For AgCl: $[C{l}^{-}{]}_{min}=K_{sp}/[A{g}^{+}]=(1.8\times 10^{-10})/0.010=1.8\times 10^{-8}M$
• For $PbC{l}_2$: $[C{l}^{-}{]}_{min}=\sqrt{K_{sp}/[P{b}^{2+}]}=\sqrt{(1.6\times 10^{-5})/0.010}=0.04M$

Since AgCl requires a much lower chloride concentration to precipitate, it will form a solid first. By adding NaCl slowly until $[C{l}^{-}]$ is just below 0.04 M, you can precipitate nearly all the $A{g}^{+}$ as AgCl while keeping $P{b}^{2+}$ in solution. The solid AgCl can then be filtered out, effectively separating the ions. This process is fundamental in qualitative analysis schemes, wastewater metal recovery, and clinical chemistry.

Common Pitfalls

1. Confusing Solubility with $K_{sp}$. A common mistake is thinking $K_{sp}$ changes when a common ion is added. It does not. $K_{sp}$ is temperature-dependent only. What changes is the molar solubility ($s$), while the ion product $[A^{+}][B^{-}]$ remains constant at the value of $K_{sp}$.

Correction: Always remember $K_{sp}$ is constant. Use it as the target value your equilibrium concentrations must multiply to reach.

1. Forgetting the Common Ion in the Initial Concentration Setup. When setting up an ICE table or equivalent, students often only include the ions from the dissolving salt.

Correction: The initial concentration of the common ion is not zero if it comes from another soluble salt. In the $Pb{I}_2$/NaI example, the initial $[I^{-}]$ is 0.10 M, not 0 M.

1. Incorrectly Adding the Change to the Common Ion. In the equilibrium expression $[I^{-}]=0.10+2s$, some might incorrectly write it as $[I^{-}]=2s$ or $[I^{-}]=0.10+s$.

Correction: The change ($2s$) must be correctly stoichiometric and added to the initial concentration from the other source.

1. Failing to Check the Validity of the Approximation. Using the approximation $0.10+2s\approx 0.10$ is valid only if $2s$ is less than 5% of 0.10. If you don't check, you might get an incorrect answer for salts with higher solubility or very low common ion concentrations.

Correction: Always calculate $s$, then check if the neglected term ($2s$) is less than 5% of the value it was added to. If not, you must solve the full equation without the approximation.

Summary

• The common ion effect is the reduction in solubility of a sparingly soluble ionic compound when a soluble salt sharing one of its ions is added to the solution. It is a direct application of Le Chatelier's principle.
• Quantitatively, the effect is governed by the constant $K_{sp}$. While molar solubility decreases, the ion product $[cation][anion]$ at equilibrium must still equal $K_{sp}$.
• To calculate new molar solubility, set up the $K_{sp}$ equation with the equilibrium concentration of the common ion expressed as (Initial Concentration from other source) + (Change from dissolution). The common ion approximation often simplifies the math.
• This principle is applied in selective precipitation to separate ions from a mixture. The ion requiring the lower concentration of precipitating agent will form a solid first, allowing for physical separation by filtration.
• Understanding this concept bridges foundational equilibrium theory with practical applications in analytical chemistry, medicine, and environmental engineering.