Solubility Product Constant (Ksp): Videos & Practice Problems

Solubility, a key chemical property, refers to a solute's ability to dissolve in a solvent, quantified by the solubility product constant (K_sp). A higher K_sp indicates 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 interdependence of these concepts in understanding ionic compounds.

Solubility Product Constant (Ksp) Concept 1

Solubility is a crucial 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 essential when analyzing the solubility 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 as follows:- Silver chloride (AgCl)- Magnesium carbonate (MgCO₃)- Calcium sulfate (CaSO₄)- Copper(II) sulfide (CuS)Among these, calcium sulfate has a Ksp of \(7.1 \times 10^{-5}\), which is the highest value when compared to the others. This means that calcium sulfate is the most soluble ionic solid in this group. Therefore, the correct answer is calcium sulfate, option C, as it has the largest Ksp value, indicating it can dissolve more readily in solution than the other substances listed.

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 represented by the equation:

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

In this reaction, the positive ions (cations) are listed first, followed by the negative ions (anions), reflecting the standard convention for writing ionic compounds. Understanding this breakdown is crucial for writing the equilibrium expression, which is derived from the concentrations of the products and reactants at equilibrium.

The equilibrium expression for this dissolution reaction can be formulated as:

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

Here, \( K_{sp} \) represents the solubility product constant, and it is important to note that the equilibrium expression only includes the concentrations of the ions in solution, while ignoring the solid ionic compound. This is because the concentration of solids and pure liquids remains constant and does not affect the equilibrium position.

In summary, mastering the breakdown of ionic solids into their respective ions is essential for accurately writing equilibrium expressions and understanding the dynamics of dissolution and precipitation in solution.

Solubility Product Constant (Ksp) Example 2

To derive the equilibrium expression for calcium nitrate, we begin 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 +2 charge, which is characteristic of Group 2A elements, and each nitrate ion (NO₃^-) carries a -1 charge. Therefore, for every formula unit of calcium nitrate that dissolves, one calcium ion and two nitrate ions are produced.

Next, we can formulate the equilibrium expression using the solubility product constant, K_sp. In equilibrium expressions, the concentration of solids is not included, as they are set to 1. Thus, 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 in front of the nitrate ions indicates that its concentration is squared in the expression. This results in the final equilibrium expression reflecting the relationship between the concentrations of the ions in solution.

In summary, the steps to derive the equilibrium expression involve dissociating the ionic solid into its ions and applying the principles of equilibrium to formulate the expression based on the concentrations of the products.

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) + PO₄^3- (aq)

Next, we write the K_sp expression based on this equilibrium equation. Since the solid does not appear in the expression, we focus on the aqueous ions:

K_sp = [Ag⁺]³ × [PO₄^3-]

To relate the concentrations of the ions to the solubility (x), we note that the concentration of silver ions will be 3x (due to the coefficient of 3 in front of Ag⁺), and the concentration of the phosphate ion will be x:

[Ag⁺] = 3x and [PO₄^3-] = x

Substituting these into the K_sp expression gives:

K_sp = (3x)³ × (x) = 27x⁴

Now, we substitute 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

Now, multiplying this result by 27 gives:

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 \(7.9 \times 10^{-10}\) 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 equilibrium expression based on this equation. The Ksp expression is:

Ksp = [Sr²⁺][F^-]²

Letting the solubility of strontium fluoride be represented by \(x\), we can express the concentrations of the ions at equilibrium. The concentration of strontium ions will be \(x\), while 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 rewrite the Ksp expression as:

Ksp = \(x(2x)^2\) = \(x(4x^2)\) = \(4x^3\)

Substituting the known Ksp value into the equation gives:

7.9 × 10^-10 = 4x³

To isolate \(x^3\), we divide both sides by 4:

x³ = \(\frac{7.9 \times 10^{-10}}{4}\) = \(1.975 \times 10^{-10}\)

Finally, to find \(x\), we take the cube root of both sides:

x = \(\sqrt[3]{1.975 \times 10^{-10}}\) ≈ \(5.8 \times 10^{-4}\) M

Thus, the solubility of strontium fluoride in molarity is approximately \(5.8 \times 10^{-4}\) 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 measures the solubility of ionic solids in a solvent. It is used to predict the extent to which a solute will dissolve in a solvent. The higher the Ksp value, the more soluble the ionic solid. Conversely, a lower Ksp value indicates lower solubility. Ksp is calculated using the concentrations of the ions in the solution at equilibrium, following the expression Ksp = [A^m+]^m[B^n-]ⁿ, where A and B are the ions formed from the dissociation of the ionic solid.

How do you calculate the Ksp value from solubility?

To calculate the Ksp value from solubility, follow these steps: 1) Write the dissociation equation for the ionic solid. 2) Express the concentrations of the ions in terms of the solubility (s). 3) Write the Ksp expression using the ion concentrations. 4) Substitute the solubility value into the Ksp expression and solve for Ksp. For example, for silver phosphate (Ag₃PO₄) with solubility s, the dissociation is Ag₃PO₄ → 3Ag⁺ + PO₄^3-. The Ksp expression is Ksp = [Ag⁺]³[PO₄^3-] = (3s)³(s) = 27s⁴.

How can 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 dissociation equation for the ionic solid. 2) Write the Ksp expression using the ion concentrations. 3) Let the solubility be represented by x. 4) Substitute x into the Ksp expression and solve for x. For example, for strontium fluoride (SrF₂) with Ksp = 7.9 × 10^-10, the dissociation is SrF₂ → Sr²⁺ + 2F^-. The Ksp expression is Ksp = [Sr²⁺][F^-]² = x(2x)² = 4x³. Solving for x gives the solubility.

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

The relationship between Ksp and the solubility of ionic solids is direct: a higher Ksp value indicates greater solubility, while a lower Ksp value indicates lower solubility. Ksp is the product of the molar concentrations of the ions at equilibrium, each raised to the power of their coefficients in the balanced dissociation equation. Therefore, the solubility of an ionic solid can be determined from its Ksp value, and vice versa, by using the appropriate equilibrium expressions and algebraic manipulations.

How do you write the equilibrium expression for a given ionic solid?

To write the equilibrium expression for a given ionic solid, follow these steps: 1) Write the balanced dissociation equation for the ionic solid. 2) Identify the ions produced and their stoichiometric coefficients. 3) Write the Ksp expression as the product of the molar concentrations of the ions, each raised to the power of their coefficients. For example, for calcium phosphate (Ca₃(PO₄)₂), the dissociation is Ca₃(PO₄)₂ → 3Ca²⁺ + 2PO₄^3-. The Ksp expression is Ksp = [Ca²⁺]³[PO₄^3-]².