Solubility Product Constant (Ksp): Videos & Practice Problems

Solubility, a key chemical property, indicates how well a solute dissolves in a solvent, represented by the solubility product constant (K_sp). A higher K_sp value signifies greater solubility of ionic solids. The dissolution process involves breaking ionic solids into their respective ions, leading to an equilibrium expression that excludes solids. K_sp can be calculated from known solubility values, while solubility can also be derived from known K_sp values, emphasizing the relationship between concentration and ionic dissociation.

Solubility Product Constant (Ksp) Concept 1

Solubility is a fundamental chemical property that describes how well a solute can dissolve in a solvent. A key concept related to solubility is the solubility product constant, denoted as K_sp. This constant serves as the equilibrium constant specifically for the solubility of ionic solids. When examining the solubility of these solids, it is important to understand that solubility can also be expressed in terms of concentration or molarity, represented by the symbol M.

The relationship between K_sp and solubility is straightforward: a higher K_sp value indicates greater solubility of the ionic solid, meaning more of it can dissolve in the solvent. Conversely, a lower K_sp value signifies that the ionic solid is less soluble, resulting in a smaller amount dissolving in the solvent. This understanding of K_sp is crucial when analyzing the solubility behavior of ionic compounds in various solutions.

Solubility Product Constant (Ksp) Example 1

To determine which substance is the most soluble among silver chloride, magnesium carbonate, calcium sulfate, and copper(II) sulfide, we can analyze their solubility product constants (Ksp). The Ksp value indicates the extent to which a compound can dissolve in water; a higher Ksp value corresponds to greater solubility.In this case, the Ksp values are compared, and the substance with the highest Ksp value is identified as the most soluble. For the given substances, calcium sulfate has a Ksp of \(7.1 \times 10^{-5}\), which is the largest among the options provided. Therefore, calcium sulfate is the most soluble ionic solid in this comparison.In summary, when evaluating solubility based on Ksp values, remember that a higher Ksp indicates a greater ability to dissolve, making it a crucial factor in determining solubility among ionic compounds.

Solubility Product Constant (Ksp) Concept 2

When an ionic solid is placed in a solvent, two competing processes occur: dissolution and precipitation. The dissolution process involves the ionic solid breaking apart into its constituent ions. For example, consider the ionic compound represented as \( \text{A}_2\text{B}_3 \). Upon dissolution, it separates into \( 2 \text{A}^{3+} \) ions and \( 3 \text{B}^{2-} \) ions in solution, which are denoted as aqueous species. This can be expressed as:

\[ \text{A}_2\text{B}_3 (s) \rightleftharpoons 2 \text{A}^{3+} (aq) + 3 \text{B}^{2-} (aq) \]

In this reaction, the positive ions (cations) are listed first, followed by the negative ions (anions), reflecting the standard convention for writing ionic compounds. The equilibrium between the dissolved ions and the solid can be described by an equilibrium expression, which is the ratio of the concentrations of the products to the reactants. It is important to note that the equilibrium expression excludes solids and liquids, focusing solely on the aqueous ions. Thus, the equilibrium expression for this dissolution can be written as:

\[ K_{sp} = \frac{[\text{A}^{3+}]^2 [\text{B}^{2-}]^3}{1} \]

Here, \( K_{sp} \) represents the solubility product constant, which quantifies the extent to which the ionic solid dissolves in the solvent. Understanding how to break down ionic solids into their respective ions is crucial for accurately writing the equilibrium expression and analyzing the solubility behavior of ionic compounds in solution.

Solubility Product Constant (Ksp) Example 2

To derive the equilibrium expression for calcium nitrate, we start by breaking down the ionic solid into its constituent aqueous ions. Calcium nitrate, represented as Ca(NO₃)₂, dissociates in solution as follows:

Ca(NO₃)₂ (s) → Ca²⁺ (aq) + 2 NO₃^- (aq)

In this dissociation, calcium (Ca) has a charge of +2, which is typical for elements in Group 2A of the periodic table. The nitrate ion (NO₃^-) carries a -1 charge, and since there are two nitrate ions, we represent this as 2 NO₃^-.

Next, we can formulate the equilibrium expression using the solubility product constant, K_sp. The general form of the equilibrium expression is given by the concentrations of the products raised to the power of their coefficients, divided by the concentration of the reactants raised to their coefficients. Since the reactant, calcium nitrate, is a solid, it is not included in the expression and is effectively set to 1.

The equilibrium expression for calcium nitrate can be written as:

K_sp = [Ca²⁺][NO₃^-]²

Here, [Ca²⁺] represents the concentration of calcium ions, and [NO₃^-] is the concentration of nitrate ions. The coefficient of 2 for the nitrate ions indicates that this concentration is squared in the expression. Thus, the final equilibrium expression captures the relationship between the solubility of calcium nitrate and the concentrations of its dissociated ions in solution.

Solubility Product Constant (Ksp) Concept 3

To calculate the solubility product constant (K_sp) for silver phosphate (Ag₃PO₄), we start by understanding the relationship between the solubility of the ionic solid and its dissociation into ions in solution. Given that the solubility of silver phosphate is 1.8 × 10⁻¹⁸ mol/L at 25 degrees Celsius, we can express the equilibrium dissociation as follows:

Ag₃PO₄ (s) ⇌ 3 Ag⁺ (aq) + PO₄³⁻ (aq)

From this equilibrium equation, we can derive the K_sp expression. Since the solid does not appear in the expression, we focus on the aqueous ions:

K_sp = [Ag⁺]³ × [PO₄³⁻]

Next, we define the concentrations of the ions in terms of the solubility (x). The coefficient of Ag⁺ is 3, leading to a concentration of 3x for silver ions, while the phosphate ion has a coefficient of 1, resulting in a concentration of x:

[Ag⁺] = 3x and [PO₄³⁻] = x

Substituting these into the K_sp expression gives:

K_sp = (3x)³ × (x) = 27x⁴

Now, substituting the given solubility value (x = 1.8 × 10⁻¹⁸) into the equation:

K_sp = 27 × (1.8 × 10⁻¹⁸)⁴

Calculating (1.8 × 10⁻¹⁸)⁴ first, we find:

(1.8 × 10⁻¹⁸)⁴ = 1.04976 × 10⁻⁷¹

Multiplying this result by 27 yields:

K_sp = 27 × 1.04976 × 10⁻⁷¹ = 2.8 × 10⁻⁷⁰

Thus, the K_sp for silver phosphate is 2.8 × 10⁻⁷⁰, which is reported with two significant figures, consistent with the significant figures of the solubility value provided.

Solubility Product Constant (Ksp) Concept 4

The solubility of an ionic solid can be calculated using its solubility product constant (Ksp). For strontium fluoride (SrF₂), the Ksp value is given as 7.9 × 10^-10 at 25 degrees Celsius. To find its solubility in molarity, we start by writing the equilibrium equation for the dissociation of strontium fluoride in water:

SrF₂(s) ⇌ Sr²⁺(aq) + 2F^-(aq)

Next, we formulate the Ksp expression based on the equilibrium concentrations of the ions:

Ksp = [Sr²⁺][F^-]²

Let the solubility of strontium fluoride be represented by x. This means that at equilibrium, the concentration of strontium ions will be x, and the concentration of fluoride ions will be 2x (since there are two fluoride ions produced for each formula unit of strontium fluoride). Thus, we can express Ksp as:

Ksp = (x)(2x)² = (x)(4x²) = 4x³

Substituting the known Ksp value into the equation gives us:

7.9 × 10^-10 = 4x³

To isolate x, we first divide both sides by 4:

x³ = 1.975 × 10^-10

Next, we take the cube root of both sides to solve for x:

x = (1.975 × 10^-10)^1/3

Calculating this gives us a final solubility of:

x ≈ 5.8 × 10^-4 M

Thus, the solubility of strontium fluoride in molarity is approximately 5.8 × 10^-4 M.

Determine the equilibrium expression of the barium nitride solid.

• A.
• B.
• C.
• D.

Answer: (C)

Manganese (V) hydroxide has a measured solubility of 3.4×10^–5 M at 25ºC. Calculate its K_sp value.

• A. 5.7×10^–9
• B. 1.5×10^–27
• C. 7.7×10^–27
• D. 4.8×10^–24

Answer: (D)

The K_sp value for strontium fluoride, SrF₂, is 7.9×10^–10 at 25ºC. Calculate its solubility in g/L.

• A. 5.8×10^–4 g/L
• B. 7.3x10^-2 g/L
• C. 4.6×10^–6 g/L
• D. 0.092 g/L

Answer: (B)

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 product of the molar concentrations of the constituent ions, each raised to the power of their coefficients in the balanced equation. Ksp is used to predict the solubility of a compound; a higher Ksp value indicates greater solubility, while a lower Ksp value indicates lesser solubility. It helps in calculating the solubility of a compound from its Ksp value and vice versa.

How do you calculate Ksp from solubility?

To calculate Ksp from solubility, follow these steps: 1) Write the balanced dissolution equation for the ionic solid. 2) Express the molar concentrations of the ions in terms of the solubility (s). 3) Write the Ksp expression using these concentrations. 4) Substitute the solubility value into the Ksp expression and solve. For example, for Ag₃PO₄ with solubility s, the equation is Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq). The Ksp expression is Ksp = [Ag⁺]³[PO₄³⁻] = (3s)³(s) = 27s⁴.

How do you determine the solubility of an ionic solid from its Ksp value?

To determine the solubility of an ionic solid from its Ksp value, follow these steps: 1) Write the balanced dissolution equation. 2) Write the Ksp expression. 3) Let the solubility be x and express the ion concentrations in terms of x. 4) Substitute the Ksp value and solve for x. For example, for SrF₂ with Ksp = 7.9 × 10⁻¹⁰, the equation is SrF₂(s) ⇌ Sr²⁺(aq) + 2F⁻(aq). The Ksp expression is Ksp = [Sr²⁺][F⁻]² = x(2x)² = 4x³. Solving 7.9 × 10⁻¹⁰ = 4x³ gives x = 5.8 × 10⁻⁴ M.

What is the relationship between Ksp and the solubility of ionic compounds?

The relationship between Ksp and the solubility of ionic compounds is direct: a higher Ksp value indicates a more soluble compound, while a lower Ksp value indicates a less soluble compound. Ksp quantifies the extent to which an ionic solid dissolves in a solvent to form its constituent ions. The solubility can be calculated from Ksp by setting up the equilibrium expression and solving for the ion concentrations. Conversely, Ksp can be determined from the known solubility of the compound.

How do you write the equilibrium expression for a dissolution reaction?

To write the equilibrium expression for a dissolution reaction, follow these steps: 1) Write the balanced chemical equation for the dissolution of the ionic solid. 2) Identify the ions produced and their stoichiometric coefficients. 3) Write the expression as the product of the molar concentrations of the ions, each raised to the power of their coefficients. For example, for the dissolution of CaF₂: CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), the equilibrium expression is Ksp = [Ca²⁺][F⁻]².