Solubilty Product Constant: Videos & Practice Problems

Solubility indicates the maximum amount of solute that can dissolve in a solvent, forming a homogeneous mixture called a solution. The solubility product constant (Ksp) relates to ionic solids; a larger Ksp signifies greater solubility. The reaction quotient (q) helps determine if a reaction is at equilibrium by comparing it to Ksp. If q equals Ksp, the system is at equilibrium. If q is greater than Ksp, the reaction shifts toward reactants, increasing precipitate. Conversely, if q is less than Ksp, the reaction shifts toward products, enhancing solubility and ion production.

Solubility Product Constant

Solubility is defined as the maximum amount of solute that can dissolve in a solvent, resulting in a homogeneous mixture known as a solution. Each ionic solid has an associated value called the solubility product constant, denoted as K_sp. This constant is crucial in understanding the solubility of ionic compounds in a solvent.

The K_sp value indicates the extent to which an ionic solid can dissolve. A larger K_sp value signifies greater solubility, meaning more of the ionic solid dissociates into its constituent ions in the solution. Conversely, a smaller K_sp value indicates lower solubility, suggesting that less of the ionic compound will dissolve in the solvent.

When analyzing the solubility of any ionic compound, it is essential to consider its K_sp value, as it provides insight into the compound's behavior in solution. Understanding how to calculate K_sp from molar solubility is a key skill in chemistry, particularly when dealing with hypothetical ionic compounds.

Solubility Product Constant

In this scenario, we are analyzing a hypothetical ionic compound represented by the formula MX₃, which has a molar solubility of 0.00562 M. To determine the solubility product constant (K_sp) for this compound, we first need to understand how it dissociates in solution. The compound dissociates into one metal ion (M³⁺) and three negative ions (X^-), leading to the equilibrium expression for K_sp.

When setting up the equilibrium, we can use an ICE (Initial, Change, Equilibrium) chart. Initially, the concentrations of the ions are zero. As the compound dissolves, the change in concentration for M³⁺ is +x, and for X^-, it is +3x. Therefore, at equilibrium, the concentrations are:

• [M³⁺] = x
• [X^-] = 3x

Since K_sp is defined as the product of the concentrations of the ions raised to the power of their coefficients in the balanced equation, we can express K_sp as:

K_sp = [M³⁺] × [X^-]³

Substituting the equilibrium concentrations into the expression gives:

K_sp = (x) × (3x)³ = x × 27x³ = 27x⁴

Given that the molar solubility (0.00562 M) represents the value of x, we can substitute this into the K_sp equation:

K_sp = 27 × (0.00562)⁴

Calculating (0.00562)⁴ yields approximately 9.9754 × 10^-10. Multiplying this by 27 results in:

K_sp ≈ 2.69 × 10^-8

This value represents the solubility product constant for the compound MX₃. Understanding how to derive K_sp from molar solubility is crucial for solving similar problems in chemistry.

In subsequent examples, one may be tasked with determining the highest molar solubility among various ionic compounds. This involves working backwards from the K_sp values to isolate the x variable, allowing for a comparison of solubility across different compounds.

Solubility Product Constant

To determine which ionic compound has the highest molar solubility in pure water, the first step is to analyze how many ions each compound dissociates into when dissolved. For example, cobalt(II) hydroxide dissociates into one cobalt(II) ion and two hydroxide ions, totaling three ions. Strontium phosphate breaks down into three strontium ions and two phosphate ions, resulting in five ions. Lead(II) chloride yields one lead(II) ion and two chloride ions, also totaling three ions. Silver cyanide dissociates into one silver ion and one cyanide ion, giving two ions. Finally, lead(II) sulfate breaks down into one lead(II) ion and one sulfate ion, resulting in two ions.

Next, we compare the compounds with the same number of ions. Cobalt(II) hydroxide and lead(II) chloride both yield three ions. By examining their solubility product constants (K_sp), we find that lead(II) chloride has a higher K_sp value of 1.60 × 10^-5, indicating it is more soluble than cobalt(II) hydroxide. Therefore, lead(II) chloride has a greater molar solubility.

For silver cyanide and lead(II) sulfate, both dissociate into two ions. The K_sp for silver cyanide is 1.0 × 10^-8, while lead(II) sulfate has a K_sp of 1.82 × 10^-8. This shows that lead(II) sulfate is more soluble, thus having a higher molar solubility than silver cyanide.

Now, we focus on strontium phosphate, lead(II) chloride, and lead(II) sulfate, which dissociate into different numbers of ions. To find the molar solubility (denoted as x) for strontium phosphate, we set up the expression based on its dissociation: K_sp = [Sr²⁺]³ × [PO₄^3-]². This results in K_sp = (3x)³ × (2x)² = 108x⁵. Given K_sp = 4.0 × 10^-28, we solve for x, yielding a molar solubility of approximately 1.30 × 10^-6 M.

For lead(II) chloride, the dissociation gives K_sp = [Pb²⁺] × [Cl^-]² = x × (2x)² = 4x³. With K_sp = 1.60 × 10^-5, solving for x results in a molar solubility of about 0.0159 M.

Lastly, for lead(II) sulfate, we have K_sp = [Pb²⁺] × [SO₄^2-] = x². Given K_sp = 1.82 × 10^-8, solving for x gives a molar solubility of approximately 1.35 × 10^-4 M.

Comparing the molar solubilities, lead(II) chloride has the highest value, followed by lead(II) sulfate, and then strontium phosphate. Thus, the compound with the highest molar solubility in pure water is lead(II) chloride, as it has the largest x value.

In summary, the strategy for determining molar solubility involves first identifying the number of ions produced by each ionic compound, comparing their K_sp values for those that dissociate into the same number of ions, and calculating the molar solubility for those that dissociate into different numbers of ions. This systematic approach allows for the identification of the most soluble compound.

Solubility Product Constant Calculations 1

Lanthanum(III) hydroxide, with a solubility product constant (K_sp) of 2.0 × 10^-21, dissolves in water to establish an equilibrium involving lanthanum ions (La³⁺) and hydroxide ions (OH^-). The dissolution can be represented by the equation:

La(OH)₃ (s) ⇌ La³⁺ (aq) + 3 OH^- (aq)

To analyze the solubility, an ICE (Initial, Change, Equilibrium) chart is utilized, focusing on the aqueous products while ignoring the solid reactant. Initially, the concentrations of La³⁺ and OH^- are zero. As the solid dissolves, the changes in concentration can be expressed as:

Initial: 0, 0

Change: +x, +3x

Equilibrium: x, 3x

The K_sp expression for this equilibrium is:

K_sp = [La³⁺][OH^-]³

Substituting the equilibrium concentrations into the K_sp expression gives:

2.0 × 10^-21 = (x)(3x)³ = 27x⁴

To find x, rearranging the equation yields:

x⁴ = (2.0 × 10^-21) / 27

Calculating this gives:

x⁴ = 7.40741 × 10^-23

Taking the fourth root of both sides results in:

x = 2.934 × 10^-6 M

This value represents the solubility of lanthanum(III) hydroxide. To find the concentration of hydroxide ions, we use the relationship OH^- = 3x:

OH^- = 3(2.934 × 10^-6) = 8.80 × 10^-6 M

Given a saturated solution volume of 2.5 liters, the moles of hydroxide ions can be calculated as:

Moles of OH^- = Concentration × Volume = (8.80 × 10^-6 M)(2.5 L) = 2.20 × 10^-5 moles

To convert moles of hydroxide ions to grams, the molar mass of OH^- is determined by adding the atomic masses of oxygen (15.99994 g/mol) and hydrogen (1.00794 g/mol), resulting in approximately 17.0073 g/mol. Thus, the mass of hydroxide ions is:

Mass = Moles × Molar Mass = (2.20 × 10^-5 moles)(17.0073 g/mol) = 3.74 × 10^-4 grams

This calculation illustrates the importance of carefully interpreting the question, as it specifically asks for the mass of hydroxide ions rather than the entire ionic compound. Understanding the relationships between solubility, concentration, and the stoichiometry of the dissolution process is crucial for solving such problems effectively.

Solubility Product Constant Calculations 1

To determine the pH of a saturated solution of aluminum hydroxide, we start with the solubility product constant (K_sp) of aluminum hydroxide, which is given as 1.9 × 10⁻¹⁰. Aluminum hydroxide dissociates in water according to the following equilibrium reaction:

Al(OH)₃ (s) ⇌ Al³⁺ (aq) + 3 OH⁻ (aq)

In this reaction, we can set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of the ions. Initially, the concentrations of Al³⁺ and OH⁻ are both 0. As the solid dissolves, we let the change in concentration of Al³⁺ be +x, leading to a change of +3x for OH⁻ due to the stoichiometry of the reaction.

At equilibrium, the concentrations are:

• [Al³⁺] = x
• [OH⁻] = 3x

Using the K_sp expression, we have:

K_sp = [Al³⁺][OH⁻]³ = x(3x)³ = 27x⁴

Setting this equal to the K_sp value, we get:

27x⁴ = 1.9 × 10⁻¹⁰

Solving for x, we find:

x⁴ = 7.03704 × 10⁻¹²

Taking the fourth root gives us:

x = 1.629 × 10⁻³ M

This value represents the concentration of Al³⁺. To find the concentration of OH⁻, we calculate:

[OH⁻] = 3x = 3 × 1.629 × 10⁻³ = 4.886 × 10⁻³ M

Next, we can find the pOH using the formula:

pOH = −log[OH⁻] = −log(4.886 × 10⁻³) = 2.31

Finally, we can calculate the pH using the relationship:

pH = 14 − pOH = 14 − 2.31 = 11.69

Thus, the pH of the saturated solution of aluminum hydroxide is 11.69.

Solubility Product Constant Calculations

To calculate the molar solubility of iron(II) carbonate in a sodium carbonate solution, we start with the solubility product constant (K_sp) for iron(II) carbonate, which is given as 2.1 × 10⁻¹¹. Iron(II) carbonate dissociates in solution into iron ions (Fe²⁺) and carbonate ions (CO₃²⁻). The dissociation can be represented as:

FeCO₃ (s) ⇌ Fe²⁺ (aq) + CO₃²⁻ (aq)

In this scenario, we are dealing with a solution that already contains carbonate ions from sodium carbonate, which dissociates into sodium ions (Na⁺) and carbonate ions (CO₃²⁻). The initial concentration of carbonate ions in the solution is 0.00167 M, which introduces the common ion effect. This effect reduces the solubility of iron(II) carbonate compared to its solubility in pure water.

Let x represent the molar solubility of iron(II) carbonate. The equilibrium concentrations of the ions in the solution will be:

• [Fe²⁺] = x
• [CO₃²⁻] = 0.00167 + x

Using the K_sp expression:

K_sp = [Fe²⁺][CO₃²⁻]

Substituting the values into the equation gives:

2.1 × 10⁻¹¹ = x(0.00167 + x)

Since K_sp values are typically very low, we can assume that x is much smaller than 0.00167 M, allowing us to simplify the equation to:

2.1 × 10⁻¹¹ ≈ x(0.00167)

Solving for x, we divide both sides by 0.00167:

x ≈ \(\frac{2.1 \times 10^{-11}}{0.00167}\)

Calculating this gives:

x ≈ 1.26 × 10⁻⁸ M

This value represents the molar solubility of iron(II) carbonate in a sodium carbonate solution, illustrating the impact of the common ion effect on solubility. Understanding this concept is crucial for predicting how the presence of common ions can influence the solubility of ionic compounds in solution.

What is the molar solubility of Fe(OH) ₃ (s) in a solution that is buffered at pH 3.50 at 25°C? The Ksp of Fe(OH)₃ is 6.3 x 10^–38 at 25°C.

• A. 1.99x10^-27 M
• B. 1.99x10^-34 M
• C. 6.64x10^-28 M
• D. 1.99x10^-6 M
• E. 1.25x10^-9 M

Answer: (D)

Arsenic trisulfide (As₂S₃) occurs naturally as the orange-yellow colored mineral orpiment. As₂S₃ is a highly insoluble substance with a Ksp value of 2.90×10⁻⁷². Calculate the solubility of As₂S₃ in g/100mL. (Molar mass of As₂S₃ = 246.04 g/mol)

• A. 2.12×10⁻¹⁷ g/100 mL
• B. 1.87×10⁻¹³ g/100 mL
• C. 6.25×10⁻¹⁵ g/100 mL
• D. 4.75×10⁻¹⁴ g/100 mL

Answer: (D)

Reaction Quotient & Precipitation

The reaction quotient, denoted as \( q \), plays a crucial role in understanding the state of a chemical reaction, particularly in relation to equilibrium. It is compared to the equilibrium constant \( K \) to determine if a reaction is at equilibrium. Additionally, when dealing with ionic solids, the solubility product constant \( K_{sp} \) is used, which is also an equilibrium constant but specifically for ionic compounds in solution.

When \( q \) is equal to \( K \) or \( K_{sp} \), it indicates that the reaction is at equilibrium. For instance, consider the dissociation of silver chloride (\( AgCl \)) into silver ions (\( Ag^+ \)) and chloride ions (\( Cl^- \)). If the initial concentrations of \( Ag^+ \) and \( Cl^- \) are both \( 1.3 \times 10^{-5} \) M, we can calculate \( q \) using the formula:

\( q = [Ag^+][Cl^-] \)

Substituting the values gives:

\( q = (1.3 \times 10^{-5})(1.3 \times 10^{-5}) = 1.77 \times 10^{-10} \)

Since this value of \( q \) matches the \( K_{sp} \) of \( 1.77 \times 10^{-10} \), we conclude that the system is at equilibrium, meaning there will be no shift in the reaction direction.

However, if future calculations yield a \( q \) that differs from \( K \) or \( K_{sp} \), this discrepancy will indicate the direction in which the reaction will shift to re-establish equilibrium. Such shifts can affect the concentrations of reactants and products, illustrating the dynamic nature of chemical equilibria.

Reaction Quotient & Precipitation

In the study of chemical equilibria, understanding the relationship between the reaction quotient (Q) and the solubility product constant (K_sp) is crucial. The reaction quotient, Q, is calculated using the current concentrations of the reactants and products, while K_sp is a constant that reflects the solubility of a compound at a specific temperature. When Q is compared to K_sp, it provides insight into the direction in which a reaction will shift to reach equilibrium.

When Q is greater than K_sp, the system is said to be supersaturated, indicating an excess of ions in solution. In this scenario, the reaction will shift towards the reactants to reduce the concentration of ions, leading to the formation of a precipitate. For example, if the concentrations of ions are 0.250 M, the calculated Q value may be 6.25 × 10^-2, which is greater than K_sp. This shift towards the reactants results in an increase in the solid precipitate while decreasing the concentration of the products.

Conversely, when Q is less than K_sp, the solution is unsaturated, meaning there are fewer ions than the solubility limit. In this case, the reaction will shift towards the products, increasing the concentration of ions in solution. For instance, if a new set of concentrations yields a Q value of 5.06 × 10^-24, which is less than K_sp, the reaction will favor the formation of more ions, thereby decreasing the amount of solid precipitate present.

At equilibrium, when Q equals K_sp, the system is in a balanced state, and the solution is described as saturated. This indicates that the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants and products.

In summary, the comparison of Q and K_sp is essential for predicting the behavior of ionic compounds in solution. A greater Q than K_sp leads to a shift towards the reactants and the formation of a precipitate, while a smaller Q than K_sp results in a shift towards the products, increasing solubility. Understanding these dynamics prepares students for more complex calculations involving the formation of precipitates and the solubility of ionic compounds.

Reaction Quotient & Precipitation Calculations 1

When mixing 0.150 liters of 0.100 molar lead(II) acetate with 0.100 liters of 0.20 molar sodium chloride, we need to determine if a precipitate will form, specifically lead(II) chloride, which has a solubility product constant (K_sp) of 1.2 × 10^-5.

Lead(II) chloride dissociates in solution into lead(II) ions (Pb²⁺) and chloride ions (Cl^-). To analyze the situation, we first calculate the concentrations of these ions after mixing the solutions. For lead(II) ions, we find the moles from lead(II) acetate:

Number of moles of Pb²⁺ = 0.150 L × 0.100 mol/L = 0.0150 moles.

Since there is one mole of Pb²⁺ per mole of lead(II) acetate, we have 0.0150 moles of Pb²⁺. The total volume after mixing is 0.150 L + 0.100 L = 0.250 L. Thus, the concentration of Pb²⁺ is:

[Pb²⁺] = 0.0150 moles / 0.250 L = 0.060 mol/L.

Next, we calculate the concentration of chloride ions from sodium chloride:

Number of moles of Cl^- = 0.100 L × 0.20 mol/L = 0.0200 moles.

Since there is one mole of Cl^- per mole of sodium chloride, we have 0.0200 moles of Cl^-. The concentration of Cl^- is:

[Cl^-] = 0.0200 moles / 0.250 L = 0.080 mol/L.

Now, we can calculate the reaction quotient (Q) for the formation of lead(II) chloride:

Q = [Pb²⁺][Cl^-]² = (0.060 mol/L)(0.080 mol/L)² = 3.84 × 10^-4.

To determine if a precipitate will form, we compare Q to K_sp. Since Q (3.84 × 10^-4) is greater than K_sp (1.2 × 10^-5), the solution is supersaturated with respect to lead(II) chloride. This indicates that the system is not at equilibrium, and the reaction will shift towards the formation of more solid lead(II) chloride to reach equilibrium.

In conclusion, because Q is greater than K_sp, a precipitate of lead(II) chloride will indeed form as the system adjusts to restore equilibrium.

Reaction Quotient & Precipitation Calculations 1

To determine the minimum pH at which iron(II) hydroxide (Fe(OH)₂) will precipitate from a solution containing iron(II) ions (Fe²⁺), we start with the solubility product constant (K_sp) of iron(II) hydroxide, which is given as 4.87 × 10⁻¹⁷. The concentration of Fe²⁺ in the solution is 0.0583 M.

The dissolution of iron(II) hydroxide can be represented by the equilibrium:

Fe(OH)₂ (s) ⇌ Fe²⁺ (aq) + 2 OH⁻ (aq)

At equilibrium, the K_sp expression is:

K_sp = [Fe²⁺][OH⁻]²

Given that the initial concentration of hydroxide ions is zero, we can denote the change in concentration of hydroxide ions as 2x, where x is the change in concentration of Fe²⁺. Thus, at equilibrium:

[Fe²⁺] = 0.0583 M

[OH⁻] = 2x

Substituting these values into the K_sp expression gives:

4.87 × 10⁻¹⁷ = (0.0583)(2x)²

Calculating the right side:

4.87 × 10⁻¹⁷ = (0.0583)(4x²)

4.87 × 10⁻¹⁷ = 0.2332x²

Solving for x²:

x² = (4.87 × 10⁻¹⁷) / 0.2332

x² = 2.08834 × 10⁻¹⁶

Taking the square root gives:

x = 1.445 × 10⁻⁸ M

Since [OH⁻] = 2x, we find:

[OH⁻] = 2(1.445 × 10⁻⁸) = 2.89 × 10⁻⁸ M

Next, we calculate the pOH:

pOH = −log(2.89 × 10⁻⁸) = 7.54

Finally, to find the pH, we use the relationship:

pH = 14 − pOH

pH = 14 − 7.54 = 6.46

Thus, the minimum pH at which iron(II) hydroxide will precipitate is 6.46. This calculation illustrates the common ion effect, where the presence of Fe²⁺ ions influences the solubility of Fe(OH)₂ in the solution.

Here’s what students ask on this topic:

What is the solubility product constant (Ksp) and how is it used?

The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of ionic solids in a solvent. It represents the maximum amount of solute that can dissolve to form a saturated solution. The Ksp value is calculated using the concentrations of the ions in the solution at equilibrium. A larger Ksp indicates greater solubility, meaning more of the ionic solid can dissolve. Conversely, a smaller Ksp indicates lower solubility. Ksp is used to predict whether a precipitate will form in a solution and to calculate the solubility of ionic compounds.

How do you calculate the solubility product constant (Ksp) from molar solubility?

To calculate the solubility product constant (Ksp) from molar solubility, follow these steps: 1) Write the balanced dissolution equation for the ionic compound. 2) Express the molar solubility (s) in terms of the ion concentrations. 3) Substitute the ion concentrations into the Ksp expression. For example, for a compound AB that dissociates into A⁺ and B^-, the dissolution equation is AB ⇌ A⁺ + B^-. If the molar solubility is s, then [A⁺] = s and [B^-] = s. The Ksp expression is Ksp = [A⁺][B^-] = s². Solve for Ksp using the given molar solubility.

What is the relationship between the reaction quotient (Q) and the solubility product constant (Ksp)?

The reaction quotient (Q) and the solubility product constant (Ksp) are used to determine the state of a solution. Q is calculated using the initial concentrations of the ions in the solution, while Ksp is the equilibrium constant for the dissolution of an ionic solid. If Q = Ksp, the solution is at equilibrium, and no net change occurs. If Q > Ksp, the solution is supersaturated, and a precipitate will form as the reaction shifts toward the reactants. If Q < Ksp, the solution is unsaturated, and more solute can dissolve as the reaction shifts toward the products.

How does temperature affect the solubility product constant (Ksp)?

Temperature can significantly affect the solubility product constant (Ksp). Generally, for most ionic compounds, an increase in temperature leads to an increase in solubility, resulting in a higher Ksp value. This is because higher temperatures provide more energy for the dissolution process, allowing more solute to dissolve. Conversely, a decrease in temperature usually decreases solubility, leading to a lower Ksp value. However, the exact effect of temperature on Ksp depends on the specific ionic compound and the enthalpy change of the dissolution process.

What happens when the reaction quotient (Q) is greater than the solubility product constant (Ksp)?

When the reaction quotient (Q) is greater than the solubility product constant (Ksp), the solution is supersaturated. This means there are more ions in the solution than can be maintained at equilibrium. As a result, the reaction will shift toward the reactants to reduce the ion concentrations and form a precipitate. This shift continues until Q equals Ksp, reestablishing equilibrium. The formation of a precipitate reduces the ion concentrations in the solution, bringing the system back to a saturated state.