3.3 Solubility Product and Precipitation Reactions

In the previous sections, we discussed the equilibria of acids, bases, and buffers. Now, we turn our attention to another important type of equilibrium: the equilibrium that exists between a slightly soluble ionic solid and its ions in a saturated solution. This concept is quantified by the solubility product constant, or K_sp, which is a powerful tool for predicting whether a precipitate will form when two solutions are mixed.

Definition

• Solubility Product Constant (K_sp): The product of the molar concentrations of the ions in a saturated solution of a slightly soluble ionic compound, each raised to the power of its coefficient in the balanced dissolution equation.
• Precipitation Reaction: A reaction in which a solid (precipitate) forms from a solution.

Simple Explanation: K_sp is a number that tells you how much of a slightly soluble salt can dissolve in water. It is the equilibrium constant for the dissolving process.

Core Concept

The core concept is dynamic equilibrium in a saturated solution. When an excess of a slightly soluble ionic compound is mixed with water, some of it dissolves, and the rest settles at the bottom. A dynamic equilibrium is established between the undissolved solid and the ions in the saturated solution.

For example, when calcium fluoride (CaF_2) is mixed with water: CaF_{2(s)} ⇌ Ca^{2+}_{(aq)} + 2F^-_{(aq)}

The equilibrium constant for this process is the solubility product constant (K_sp). Because the concentration of the pure solid is constant, it is not included in the expression. The K_sp expression is: K_sp = [Ca^{2+}][F^-]²

This means that at a given temperature, the product of the ion concentrations (each raised to its stoichiometric coefficient) in a saturated solution is a constant value.

Detailed Explanation

The General K_sp Expression

For a general slightly soluble ionic compound A_mB_n that dissolves according to the equation: A_mB_{n(s)} ⇌ mA^{n+}_{(aq)} + nB^{m-}_{(aq)}

The solubility product constant is expressed as: K_sp = [A^{n+}]^m [B^{m-}]ⁿ

Calculating Ion Concentrations from K_sp

If you know the K_sp of a salt, you can calculate the concentrations of its ions in a saturated solution. The textbook provides the example of silver bromide (AgBr), which has a K_sp of 1.60 × 10^-10.

1. Write the dissociation equation: AgBr_{(s)} ⇌ Ag^+_{(aq)} + Br^-_{(aq)} 2. Set up the K_sp expression: K_sp = [Ag^+][Br^-] 3. Let the solubility (moles dissolved per dm³) be x: If x moles of AgBr dissolve, then x moles of Ag^+ and x moles of Br^- are produced. So, [Ag^+] = x and [Br^-] = x. 4. Solve for x: 1.60 × 10^-10 = (x)(x) = x² x = √1.60 × 10^-10 = 1.26 × 10^-5 M 5. Conclusion: The concentration of Ag^+ and Br^- in a saturated solution of AgBr is 1.26 × 10^-5 M.

Example: Calculating the Concentration of AgCl

The textbook also provides a step-by-step example for silver chloride (AgCl), with K_sp = 1.8 × 10^-10 at 25°C.

1. Dissociation: AgCl_{(s)} ⇌ Ag^+_{(aq)} + Cl^-_{(aq)} 2. K_sp expression: K_sp = [Ag^+][Cl^-] 3. Concentration calculation: Let the solubility of AgCl be x mol/dm³. Then, K_sp = [x][x] = x². 1.8 × 10^-10 = x² x = √1.8 × 10^-10 = 1.34 × 10^-5 mol/dm³ The solubility (or concentration of Ag^+ and Cl^-) in water is approximately 1.34 × 10^-5 mol/dm³.

Key Steps for Calculating Ion Concentrations from K_sp

• Write the dissociation equation for the salt.
• Set up the K_sp expression.
• Solve for ion concentrations based on stoichiometry.

Intermediate Level

The solubility product constant, K_sp, is an equilibrium constant for the dissolution of a slightly soluble ionic compound. It applies only to saturated solutions, where the rate of dissolution equals the rate of precipitation. The K_sp expression is the product of the ion concentrations (each raised to the power of its coefficient in the balanced equation). Importantly, the concentration of the pure solid is not included because it is constant. If the product of the ion concentrations (the reaction quotient, Q) is less than K_sp, the solution is unsaturated, and no precipitate will form. If Q = K_sp, the solution is saturated. If Q > K_sp, the solution is supersaturated, and a precipitate will form until Q = K_sp.

Important Points

• - K_sp: Solubility product constant, an equilibrium constant for the dissolution of a slightly soluble salt.
• - Dynamic Equilibrium: A saturated solution is in dynamic equilibrium, where the rate of dissolution equals the rate of precipitation.
• - K_sp Expression: The product of the molar concentrations of the ions, each raised to the power of its coefficient in the balanced dissolution equation.
• - Pure Solids are not included in the K_sp expression (their concentration is constant).
• - K_sp is temperature-dependent.
• - The higher the K_sp value of a compound, the more soluble it is in water.

Remember

• - K_sp: Measures the extent of dissolution of a slightly soluble salt.
• - K_sp = [ion 1]^coefficient × [ion 2]^coefficient.
• - Pure solids = 1 (not included in the expression).
• - Higher K_sp = more soluble.
• - Example: K_sp(AgCl) = [Ag^+][Cl^-].
• - Example: K_sp(CaF_2) = [Ca^{2+}][F^-]².
• - Solubility is the amount of salt that dissolves per liter; it is not the same as K_sp, although they are related.

Common Mistakes

• - Including the concentration of the solid in the K_sp expression. The concentration of a pure solid is constant and is taken as 1. For example, for AgCl_{(s)} ⇌ Ag^+ + Cl^-, the K_sp is just [Ag^+][Cl^-], not [Ag^+][Cl^-]/[AgCl].
• - Confusing K_sp with solubility. K_sp is an equilibrium constant (unitless), while solubility is a concentration (e.g., mol/L). They are related, but they are not the same thing.
• - Forgetting to raise the ion concentrations to the correct powers. The exponent is the stoichiometric coefficient in the balanced equation. For CaF_2, it's [F^-]², not just [F^-].
• - Assuming that K_sp changes with concentration. K_sp is an equilibrium constant that depends only on temperature, not on the initial concentrations of the ions.
• - Using the wrong stoichiometry to calculate solubility from K_sp. The relationship between K_sp and molar solubility depends on the number of ions produced per formula unit.

Quick Revision

The solubility product constant (K_sp) is the equilibrium constant for the dissolution of a slightly soluble ionic compound. It is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation. The concentration of the pure solid is not included because it is constant. K_sp values are temperature-dependent and can be used to calculate the concentrations of ions in a saturated solution, or to predict whether a precipitate will form when solutions are mixed. The higher the K_sp value, the more soluble the compound.

60-Second Revision

The solubility product constant (K_sp) is an equilibrium constant for the dissolution of a slightly soluble salt. It is the product of the ion concentrations in a saturated solution, each raised to its stoichiometric coefficient. Pure solids are not included. The higher the K_sp, the more soluble the salt. K_sp is used to calculate ion concentrations in saturated solutions and to predict whether a precipitate will form when solutions are mixed.

Frequently Asked Questions (FAQs)

Q1. What is the solubility product constant (K_sp)? Answer: The solubility product constant (K_sp) is the equilibrium constant for the dissolution of a slightly soluble ionic compound. It is defined as the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced chemical equation.

Q2. Write the equation and the equilibrium expression for the dissolving of barium sulphate (BaSO_4). Also calculate the concentration of each ion if K_sp = 1 × 10^-10. Answer: The dissociation equation is: BaSO_{4(s)} ⇌ Ba^{2+}_{(aq)} + SO_4^{2-}_{(aq)} The K_sp expression is: K_sp = [Ba^{2+}][SO_4^{2-}] Let the solubility be x. Then [Ba^{2+}] = x and [SO_4^{2-}] = x. 1 × 10^-10 = x² x = √1 × 10^-10 = 1 × 10^-5 M So, the concentration of each ion is 1 × 10^-5 M.

Q3. Calculate the concentration of Pb^{2+} and F^- ions in a saturated solution of PbF_2 (K_sp = 4 × 10^-8). Answer: The dissociation equation is: PbF_{2(s)} ⇌ Pb^{2+}_{(aq)} + 2F^-_{(aq)} The K_sp expression is: K_sp = [Pb^{2+}][F^-]² Let the solubility of PbF_2 be x. Then [Pb^{2+}] = x and [F^-] = 2x. 4 × 10^-8 = (x)(2x)² = 4x³ x³ = 4 × 10^-84 = 1 × 10^-8 x = √[3]1 × 10^-8 = 2.15 × 10^-3 M So, [Pb^{2+}] = 2.15 × 10^-3 M and [F^-] = 2 × 2.15 × 10^-3 = 4.3 × 10^-3 M.

Q4. Why is the concentration of a pure solid not included in the K_sp expression? Answer: The concentration of a pure solid is constant and does not change with the amount of solid present. Its "concentration" is its density divided by its molar mass, which is a fixed value. Therefore, it is incorporated into the equilibrium constant and is not included in the K_sp expression.

Q5. What does a higher K_sp value indicate about a compound's solubility? Answer: A higher K_sp value indicates that the compound is more soluble. It means that at equilibrium, there are higher concentrations of ions in the saturated solution. However, comparing the solubilities of different salts based solely on K_sp is only valid for salts with the same stoichiometry (e.g., both AB type).

Q6. How can K_sp be used to predict whether a precipitate will form? Answer: To predict precipitation, you calculate the reaction quotient, Q, using the initial concentrations of the ions (before any precipitation occurs). If Q < K_sp, the solution is unsaturated, and no precipitate will form. If Q = K_sp, the solution is saturated. If Q > K_sp, the solution is supersaturated, and a precipitate will form until Q = K_sp.

Q7. What is the effect of a common ion on the solubility of a slightly soluble salt? Answer: The presence of a common ion (an ion that is already part of the salt) decreases the solubility of the salt. This is an application of Le Chatelier's Principle: the equilibrium shifts to the left to remove the added common ion, causing more of the salt to precipitate out of solution.

Q8. Is K_sp temperature-dependent? Answer: Yes, K_sp is an equilibrium constant, so it depends on temperature. The solubility of most solids increases with temperature, which means their K_sp values also increase with temperature.

Q9. What is the difference between solubility and K_sp? Answer: Solubility is the amount of a substance that dissolves in a given volume of solvent to form a saturated solution (usually expressed in mol/dm³ or g/100 mL). K_sp is the equilibrium constant for the dissolution process. While they are related, they are not the same. The relationship between them depends on the stoichiometry of the dissolution.

Multiple Choice Questions (MCQs)

Q1. What does the solubility product constant (K_sp) represent? A. The concentration of a saturated solution. B. The product of ion concentrations in a saturated solution. C. The solubility of a compound. D. The rate of dissolution. Correct Answer: B Explanation: K_sp is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced dissolution equation.

Q2. For the dissolution of AgCl_{(s)} ⇌ Ag^+ + Cl^-, what is the correct K_sp expression? A. K_sp = [Ag^+][Cl^-] / [AgCl] B. K_sp = [Ag^+][Cl^-] C. K_sp = [AgCl] / ([Ag^+][Cl^-]) D. K_sp = [Ag^+] + [Cl^-] Correct Answer: B Explanation: The concentration of the pure solid (AgCl) is constant and is not included in the K_sp expression. The K_sp is the product of the ion concentrations.

• Q3. If the K_sp of a salt is 1.0 × 10^-12, what is its molar solubility if it dissociates into one cation and one anion (e.g., AB type)?
• A. 1.0 × 10^-6 M
• B. 1.0 × 10^-12 M
• C. 1.0 × 10^-4 M
• D. 1.0 × 10^-24 M
• Correct Answer: A
• Explanation: For an AB type salt, K_sp = s², so s = √K_sp = √1.0 × 10^-12 = 1.0 × 10^-6 M.

• Q4. For a salt of the type AB_2, what is the relationship between K_sp and molar solubility (s)?
• A. K_sp = s²
• B. K_sp = 2s²
• C. K_sp = 4s³
• D. K_sp = 27s⁴
• Correct Answer: C
• Explanation: For AB_{2(s)} ⇌ A^{2+} + 2B^-, K_sp = [A^{2+}][B^-]² = (s)(2s)² = 4s³.

• Q5. What happens to the solubility of a slightly soluble salt when a common ion is added to the solution?
• A. It increases.
• B. It decreases.
• C. It remains the same.
• D. It becomes zero.
• Correct Answer: B
• Explanation: According to Le Chatelier's Principle, adding a common ion shifts the dissolution equilibrium to the left, decreasing the solubility of the salt.

• Q6. The K_sp of AgCl is 1.8 × 10^-10. What is the concentration of Ag^+ in a saturated solution of AgCl?
• A. 1.8 × 10^-10 M
• B. 1.34 × 10^-5 M
• C. 9.0 × 10^-11 M
• D. 3.6 × 10^-10 M
• Correct Answer: B
• Explanation: For AgCl, K_sp = s². So, s = √1.8 × 10^-10 = 1.34 × 10^-5 M. This is the concentration of Ag^+.

Q7. Why is the concentration of a pure solid not included in the K_sp expression? A. Because it is zero. B. Because it is a constant. C. Because it does not participate in the reaction. D. Because it is a catalyst. Correct Answer: B Explanation: The concentration (or activity) of a pure solid is constant and does not depend on the amount of solid present. It is incorporated into the equilibrium constant.

Q8. Which of the following salts has the highest K_sp value (assuming all have the same stoichiometry)? A. A salt with K_sp = 1 × 10^-10 B. A salt with K_sp = 1 × 10^-8 C. A salt with K_sp = 1 × 10^-6 D. A salt with K_sp = 1 × 10^-12 Correct Answer: C Explanation: A higher K_sp value indicates greater solubility (for salts with the same stoichiometry). 1 × 10^-6 is the largest value among the options.

• Q9. What is the name of the equilibrium constant for the dissolution of a slightly soluble ionic compound?
• A. K_a
• B. K_b
• C. K_w
• D. K_sp
• Correct Answer: D
• Explanation: The equilibrium constant for the dissolution of a slightly soluble salt is called the solubility product constant, K_sp.

Q10. If Q > K_sp, what will happen? A. The solution is unsaturated, and no precipitate forms. B. The solution is saturated, and no precipitate forms. C. A precipitate will form until Q = K_sp. D. The solid will dissolve until Q = K_sp. Correct Answer: C Explanation: If the reaction quotient Q is greater than K_sp, the solution is supersaturated, and a precipitate will form to reduce the ion concentrations until Q = K_sp.

Multiple Choice Questions (MCQs)

Q1. What is the solubility product constant (K_sp)? A. The product of ion concentrations in a saturated solution. B. The concentration of a saturated solution. C. The solubility of a compound. D. The rate of dissolution. Correct Answer: A Explanation: K_sp is the product of the molar concentrations of the ions in a saturated solution, each raised to the power of its coefficient in the balanced equation.

Q2. For the dissolution of CaF_2_{(s)} ⇌ Ca^{2+} + 2F^-, what is the correct K_sp expression? A. K_sp = [Ca^{2+}][F^-] B. K_sp = [Ca^{2+}][F^-]² C. K_sp = [Ca^{2+}]²[F^-] D. K_sp = [CaF_2] / ([Ca^{2+}][F^-]²) Correct Answer: B Explanation: The K_sp expression is the product of the ion concentrations, each raised to the power of its coefficient in the balanced equation. For CaF_2, it is [Ca^{2+}][F^-]².

• Q3. If the K_sp of a salt is 1.0 × 10^-10, what is its molar solubility if it dissociates into one cation and one anion (e.g., AB type)?
• A. 1.0 × 10^-5 M
• B. 1.0 × 10^-10 M
• C. 1.0 × 10^-6 M
• D. 1.0 × 10^-8 M
• Correct Answer: A
• Explanation: For an AB type salt, K_sp = s², so s = √K_sp = √1.0 × 10^-10 = 1.0 × 10^-5 M.

• Q4. What happens to the solubility of a slightly soluble salt when a common ion is added?
• A. It increases.
• B. It decreases.
• C. It remains the same.
• D. It becomes zero.
• Correct Answer: B
• Explanation: According to Le Chatelier's Principle, adding a common ion shifts the dissolution equilibrium to the left, decreasing the solubility of the salt.

• Q5. The K_sp of AgCl is 1.8 × 10^-10. What is the concentration of Ag^+ in a saturated solution of AgCl?
• A. 1.8 × 10^-10 M
• B. 1.34 × 10^-5 M
• C. 9.0 × 10^-11 M
• D. 3.6 × 10^-10 M
• Correct Answer: B
• Explanation: For AgCl, K_sp = s². So, s = √1.8 × 10^-10 = 1.34 × 10^-5 M. This is the concentration of Ag^+.

Q6. What is the effect of temperature on K_sp? A. K_sp is independent of temperature. B. K_sp always decreases with increasing temperature. C. K_sp always increases with increasing temperature. D. K_sp may increase or decrease with temperature, depending on the salt. Correct Answer: D Explanation: K_sp is an equilibrium constant, so it depends on temperature. For most salts, solubility increases with temperature, so K_sp increases. However, for some salts, the opposite is true.

• Q7. For a salt of the type AB_2, what is the relationship between K_sp and molar solubility (s)?
• A. K_sp = s²
• B. K_sp = 2s²
• C. K_sp = 4s³
• D. K_sp = 27s⁴
• Correct Answer: C
• Explanation: For AB_{2(s)} ⇌ A^{2+} + 2B^-, K_sp = [A^{2+}][B^-]² = (s)(2s)² = 4s³.

Q8. Which of the following salts has the highest solubility if all have the same K_sp value? A. AB type B. AB_2 type C. AB_3 type D. All have the same solubility. Correct Answer: A Explanation: For a given K_sp, the salt with the fewest ions per formula unit (AB type) has the highest molar solubility. For example, if K_sp = 1 × 10^-12, the solubility of AB is 1 × 10^-6 M, while that of AB_2 is 6.3 × 10^-5 M (wait, let's recalculate: for AB, s = √10^-12 = 10^-6; for AB_2, s = √[3]10^-12/4 = √[3]2.5 × 10^-13 = 6.3 × 10^-5. Actually, the AB type has lower solubility? Let's check: For AB, s = 10^-6. For AB_2, s = 6.3 × 10^-5. So AB_2 is more soluble. The correct answer is not A. Wait, I need to be careful. For a given K_sp, the salt with fewer ions per formula unit has higher molar solubility. For AB, s = √K_sp. For AB_2, s = √[3]K_sp/4. For K_sp = 10^-12, √10^-12 = 10^-6, while √[3]10^-12/4 = √[3]2.5 × 10^-13 ≈ 6.3 × 10^-5. So AB_2 has higher solubility. The question asks which has the highest solubility for the same K_sp. The salt with the smallest number of ions per formula unit (AB) has the lowest solubility? Actually, let's derive: For AB, s = √K_sp. For AB_2, s = (K_sp/4)¹³. For AB_3, s = (K_sp/27)¹⁴. Comparing these, for the same K_sp, the salt with more ions per formula unit has higher molar solubility. So the correct answer is C (AB_3). Let's correct the MCQ: The correct answer is C.

• Q9. What is the reaction quotient (Q) used for in precipitation reactions?
• A. To calculate K_sp.
• B. To determine if a precipitate will form by comparing Q to K_sp.
• C. To calculate the molar mass of the precipitate.
• D. To measure the temperature of the solution.
• Correct Answer: B
• Explanation: The reaction quotient Q is calculated using the initial ion concentrations. If Q > K_sp, a precipitate will form.

Q10. Why is the concentration of a pure solid not included in the K_sp expression? A. Because it is zero. B. Because it is a constant. C. Because it does not participate in the reaction. D. Because it is a catalyst. Correct Answer: B Explanation: The concentration (or activity) of a pure solid is constant and does not depend on the amount of solid present. It is incorporated into the equilibrium constant.