Question

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{CaF}_{2}(s), K_{\mathrm{sp}}=4.0 \times 10^{-11}\), or \(\mathrm{BaF}_{2}(s), K_{\mathrm{sp}}=2.4 \times 10^{-5}\) b. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s), K_{\infty}=1.3 \times 10^{-32}\), or \(\mathrm{FePO}_{4}(s), K_{\mathrm{sp}}=1.0 \times 10^{-22}\)

Short answer

a. The molar solubility of CaF2 is \(x = 2.1 \times 10^{-4}\) M and the molar solubility of BaF2 is \(y = 6.9 \times 10^{-3}\) M. Thus, CaF2(s) has the smallest molar solubility. b. The molar solubility of Ca3(PO4)2 is \(p = 1.6 \times 10^{-8}\) M and the molar solubility of FePO4 is \(q = 1.0 \times 10^{-11}\) M. Thus, FePO4(s) has the smallest molar solubility.

Step-by-step solution

Solving for molar solubility of CaF2

Given the balanced chemical equation for the dissolution of CaF2: \[ \mathrm{CaF}_{2}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^-(aq) \] Let the molar solubility of CaF2 be 'x'. Hence, at equilibrium: \[ [\mathrm{Ca}^{2+}] = x \] \[ [\mathrm{F}^-] = 2x \] The Ksp expression for CaF2 is given by: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 4.0 \times 10^{-11} = (x)(2x)^2 \] Solve for 'x', which gives the molar solubility of CaF2.

Solving for molar solubility of BaF2

Given the balanced chemical equation for the dissolution of BaF2: \[ \mathrm{BaF}_{2}(s) \rightarrow \mathrm{Ba}^{2+}(aq) + 2\mathrm{F}^-(aq) \] Let the molar solubility of BaF2 be 'y'. Hence, at equilibrium: \[ [\mathrm{Ba}^{2+}] = y \] \[ [\mathrm{F}^-] = 2y \] The Ksp expression for BaF2 is given by: \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^-]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 2.4 \times 10^{-5} = (y)(2y)^2 \] Solve for 'y', which gives the molar solubility of BaF2. Upon calculating the molar solubility of CaF2(s) and BaF2(s), we can determine which one has the smaller molar solubility.

Solving for molar solubility of Ca3(PO4)2

Given the balanced chemical equation for the dissolution of Ca3(PO4)2: \[ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightarrow 3\mathrm{Ca}^{2+}(aq) + 2\mathrm{PO}_4^{3-}(aq) \] Let the molar solubility of Ca3(PO4)2 be 'p'. Hence, at equilibrium: \[ [\mathrm{Ca}^{2+}] = 3p \] \[ [\mathrm{PO}_4^{3-}] = 2p \] The Ksp expression for Ca3(PO4)2 is given by: \[ K_{sp} = [\mathrm{Ca}^{2+}]^3[\mathrm{PO}_4^{3-}]^2 \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 1.3 \times 10^{-32}=(3p)^3(2p)^2 \] Solve for 'p', which gives the molar solubility of Ca3(PO4)2.

Solving for molar solubility of FePO4

Given the balanced chemical equation for the dissolution of FePO4: \[ \mathrm{FePO}_{4}(s) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{PO}_4^{3-}(aq) \] Let the molar solubility of FePO4 be 'q'. Hence, at equilibrium: \[ [\mathrm{Fe}^{3+}] = q \] \[ [\mathrm{PO}_4^{3-}] = q \] The Ksp expression for FePO4 is given by: \[ K_{sp} = [\mathrm{Fe}^{3+}][\mathrm{PO}_4^{3-}] \] Plug in the known Ksp value and the above expressions for the equilibrium concentrations: \[ 1.0 \times 10^{-22} = (q)(q) \] Solve for 'q', which gives the molar solubility of FePO4. Upon calculating the molar solubility of Ca3(PO4)2(s) and FePO4(s), we can determine which one has the smaller molar solubility.

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is a crucial concept in chemistry that helps to measure how much a substance dissolves in water. This constant is specifically associated with sparingly soluble salts. When a solid dissolves in water, it breaks down into its constituent ions, and Ksp quantifies this process in terms of ion concentrations. Each salt has its own unique Ksp value.

Ksp is calculated using the concentrations of the resulting ions at equilibrium, according to the balanced dissolution equation. For example, the dissolution of calcium fluoride, CaF\( _2 \), into calcium ions, Ca\( ^{2+} \), and fluoride ions, F\( ^- \), can be expressed with its Ksp equation as follows:

• K\( _{sp} \) for CaF\( _2 \): \( [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2 \).

When comparing two salts, their respective Ksp values tell us which one dissolves more in water. A smaller Ksp indicates lower solubility under similar conditions, meaning fewer ions are present in the solution at equilibrium.

Understanding Ksp helps predict whether a precipitate will form in a chemical reaction and is especially important in analyzing solutions in laboratory and industrial applications.

Equilibrium Concentration

Equilibrium concentration describes the amount of ions present in a solution when a solid is in a state of dynamic equilibrium with its dissolved ions in a saturated solution. At equilibrium, the rate at which the solid dissolves equals the rate at which the ions recombine to form the solid, creating a stable concentration of ions.

In dissolution reactions, like that of barium fluoride, BaF\(_2\), equilibrium means that any shift in concentration will adjust until the solution stabilizes. When calculating molar solubility, as shown for BaF\(_2\), these concentrations determine the values substituted back into the Ksp expression:

• For BaF\(_2\): \([\mathrm{Ba}^{2+}] = y \), \([\mathrm{F}^-] = 2y \).

These expressions help us find the molar solubility, which indicates how much solute can dissolve in the solvent to reach saturation.

Grasping equilibrium concentrations is essential for estimating reaction outcomes, creating standard solutions, and anticipating material behaviour in varied chemical contexts.

Chemical Dissolution Reactions

Chemical dissolution reactions describe the process by which a solid (usually a salt) dissolves in a solvent (typically water), leading to the formation of ions in solution. These reactions are vital for understanding how substances interact with their environment, especially in fields like biochemistry, geology, and environmental science.

In a dissolution reaction, the solid involved breaks into its constituent ions. For instance, dissolving iron(III) phosphate, FePO\(_4\), produces iron(III) ions, Fe\( ^{3+} \), and phosphate ions, PO\(_4^{3-}\), as follows:

• \( \mathrm{FePO}_{4}(s) \rightarrow \mathrm{Fe}^{3+}(aq) + \mathrm{PO}_4^{3-}(aq) \)

These reactions are essential for determining solubility, as they provide the basis for calculating Ksp and equilibrium concentrations. Each reaction must be balanced and clearly show the transformations of substances from solid to ionic form.

Learning about these reactions helps predict and manipulate solubility, one of the key properties affecting everything from nutrient absorption in plants to pollutant release in ecosystems.