Question

What are \(\left[\mathrm{Ba}^{2+}\right]\) and \(\left[\mathrm{F}^{-}\right]\) in a saturated solution of \(\mathrm{BaF}_{2}(\mathrm{~s})\) ? The \(K_{\text {sp }}\) of \(\mathrm{BaF}_{2}\) (s) is \(1.8 \times 10^{-7}\).

Short answer

\([\mathrm{Ba}^{2+}] = 3.56 \times 10^{-3}\, \text{mol/L}\) and \([\mathrm{F}^{-}] = 7.12 \times 10^{-3}\, \text{mol/L}\).

Step-by-step solution

Understand the Dissolution Equation

The dissolution of barium fluoride, \(\text{BaF}_2\), in water can be represented by the equilibrium reaction: \[\text{BaF}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2\text{F}^{-} (aq)\]. The solubility product constant, \(K_{sp}\), is given for this equilibrium system.

Set Up the Expression for Ksp

For the dissolution, \(K_{sp}\) is given by the expression: \[K_{sp} = [\text{Ba}^{2+}][\text{F}^{-}]^2\]. This represents the product of the ion concentrations at equilibrium.

Assign Solubility Variables

Let \(s\) be the solubility of \(\text{BaF}_2\) in mol/L. At equilibrium, the concentration of \([\text{Ba}^{2+}]\) is \(s\) and \([\text{F}^{-}]\) is \(2s\) because two fluoride ions are produced for each formula unit of \(\text{BaF}_2\) that dissolves.

Substitute Variables into Ksp Expression

Substitute \([\text{Ba}^{2+}] = s\) and \([\text{F}^{-}] = 2s\) into the \(K_{sp}\) expression: \[K_{sp} = (s)(2s)^2 = 4s^3\].

Solve for Solubility, s

Set \[4s^3 = K_{sp}\] where \(K_{sp} = 1.8 \times 10^{-7}\). Solve for \(s\): \[s^3 = \frac{1.8 \times 10^{-7}}{4} = 4.5 \times 10^{-8}\]. Take the cube root of the result to find \(s\).

Calculate Ion Concentrations

The cube root of \(4.5 \times 10^{-8}\) gives \(s \approx 3.56 \times 10^{-3}\, \text{mol/L}\). Therefore, \([\text{Ba}^{2+}] = s = 3.56 \times 10^{-3}\, \text{mol/L}\) and \([\text{F}^{-}] = 2s = 7.12 \times 10^{-3}\, \text{mol/L}\).

Key concepts

Equilibrium Reaction

The concept of an equilibrium reaction is key to understanding solubility in chemistry. When a solid dissolves in a liquid to form a solution, an equilibrium is established between the dissolved ions and the undissolved solid. In the case of barium fluoride (BaF₂), the dissolution process can be described by the equilibrium reaction:\[ \mathrm{BaF}_2 (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + 2\mathrm{F}^{-} (aq) \]This equation shows that solid barium fluoride dissociates into barium ions \(\mathrm{Ba}^{2+}\) and fluoride ions \(\mathrm{F}^{-}\). At equilibrium, the rate at which the solid dissolves equals the rate at which the ions combine to re-form the solid. It's a dynamic balance, not a complete dissolving of all solid material.

Ion Concentration

Ion concentration is central to the concept of equilibrium reactions in solubility. When barium fluoride dissolves, it releases ions into the solution at specific concentrations. According to the dissolution equation, one molecule of \(\mathrm{BaF}_2\) produces one \(\mathrm{Ba}^{2+}\) ion and two \(\mathrm{F}^{-}\) ions.

• \([\mathrm{Ba}^{2+}] = s\): The concentration of barium ions is the same as the solubility \(s\) of \(\mathrm{BaF}_2\).
• \([\mathrm{F}^{-}] = 2s\): The concentration of fluoride ions is double \(s\), because each unit of \(\mathrm{BaF}_2\) yields two fluoride ions.

Equilibrium ion concentrations don't change unless the system's conditions are altered, like temperature or pressure. This constancy allows calculations involving equilibrium reactions to reliably predict ion concentrations.

Dissolution Equation

A dissolution equation describes how a solid substance dissolves in a solvent to form its constituent ions. For \(\mathrm{BaF}_2\), the dissolution equation is:\[ \mathrm{BaF}_2 (s) \rightleftharpoons \mathrm{Ba}^{2+} (aq) + 2\mathrm{F}^{-} (aq) \]This equation helps us understand the stoichiometric relationships in the solution. It signifies that from one \(\mathrm{BaF}_2\) unit, we get one barium ion and two fluoride ions once equilibrium is reached. Such equations are crucial for setting up expressions to calculate solubility and ion concentrations.

Solubility Calculation

Solubility calculation involves determining the maximum amount of a solute that can dissolve in a solvent at equilibrium. To find the ion concentration in a saturated solution, we use the solubility product constant \(K_{sp}\). For \(\mathrm{BaF}_2\), \(K_{sp}\) is calculated as follows:1. Assign solubility \(s\) to \([\mathrm{Ba}^{2+}]\). This represents the molar solubility of the solid.2. Recognize that \([\mathrm{F}^{-}] = 2s\) because the dissolution produces two fluoride ions for every barium fluoride.3. Insert these into the \( K_{sp} \) expression: \[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2 = s(2s)^2 = 4s^3 \]4. Given \(K_{sp} = 1.8 \times 10^{-7}\), solve for \(s\): \[ s^3 = \frac{1.8 \times 10^{-7}}{4} = 4.5 \times 10^{-8} \] 5. Find \(s\) by taking the cube root: \[ s \approx 3.56 \times 10^{-3} \, \text{mol/L} \]This calculation determines the maximum concentration of ions in a solution at equilibrium, informing us about the solubility limit of \(\mathrm{BaF}_2\).