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Predicting a precipitate

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Intros
Lessons
  1. Predicting if a solution will form a precipitate.
  2. Recalling Ksp.
  3. The ionic product Q (with worked example).
  4. Predicting precipitates using Ksp and Q.
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Examples
Lessons
  1. Use the solubility product expression to predict a precipitate.1
    1. A solution was made by combining 75 mL of 0.2M Mg2+(aq) and 110 mL of 0.08M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Mg(OH)2 will form in the resultant mixture.
    2. A solution was made by combining 20 mL of 0.01M Fe3+(aq) and 25 mL of 0.01M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Fe(OH)3 will form in the resultant mixture.
    3. A solution was made combining 50 mL of 0.003M Ca2+(aq) and 40 mL of 0.008M SO42-(aq). Use these values and the solubility product expression to predict whether a precipitate of CaSO4 will form in the resultant mixture.

      1 Source for Ksp solubility constant data at 250C: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Topic Notes
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Introduction

Predicting precipitate formation is a crucial skill in chemistry, essential for understanding various chemical reactions and processes. Our introduction video provides a comprehensive overview of this topic, laying the foundation for a deeper understanding. This article delves into the key concepts of precipitate prediction, focusing on the ionic product Q and its relationship to the solubility product Ksp. We'll explore how these fundamental principles interact and guide the formation of precipitates in solution. By mastering these concepts, you'll be able to accurately predict when and under what conditions precipitates will form. This knowledge is invaluable in fields ranging from environmental science to industrial chemistry. Throughout this article, we'll break down the process of using Q and Ksp to make precise predictions about precipitate formation, equipping you with the tools to analyze and understand complex chemical reactions.

Understanding the Solubility Product Constant (Ksp)

The solubility product constant, commonly known as Ksp, is a fundamental concept in chemistry that plays a crucial role in understanding the behavior of slightly soluble ionic compounds in aqueous solutions. This constant provides valuable insights into the solubility of a substance and its tendency to form precipitates under specific conditions. To fully grasp the significance of Ksp, it's essential to explore its relationship with saturated solutions and equilibrium.

A saturated solution is a state where the maximum amount of solute has dissolved in a solvent at a given temperature. In this condition, the solution reaches a dynamic equilibrium between the dissolved ions and the undissolved solid. The Ksp value represents this equilibrium state mathematically, expressing the product of the concentrations of ions in a saturated solution raised to the power of their stoichiometric coefficients.

For instance, consider the dissociation of a slightly soluble salt AB in water:

AB(s) A+(aq) + B-(aq)

The Ksp for this reaction would be expressed as:

Ksp = [A+][B-]

Where [A+] and [B-] represent the molar concentrations of the ions at equilibrium. This equation demonstrates how Ksp directly relates to ion concentrations in a saturated solution. A lower Ksp value indicates lower solubility, while a higher Ksp suggests greater solubility.

The significance of Ksp extends beyond merely describing solubility. It serves as a powerful tool for predicting whether a precipitate will form when solutions containing the relevant ions are mixed. This predictive capability is invaluable in various fields, including environmental science, water treatment, and industrial processes.

To illustrate how Ksp can be used to predict precipitate formation, let's consider an example from a video demonstration. Imagine we have two solutions: one containing lead(II) ions (Pb2+) and another containing iodide ions (I-). The Ksp of lead(II) iodide (PbI2) at a specific temperature is known. By comparing the ion product (Q) of the mixed solution to the Ksp value, we can determine whether a precipitate will form:

- If Q < Ksp: The solution is unsaturated, and no precipitate forms.

- If Q = Ksp: The solution is exactly saturated, at equilibrium.

- If Q > Ksp: The solution is supersaturated, and a precipitate will form.

This comparison allows chemists to predict and control precipitation reactions in various applications, from water purification to the synthesis of specific compounds.

Understanding the concept of dynamic equilibrium is crucial when working with Ksp. In a saturated solution, the rate of dissolution equals the rate of crystallization, creating a dynamic balance. This equilibrium state is characterized by constant macroscopic properties, despite ongoing microscopic changes. The Ksp value quantifies this equilibrium, providing a measure of the extent to which a compound dissociates in solution.

The practical applications of Ksp are vast and varied. In environmental science, it helps predict the mobility of pollutants in water systems. In medicine, it's used to understand the formation of kidney stones and develop treatments. Industries utilize Ksp calculations to optimize processes involving precipitation or dissolution, such as in ore extraction or water softening.

In conclusion, the solubility product constant (Ksp) is a powerful tool in chemistry, offering insights into the behavior of slightly soluble compounds in solution. By understanding Ksp, its relationship to saturated solutions and equilibrium, and its role in predicting precipitate formation, chemists can manipulate and control a wide range of chemical processes. This knowledge not only enhances our understanding of natural phenomena but also enables the development of innovative solutions to real-world problems across various scientific and industrial domains.

The Ionic Product Q: Definition and Calculation

The ionic product Q is a crucial concept in chemistry, particularly when dealing with solubility equilibria. It represents the product of ion concentrations in a solution at any given moment, whether the system is at equilibrium or not. This distinguishes Q from the solubility product constant (Ksp), which specifically applies to a system at equilibrium.

Understanding the difference between Q and Ksp is essential for predicting whether a precipitate will form or dissolve in a solution. When Q is less than Ksp, the solution is undersaturated, and more solid can dissolve. When Q equals Ksp, the solution is at equilibrium. However, when Q exceeds Ksp, the solution is supersaturated, and precipitation occurs.

Calculating the ionic product Q involves a step-by-step process using ion concentrations. Let's break down this calculation using the example of calcium fluoride (CaF2) from the video:

  1. Identify the ions involved: In this case, we have calcium (Ca2+) and fluoride (F-) ions.
  2. Write the dissociation equation: CaF2 Ca2+ + 2F-
  3. Determine the ion concentrations: Let's say we have [Ca2+] = 0.010 M and [F-] = 0.020 M.
  4. Apply the Q formula: For CaF2, Q = [Ca2+][F-]2
  5. Substitute the values: Q = (0.010)(0.020)2
  6. Calculate: Q = 4.0 × 10-6

It's crucial to note that when combining solutions, volume changes can significantly affect ion concentrations and, consequently, the value of Q. For instance, if we mix equal volumes of 0.020 M CaCl2 and 0.020 M NaF:

  1. Calculate the new volume: Total volume = Volume of CaCl2 + Volume of NaF
  2. Determine new concentrations: [Ca2+] = (Initial [Ca2+] × Initial volume) / Total volume
  3. Repeat for [F-]: [F-] = (Initial [F-] × Initial volume) / Total volume
  4. Apply the Q formula with these new concentrations

Considering volume changes is vital because dilution affects ion concentrations, which directly impact the ionic product Q. Neglecting this factor can lead to inaccurate predictions about precipitation or dissolution.

In practice, calculating Q helps chemists predict and control precipitation reactions in various applications, such as water treatment, mineral processing, and pharmaceutical formulations. By comparing Q to Ksp, one can determine whether a solution will form a precipitate, remain stable, or dissolve an existing solid.

To further illustrate the importance of Q, consider a scenario where you're trying to remove calcium ions from hard water. By adding fluoride ions and calculating Q, you can determine the optimal concentration needed to precipitate calcium fluoride effectively. This process is fundamental in water softening techniques.

Moreover, the concept of Q extends beyond simple precipitation reactions. In more complex systems, such as those involving multiple equilibria or redox reactions, understanding and calculating Q becomes even more critical for predicting system behavior and optimizing reaction conditions.

In conclusion, mastering the calculation of the ionic product Q is essential for chemists and chemical engineers. It provides valuable insights into solution behavior, helps predict precipitation outcomes, and guides the design of processes involving ionic species. By carefully considering ion concentrations and volume changes, professionals can leverage the power of Q to solve real-world problems in diverse fields, from environmental science to materials engineering.

Comparing Q and Ksp: Predicting Precipitate Formation

Understanding the relationship between the reaction quotient (Q) and the solubility product constant (Ksp) is crucial in predicting precipitate formation in chemical reactions. There are three possible scenarios when comparing Q and Ksp, each with distinct implications for precipitate formation. Let's explore these scenarios in detail.

Scenario 1: Q < Ksp

When Q is less than Ksp, it indicates that the solution is unsaturated. In this scenario, no precipitate will form. The concentrations of ions in the solution are below the saturation point, meaning that all the dissolved ions remain in solution. The system has not reached equilibrium, and more solute can still dissolve if added to the solution.

Scenario 2: Q = Ksp

When Q is equal to Ksp, the solution is exactly saturated. This represents a state of equilibrium between the dissolved ions and the solid precipitate. At this point, the rate of dissolution is equal to the rate of precipitation. No net change occurs in the amount of precipitate or dissolved ions. While this scenario is theoretically possible, it is rarely observed in practice due to the precise conditions required.

Scenario 3: Q > Ksp

When Q exceeds Ksp, the solution is supersaturated. This condition leads to precipitate formation. The ion concentrations in the solution are higher than the solubility limit, causing the excess ions to come out of solution and form a solid precipitate. The system will move towards equilibrium by reducing the concentration of ions in solution through precipitation until Q becomes equal to Ksp.

To illustrate how to compare Q and Ksp and predict precipitate formation, let's consider an example from the video. Suppose we have a solution containing lead(II) ions (Pb²) and iodide ions (I), and we want to determine if lead(II) iodide (PbI) will precipitate.

Given:

  • Ksp for PbI = 7.9 × 10
  • [Pb²] = 1.0 × 10² M
  • [I] = 2.0 × 10³ M

Step 1: Write the balanced equation for the dissociation of PbI

PbI Pb² + 2I

Step 2: Calculate Q using the given concentrations

Q = [Pb²][I]² = (1.0 × 10²)(2.0 × 10³)² = 4.0 × 10

Step 3: Compare Q to Ksp

Q (4.0 × 10) > Ksp (7.9 × 10)

Since Q is greater than Ksp, we can conclude that the solution is supersaturated, and a precipitate of PbI will form. The system will move towards equilibrium by precipitating PbI until the ion concentrations decrease to the point where Q equals Ksp.

This example demonstrates the practical application of comparing Q and Ksp to predict precipitate formation. By understanding these three scenarios, chemists can anticipate the behavior of solutions and design experiments or processes accordingly. This knowledge is particularly valuable in fields such as environmental science, where predicting the solubility of pollutants is crucial, or in industrial applications where controlling precipitation is essential for product quality or waste management.

Practical Applications and Examples

Predicting precipitate formation is a crucial skill in chemistry with numerous real-world applications. Let's explore additional examples and their step-by-step solutions, along with practical uses in water treatment, geology, and industrial processes.

Example 1: Mixing silver nitrate (AgNO3) and sodium chloride (NaCl) solutions

Step 1: Write the balanced equation: AgNO3 + NaCl AgCl + NaNO3

Step 2: Identify the spectator ions (NO3- and Na+) and the potentially reactive ions (Ag+ and Cl-)

Step 3: Write the net ionic equation: Ag+ + Cl- AgCl (s)

Step 4: Check solubility rules: AgCl is insoluble, confirming precipitate formation

Example 2: Combining barium chloride (BaCl2) and sodium sulfate (Na2SO4) solutions

Step 1: Write the balanced equation: BaCl2 + Na2SO4 BaSO4 + 2NaCl

Step 2: Identify spectator ions (Na+ and Cl-) and potentially reactive ions (Ba2+ and SO42-)

Step 3: Write the net ionic equation: Ba2+ + SO42- BaSO4 (s)

Step 4: Check solubility rules: BaSO4 is insoluble, confirming precipitate formation

In water treatment, precipitate formation plays a vital role in removing contaminants. For instance, adding aluminum sulfate (Al2(SO4)3) to water causes the formation of aluminum hydroxide (Al(OH)3) precipitate, which attracts and settles out impurities. This process, known as coagulation-flocculation, is essential for producing clean drinking water.

Geology relies heavily on precipitation reactions to understand mineral formation. For example, the creation of limestone caves involves the precipitation of calcium carbonate (CaCO3) from calcium-rich groundwater. As carbon dioxide (CO2) is released from the water, it triggers the precipitation reaction: Ca2+ + CO32- CaCO3 (s)

Industrial processes also utilize precipitate formation extensively. In metal recovery, valuable metals can be selectively precipitated from solution. For instance, copper can be recovered from mining waste by adding iron, causing a displacement reaction: Cu2+ + Fe Cu (s) + Fe2+

Another industrial application is in wastewater treatment, where heavy metals are removed through precipitation. Adding hydroxide ions to a solution containing metal ions causes the formation of insoluble metal hydroxides, which can then be filtered out. For example: Pb2+ + 2OH- Pb(OH)2 (s)

The ability to predict and control precipitate formation is crucial in various chemical processes. In the production of pharmaceuticals, precipitate formation can be used to purify compounds or create specific drug formulations. In the food industry, controlled precipitation is employed in the production of certain dairy products, such as cheese and yogurt.

Understanding precipitate formation is also essential in preventing unwanted reactions. In oil and gas pipelines, scale formation due to the precipitation of minerals like calcium carbonate can cause significant problems. By predicting and mitigating these reactions, engineers can prevent costly damage and maintain efficient operations.

In environmental science, precipitate formation plays a role in natural water purification processes. As rainwater percolates through soil and rock, various minerals precipitate out, affecting water quality and composition. This knowledge is crucial for understanding groundwater chemistry and managing water resources.

In conclusion, the ability to predict and understand precipitate formation is a valuable skill with wide-ranging applications. From ensuring clean drinking water and recovering valuable metals to understanding geological processes and optimizing industrial operations, this fundamental chemical concept continues to play a crucial role in numerous fields, highlighting the importance of mastering these principles in chemistry education and research.

Common Mistakes and Troubleshooting

When predicting precipitate formation, students often encounter several common errors that can lead to incorrect results. Understanding these mistakes and learning how to avoid them is crucial for mastering this important chemistry concept. One frequent error is unit conversion mistakes, particularly when dealing with molar solubility and Ksp values. Students must ensure they consistently use the correct units throughout their calculations to avoid discrepancies in their final answers.

Another common pitfall is forgetting to consider volume changes during reactions. When mixing solutions, the total volume often increases, which can affect concentration calculations. Students should always account for these volume changes during reactions to accurately determine the final concentrations of ions in solution. Additionally, misinterpreting the relationship between Q (reaction quotient) and solubility product constant Ksp is a frequent source of confusion. Remember that precipitation occurs when Q exceeds Ksp, not when they are equal.

To avoid these errors, students should develop a systematic approach to solving precipitate prediction problems. Start by clearly identifying the given information and the question at hand. Next, write out the balanced chemical equation and determine the Ksp expression. Calculate initial concentrations, accounting for any dilution effects. Then, set up the Q expression and compare it to Ksp. Finally, draw conclusions based on this comparison.

A troubleshooting guide for solving precipitate prediction problems can be invaluable. If your answer seems off, first check your unit conversions and ensure all concentrations are expressed in mol/L. Verify that you've accounted for any volume changes during reactions. Double-check your Ksp expression to make sure it matches the balanced equation. If using an ICE table, confirm that you've correctly identified the initial concentrations and applied the stoichiometric relationships.

Common calculation errors include forgetting to square or cube ion concentrations in the Ksp expression when necessary. Always review the coefficients in the balanced equation to determine the correct exponents. If you're still struggling, try working backwards from the solubility product constant Ksp to determine the maximum possible concentration of ions before precipitation occurs. This can help you develop intuition about the relationship between ion concentrations and precipitate formation.

By being aware of these common errors and following a structured problem-solving approach, students can significantly improve their accuracy in predicting precipitate formation. Regular practice and careful attention to detail will help reinforce these concepts and build confidence in tackling even complex precipitation problems.

Conclusion

In this article, we've explored the crucial concepts of Q (reaction quotient) and Ksp (solubility product constant) and their pivotal role in predicting precipitate formation. Understanding the relationship between Q and Ksp is essential for accurately determining whether a precipitate will form in a given solution. The introductory video provided a valuable foundation for grasping these concepts, making complex chemical principles more accessible. To truly master precipitate prediction, it's crucial to practice solving precipitate prediction problems. This hands-on approach will reinforce your understanding and improve your problem-solving skills. We encourage you to seek out additional resources and practice problems to further enhance your knowledge in this area. By mastering the interplay between Q and Ksp (solubility product constant), you'll be well-equipped to tackle more advanced chemistry topics and real-world applications. Remember, consistent practice and exploration of diverse scenarios will solidify your understanding of precipitate prediction.

Predicting if a solution will form a precipitate

Predicting if a solution will form a precipitate. Recalling Ksp.

Step 1: Understanding the Solubility Product Constant (Ksp)

The solubility product constant, Ksp, is a crucial concept in predicting whether a precipitate will form in a solution. Ksp is a constant value for a given compound at a specific temperature. It represents the maximum product of the ion concentrations in a saturated solution of that compound. If the ion product exceeds this value, the solution becomes supersaturated, and a precipitate will form. Conversely, if the ion product is less than Ksp, the solution remains unsaturated, and no precipitate forms.

Step 2: Introducing the Ionic Product (Q)

To predict precipitation, we need to calculate the ionic product, Q, which is the product of the concentrations of the ions in the solution at any given moment. The ionic product is compared to the solubility product constant, Ksp, to determine whether a precipitate will form. If Q is greater than Ksp, the solution is supersaturated, and a precipitate will form. If Q is less than Ksp, the solution is unsaturated, and no precipitate will form.

Step 3: Calculating the Ionic Product (Q)

To calculate Q, you need to know the concentrations of the ions in the solution. Multiply these concentrations together to get the ionic product. For example, if you have a solution containing ions A+ and B-, and their concentrations are [A+] and [B-], respectively, then Q = [A+] * [B-]. This product will be compared to the Ksp value of the compound AB.

Step 4: Comparing Q with Ksp

Once you have calculated the ionic product, Q, compare it with the solubility product constant, Ksp. There are three possible outcomes:

  • If Q < Ksp, the solution is unsaturated, and no precipitate will form.
  • If Q = Ksp, the solution is exactly saturated, and no additional precipitate will form.
  • If Q > Ksp, the solution is supersaturated, and a precipitate will form.

Step 5: Practical Application and Example

Let's consider a practical example to illustrate these steps. Suppose you have a solution containing 0.01 M of ion A+ and 0.02 M of ion B-. The Ksp for the compound AB is 1.0 x 10-4. First, calculate the ionic product: Q = [A+] * [B-] = 0.01 * 0.02 = 2.0 x 10-4. Next, compare Q with Ksp. In this case, Q (2.0 x 10-4) is greater than Ksp (1.0 x 10-4), indicating that the solution is supersaturated and a precipitate will form.

Step 6: Conclusion

By understanding and applying the concepts of Ksp and Q, you can predict whether a precipitate will form in a solution. This involves calculating the ionic product and comparing it with the solubility product constant. If the ionic product exceeds the solubility product constant, a precipitate will form, indicating that the solution is supersaturated. This method is a powerful tool in chemistry for predicting and understanding precipitation reactions.

FAQs

  1. How do you predict whether a precipitate will form?

    To predict precipitate formation, compare the reaction quotient (Q) with the solubility product constant (Ksp). Calculate Q using the initial ion concentrations. If Q > Ksp, a precipitate will form; if Q < Ksp, no precipitate forms; if Q = Ksp, the solution is at equilibrium.

  2. What are the requirements to form a precipitate?

    For a precipitate to form, the solution must be supersaturated. This occurs when the product of ion concentrations (Q) exceeds the solubility product constant (Ksp). Additionally, the compound must be insoluble or slightly soluble in the given solvent.

  3. Under what conditions will a precipitate form?

    A precipitate forms when the solution is supersaturated (Q > Ksp). This can happen when two solutions containing the precipitate's ions are mixed, when the solvent evaporates, or when the temperature changes, affecting solubility.

  4. How do you determine if a precipitate will form in a solution?

    To determine if a precipitate will form, follow these steps: 1) Write the balanced equation, 2) Calculate initial ion concentrations, 3) Determine Q, 4) Compare Q to Ksp. If Q > Ksp, a precipitate will form. Always consider volume changes and use consistent units in your calculations.

  5. What role does Ksp play in predicting precipitate formation?

    Ksp (solubility product constant) is crucial in predicting precipitate formation. It represents the equilibrium constant for a saturated solution of a sparingly soluble ionic compound. By comparing Q to Ksp, we can determine if a solution is undersaturated (Q < Ksp), saturated (Q = Ksp), or supersaturated (Q > Ksp), thus predicting precipitate formation.

Prerequisite Topics for Predicting a Precipitate

Understanding how to predict a precipitate in a chemical reaction is a crucial skill in chemistry. However, to master this concept, it's essential to have a solid foundation in several prerequisite topics. These topics provide the necessary background knowledge and skills to comprehend the intricacies of precipitation reactions.

One fundamental prerequisite is understanding tables of values of linear relationships. This algebraic skill is vital when dealing with chemical reactions and writing equations. It helps in organizing and interpreting data related to reactants and products, which is crucial when predicting precipitates.

Another key concept is the solubility constant, also known as the solubility product constant (Ksp). This topic is directly related to predicting precipitates, as it quantifies the solubility of a substance in a solution. Understanding Ksp is essential for determining whether a precipitate will form in a given reaction.

The concept of dynamic equilibrium in chemistry is also crucial. It helps explain the balance between dissolution and precipitation in a saturated solution. This understanding is vital when considering the conditions under which a precipitate might form or dissolve.

Knowledge of state symbols and phase changes is equally important. This topic covers the different states of matter and how they change during chemical reactions. Understanding volume changes in chemical reactions is particularly relevant when predicting precipitates, as the formation of a solid from a solution involves a phase change.

Lastly, while it might seem unrelated at first, calculating cell potential in voltaic cells can provide valuable insights. The principles used in predicting redox reactions can be applied to understanding precipitation reactions, as both involve the transfer of electrons and the formation of new compounds.

By mastering these prerequisite topics, students can develop a comprehensive understanding of the factors that influence precipitation reactions. This knowledge allows for more accurate predictions of when and how precipitates will form in various chemical scenarios. The interplay between solubility, equilibrium, and reaction conditions becomes clearer, enabling students to approach more complex problems with confidence.

In conclusion, while predicting a precipitate might seem like a standalone topic, it's deeply interconnected with various fundamental concepts in chemistry and mathematics. By building a strong foundation in these prerequisite areas, students can enhance their problem-solving skills and gain a deeper appreciation for the intricacies of chemical reactions and precipitate formation.

In this lesson, we will learn:

  • To recall the ionic product Q and how it relates to the solubility product Ksp.
  • How to calculate the ionic product and use it to predict whether a precipitate will form in solution.

Notes:

  • As mentioned in the lesson Solubility product; the value of Ksp for a saturated solution of a given compound is constant. It doesn’t matter how the ion concentrations are made up (whether by one solution or multiple combined); if the solution is saturated then the product of the individual ion concentrations will be equal to Ksp.
    • Using this fact, we can predict if a precipitate forms in the solution or not before making it by using the Ksp value. This Ksp product value is the product of the ion concentrations required to make a saturated solution (for the equilibrium to be established).
    • If this product of ion concentration is smaller than Ksp, the solution will not be saturated and the equilibrium will not be established.
  • The ‘product of ion concentration’ is known simply as the ionic product with label Q. As a product, Q is calculated by multiplying the concentrations of ions together. For aqueous ions Mx+ and Xm- forming precipitate MmXx:

    Q = [M+]m [X-]x

  • Compare this to Ksp which is the ion concentration needed to form a saturated solution. Using these two terms, three situations can be described when two solutions are mixed to create a new ionic species.
    • If Q is smaller than Ksp, then there are less ions in solution than necessary to form a saturated solution. Without a saturated solution, a precipitate will not form.
    • If Q is equal to Ksp, then there are just enough ions in solution necessary for a saturated solution to form. In this scenario, a solution that has just reached saturation will form. This means that any further addition of aqueous ions will form a precipitate, as the equilibrium shifts to the left to favor the undissolved state.
    • If Q is greater to Ksp, then there are more ions in solution than are necessary to form a saturated solution. In this scenario, an equilibrium will be established to form a precipitate with the excess ions; a precipitate will form.

  • For example: a solution made by combining 100 mL of 0.1 M Ca2+(aq) and 75mL of 0.2 M F-(aq) :

    Q = [Ca2+][F-]2\qquad To Form CaF2

    These solutions dilute each other. Find the new concentration by dividing original volume by combined volume.

    [Ca2+]sol = 0.1 M * 0.1 L0.175 L\large \frac{0.1 \, L}{0.175 \, L} = 0.057 M

    [F-]sol = 0.2 M * 0.075 L0.175 L\large \frac{0.075 \, L}{0.175 \, L} = 0.086 M

    Q = [0.057] x [0.086]2 \, = 4.22*10-4

    Ksp (CaF2) = 3.45*10-11 M


    Q > Ksp therefore a precipitate will form.