Common Ion Effect On Solubility Pogil Answers

The common ion effect is a phenomenon in chemistry that describes the decrease in solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution. This principle is a specific application of Le Chatelier's principle and is crucial for understanding and predicting the behavior of solutions in various chemical systems. This article delves into the common ion effect, providing detailed explanations, examples, and answers to frequently asked questions to enhance understanding.

Understanding the Common Ion Effect

The common ion effect is a direct consequence of Le Chatelier's principle, which states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. In the context of solubility, the stress is the addition of a common ion, which causes the equilibrium to shift towards the precipitation of the sparingly soluble salt, thereby reducing its solubility.

Solubility Equilibrium

To understand the common ion effect, it's essential to grasp the concept of solubility equilibrium. When a sparingly soluble salt is added to water, it dissolves to a small extent, establishing an equilibrium between the solid salt and its constituent ions in solution. For example, consider the dissolution of silver chloride (AgCl):

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The equilibrium expression for this reaction is given by the solubility product constant, Ksp:

Ksp = [Ag+][Cl-]

Ksp is a measure of the maximum concentration of ions that can exist in equilibrium with the solid salt. A smaller Ksp value indicates lower solubility.

The Effect of a Common Ion

Now, let's consider what happens when a soluble salt containing a common ion is added to this solution. Suppose we add sodium chloride (NaCl) to the AgCl solution. NaCl is a strong electrolyte and dissociates completely into sodium ions (Na+) and chloride ions (Cl-):

NaCl(s) → Na+(aq) + Cl-(aq)

The addition of Cl- ions from NaCl increases the concentration of Cl- in the AgCl solution. According to Le Chatelier's principle, the system will shift to relieve this stress by reducing the concentration of Ag+ ions, causing more AgCl to precipitate out of the solution. This results in a decrease in the solubility of AgCl.

Illustrative Examples

Example 1: Silver Chloride (AgCl) and Sodium Chloride (NaCl)

As discussed, the solubility of AgCl is reduced when NaCl is added. Let's calculate the solubility of AgCl in pure water and in a solution of NaCl to quantify this effect.

Solubility of AgCl in Pure Water

The Ksp of AgCl is approximately 1.8 x 10-10. In pure water, the dissolution of AgCl can be represented as:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

Let s be the solubility of AgCl in moles per liter (M). Then, at equilibrium:

[Ag+] = s
[Cl-] = s

Therefore, the Ksp expression is:

Ksp = [Ag+][Cl-] = s * s = s^2

Solving for s:

s = √(Ksp) = √(1.8 x 10-10) ≈ 1.34 x 10-5 M

So, the solubility of AgCl in pure water is approximately 1.34 x 10-5 M.

Solubility of AgCl in 0.1 M NaCl Solution

Now, let's calculate the solubility of AgCl in a 0.1 M NaCl solution. The initial concentration of Cl- from NaCl is 0.1 M. The dissolution of AgCl still contributes to the Cl- concentration, but we can assume that the change in Cl- concentration due to AgCl dissolution is negligible compared to the initial concentration from NaCl.

At equilibrium:

[Ag+] = s
[Cl-] = 0.1 + s ≈ 0.1 (since s is very small)

The Ksp expression is:

Ksp = [Ag+][Cl-] = s * (0.1) = 1.8 x 10-10

Solving for s:

s = (1.8 x 10-10) / 0.1 = 1.8 x 10-9 M

Therefore, the solubility of AgCl in a 0.1 M NaCl solution is approximately 1.8 x 10-9 M.

Comparing the two solubilities, it is evident that the solubility of AgCl is significantly reduced in the presence of the common ion (Cl-) from NaCl.

Example 2: Lead(II) Chloride (PbCl2) and Sodium Chloride (NaCl)

Consider the dissolution of lead(II) chloride (PbCl2):

PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)

The Ksp for PbCl2 is approximately 1.6 x 10-5.

Solubility of PbCl2 in Pure Water

In pure water, the equilibrium concentrations are:

[Pb2+] = s
[Cl-] = 2s

The Ksp expression is:

Ksp = [Pb2+][Cl-]^2 = s * (2s)^2 = 4s^3

Solving for s:

s = ∛(Ksp / 4) = ∛((1.6 x 10-5) / 4) ≈ 0.0159 M

The solubility of PbCl2 in pure water is approximately 0.0159 M.

Solubility of PbCl2 in 0.1 M NaCl Solution

In a 0.1 M NaCl solution, the initial Cl- concentration is 0.1 M. At equilibrium:

[Pb2+] = s
[Cl-] = 0.1 + 2s ≈ 0.1 (since s is small)

The Ksp expression is:

Ksp = [Pb2+][Cl-]^2 = s * (0.1)^2 = 1.6 x 10-5

Solving for s:

s = (1.6 x 10-5) / (0.1)^2 = 0.0016 M

The solubility of PbCl2 in a 0.1 M NaCl solution is approximately 0.0016 M. Again, the solubility is reduced due to the common ion effect.

Factors Affecting the Common Ion Effect

Several factors can influence the magnitude of the common ion effect:

1. Concentration of the Common Ion: The higher the concentration of the common ion, the greater the decrease in solubility of the sparingly soluble salt.
2. Ksp Value: Salts with smaller Ksp values are more susceptible to the common ion effect. This is because even a small increase in the common ion concentration can significantly shift the equilibrium.
3. Temperature: The solubility of salts generally increases with temperature, affecting the Ksp value. The common ion effect will still be observed, but the overall solubility may be higher at higher temperatures.
4. Presence of Other Ions: The presence of other ions in the solution can also influence solubility through complex ion formation or changes in ionic strength, which can affect the activity coefficients of the ions.

Applications of the Common Ion Effect

The common ion effect has several practical applications in various fields:

1. Quantitative Analysis: In gravimetric analysis, the common ion effect is used to ensure complete precipitation of an analyte. By adding an excess of a common ion, the solubility of the precipitate is minimized, leading to more accurate results.
2. Pharmaceuticals: In drug formulation, the solubility of a drug can be controlled by adjusting the concentration of a common ion in the solution. This can affect the drug's absorption and bioavailability.
3. Environmental Chemistry: The common ion effect is important in understanding the behavior of pollutants in natural waters. For example, the solubility of heavy metal salts in contaminated soils can be affected by the presence of common ions from fertilizers or industrial waste.
4. Industrial Processes: In various industrial processes, controlling the solubility of salts is crucial for product purification and waste management. The common ion effect can be used to selectively precipitate certain ions from a solution.

Common Mistakes to Avoid

When working with the common ion effect, it is essential to avoid common mistakes:

• Forgetting to Consider Stoichiometry: Ensure that you correctly account for the stoichiometry of the dissolution reaction when setting up the Ksp expression. For example, for PbCl2, the chloride ion concentration is twice the lead(II) ion concentration in pure water.
• Assuming Negligible Change in Common Ion Concentration: In some cases, the contribution of the sparingly soluble salt to the common ion concentration may not be negligible. Always check if the assumption ([common ion]initial >> s) is valid.
• Ignoring Other Equilibria: The presence of other equilibria in the solution, such as complex ion formation or acid-base reactions, can affect the solubility of the salt. Make sure to consider all relevant equilibria when solving problems.
• Misunderstanding Le Chatelier's Principle: The common ion effect is a direct application of Le Chatelier's principle. Make sure you understand the principle and how it applies to solubility equilibria.

Practice Problems

To reinforce understanding of the common ion effect, here are some practice problems:

1. Calculate the solubility of silver bromide (AgBr) in pure water and in a 0.05 M solution of potassium bromide (KBr). The Ksp of AgBr is 5.0 x 10-13.
2. Determine the solubility of calcium fluoride (CaF2) in pure water and in a 0.1 M solution of sodium fluoride (NaF). The Ksp of CaF2 is 3.9 x 10-11.
3. What is the solubility of barium sulfate (BaSO4) in a 0.02 M solution of sodium sulfate (Na2SO4)? The Ksp of BaSO4 is 1.1 x 10-10.
4. Compare the solubility of lead(II) iodide (PbI2) in pure water and in a 0.01 M solution of lead(II) nitrate (Pb(NO3)2). The Ksp of PbI2 is 7.1 x 10-9.
5. Calculate the molar solubility of magnesium hydroxide (Mg(OH)2) in pure water and in a solution buffered at pH 12. The Ksp of Mg(OH)2 is 5.6 x 10-12.

Common Ion Effect POGIL (Process Oriented Guided Inquiry Learning) Activities and Answers

POGIL activities are designed to help students actively construct their understanding of chemical concepts through guided inquiry. Here’s a look at how the common ion effect might be addressed in a POGIL activity, along with potential answers to typical questions:

Sample POGIL Activity: The Common Ion Effect

Introduction: Consider the dissolution of a sparingly soluble salt, such as calcium oxalate (CaC2O4), in water.

CaC2O4(s) ⇌ Ca2+(aq) + C2O42-(aq)

Model 1: Solubility of Calcium Oxalate in Pure Water

The solubility product constant (Ksp) for calcium oxalate is 2.3 x 10-9.

Questions:

1. Write the expression for the Ksp of calcium oxalate. Answer: Ksp = [Ca2+][C2O42-]
2. If the solubility of calcium oxalate in pure water is s M, express the equilibrium concentrations of Ca2+ and C2O42- in terms of s. Answer: [Ca2+] = s, [C2O42-] = s
3. Calculate the solubility of calcium oxalate in pure water. Answer: Ksp = s2; s = √(Ksp) = √(2.3 x 10-9) ≈ 4.8 x 10-5 M

Model 2: Solubility of Calcium Oxalate in a Solution Containing a Common Ion

Now, consider the solubility of calcium oxalate in a 0.01 M solution of sodium oxalate (Na2C2O4). Sodium oxalate is a soluble salt that dissociates completely:

Na2C2O4(s) → 2Na+(aq) + C2O42-(aq)

Questions:

1. What is the concentration of oxalate ions (C2O42-) in the 0.01 M sodium oxalate solution before any calcium oxalate dissolves? Answer: 0.01 M
2. Let s' be the solubility of calcium oxalate in the sodium oxalate solution. Express the equilibrium concentrations of Ca2+ and C2O42- in terms of s', considering the initial concentration of oxalate ions from sodium oxalate. Answer: [Ca2+] = s', [C2O42-] = 0.01 + s'
3. Write the Ksp expression for calcium oxalate in the sodium oxalate solution. Answer: Ksp = [Ca2+][C2O42-] = s'(0.01 + s')
4. Assuming that s' is much smaller than 0.01, simplify the Ksp expression and calculate the solubility of calcium oxalate in the sodium oxalate solution. Answer: Ksp ≈ s'(0.01); s' = Ksp/0.01 = (2.3 x 10-9)/0.01 = 2.3 x 10-7 M
5. Compare the solubility of calcium oxalate in pure water and in the sodium oxalate solution. What do you observe? Answer: The solubility of calcium oxalate is much lower in the sodium oxalate solution (2.3 x 10-7 M) compared to pure water (4.8 x 10-5 M).

Model 3: Le Chatelier's Principle and the Common Ion Effect

Le Chatelier's principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Questions:

1. In the context of the common ion effect, what is the "stress" applied to the calcium oxalate equilibrium when sodium oxalate is added? Answer: The addition of oxalate ions increases the concentration of C2O42-, which is the stress.
2. According to Le Chatelier's principle, in which direction will the equilibrium shift to relieve this stress? Answer: The equilibrium will shift to the left, favoring the formation of solid calcium oxalate and reducing the concentration of calcium ions in solution.
3. Explain how the common ion effect is an example of Le Chatelier's principle. Answer: The addition of a common ion causes the equilibrium to shift in a direction that reduces the solubility of the sparingly soluble salt, thereby relieving the stress caused by the increased concentration of the common ion.

Conclusion Questions:

1. Define the common ion effect in your own words. Answer: The common ion effect is the decrease in solubility of a sparingly soluble salt when a soluble salt containing a common ion is added to the solution.
2. Explain why the common ion effect occurs. Answer: The common ion effect occurs because the addition of a common ion shifts the solubility equilibrium according to Le Chatelier's principle, reducing the solubility of the sparingly soluble salt.
3. Give an example of how the common ion effect can be used in a practical application. Answer: The common ion effect can be used in quantitative analysis to ensure complete precipitation of an analyte by adding an excess of a common ion.

These questions encourage students to explore the concept actively, applying Le Chatelier's principle to understand and predict the changes in solubility due to the presence of common ions.

Conclusion

The common ion effect is a fundamental concept in chemistry that explains how the solubility of a sparingly soluble salt is affected by the presence of a common ion. It is a direct application of Le Chatelier's principle and has significant implications in various fields, including analytical chemistry, pharmaceuticals, environmental science, and industrial processes. By understanding the principles and factors influencing the common ion effect, one can effectively predict and control the solubility of salts in different chemical systems.