Barium fluoride (\(\text{BaF}_2\)) is a white crystalline solid and an inorganic compound used in materials science and chemistry. It is considered sparingly soluble in water. While many ionic compounds dissolve completely, barium fluoride dissolves only to a very small extent. This classification means that a tiny amount separates into its constituent ions when placed in water.
Understanding Chemical Solubility
Solubility describes the ability of a solute, like barium fluoride, to dissolve in a solvent, such as water, forming a uniform solution. Chemists classify solubility using categories like soluble, sparingly soluble, and insoluble. Substances are generally considered soluble if they dissolve beyond a threshold of more than \(1 \text{ gram}\) per \(100 \text{ milliliters}\) of water. Barium fluoride falls outside this range, dissolving at a rate of only about \(1.61 \text{ grams}\) per liter of water at \(25^\circ \text{C}\).
The general principle of “like dissolves like” explains why polar solvents, such as water, are effective at dissolving polar or ionic solutes. Water molecules are highly polar and can effectively pull apart the ions of many salts. The behavior of barium fluoride is an exception to common solubility rules for fluoride salts, requiring a deeper look into the energetic competition during dissolution.
The Chemical Forces Behind Barium Fluoride’s Behavior
Dissolving an ionic solid involves a competition between two energetic forces: lattice energy and hydration energy. Lattice energy is the energy required to break the strong electrostatic bonds holding the solid crystal structure together, separating the positive barium ions (\(\text{Ba}^{2+}\)) from the negative fluoride ions (\(\text{F}^{-}\)). Hydration energy is the energy released when water molecules surround and stabilize these separated ions in the solution.
For a substance to be readily soluble, the hydration energy released must be sufficient to overcome the lattice energy. In the case of barium fluoride, the lattice energy significantly outweighs the hydration energy. This imbalance is the primary reason for its low solubility.
The barium ion (\(\text{Ba}^{2+}\)) is a relatively large cation compared to other Group 2 metal ions. Because the ion is larger, its charge is spread over a greater volume, leading to a weaker attraction for polar water molecules and a lower hydration energy. This lower stabilizing energy is insufficient to break the strong crystal structure, which is held together by the \(+2\) charge of the barium ion and the small size of the fluoride ion.
Measuring Insolubility: The Solubility Product Constant (\(\text{K}_{sp}\))
Scientists use the Solubility Product Constant (\(\text{K}_{sp}\)) to quantitatively measure how much of a sparingly soluble substance dissolves. The \(\text{K}_{sp}\) is an equilibrium constant describing the balance between the solid compound and its dissolved ions in a saturated solution.
For barium fluoride, the dissolution expression is \(\text{BaF}_{2}(\text{s}) \rightleftharpoons \text{Ba}^{2+}(\text{aq}) + 2\text{F}^{-}(\text{aq})\), and the \(\text{K}_{sp}\) is \([\text{Ba}^{2+}][\text{F}^{-}]^2\). A very small \(\text{K}_{sp}\) value confirms that the equilibrium strongly favors the undissolved solid. The \(\text{K}_{sp}\) for barium fluoride at \(25^\circ \text{C}\) is approximately \(1.7 \times 10^{-6}\).