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Understanding the dissolution equation is crucial as it provides a clear picture of how a compound dissociates in water. For lithium fluoride (\(\mathrm{LiF}\)), the dissolution equation is:

• \[ \mathrm{LiF}_{(s)} \rightleftharpoons \mathrm{Li^+}_{(aq)} + \mathrm{F^-}_{(aq)} \]

Here, the arrow indicates that the solid \(\mathrm{LiF}\) dissolves in water to form ions. Each unit of \(\mathrm{LiF}\) separates into one lithium ion (\(\mathrm{Li^+}\)) and one fluoride ion (\(\mathrm{F^-}\)). This equation is key to future calculations involving the ionic concentration and the solubility product constant (\(K_{sp}\)). It shows a 1:1 ratio of dissociation, meaning the amount of lithium ions will equal the fluoride ions in solution.
Recognizing this simple dissociation helps streamline the process of determining molarity and \(K_{sp}\) by clarifying the relationships between the solute and the resulting ions.

Molarity is the concentration of solute in a solution, and in this context, it measures the concentration of ions from \(\mathrm{LiF}\) dissolving in water. Let's break it down:

• First, we need the molar mass of \(\mathrm{LiF}\), which consists of lithium (6.94 g/mol) and fluoride (19.00 g/mol). Adding these gives 25.94 g/mol.

This tells us how much \(\mathrm{LiF}\) equals one mole. With 0.13 grams dissolved in 100 mL, the moles of \(\mathrm{LiF}\) can be calculated:

• Moles = \(\frac{0.13 \text{ g}}{25.94 \text{ g/mol}} \approx 0.00501 \text{ mol}\)

The 100 mL solution converts to 0.1 liters. This allows for the calculation of molarity as follows:

• Molarity = \(\frac{0.00501 \text{ mol}}{0.100 \text{ L}} = 0.0501 \text{ M}\)

For \(\mathrm{LiF}\), this means 0.0501 M of both \(\mathrm{Li^+}\) and \(\mathrm{F^-}\), as each \(\mathrm{LiF}\) splits into one of each ion.

The solubility product constant, \(K_{sp}\), indicates the extent of a compound's dissolution in water. For \(\mathrm{LiF}\), \(K_{sp}\) is calculated using the ionic concentrations from the dissolution.

• Write down the expression: \(K_{sp} = [\mathrm{Li^+}][\mathrm{F^-}]\).
• Substitute the molarity of \(\mathrm{Li^+}\) and \(\mathrm{F^-}\) calculated as 0.0501 M each.
• \(K_{sp} = (0.0501) \times (0.0501) = 0.00251\).

Thus, the \(K_{sp}\) value is 0.00251 at 20°C. This low value reflects \(\mathrm{LiF}\)'s limited solubility, showcasing how \(K_{sp}\) values help compare solubilities across different compounds. Remember that higher \(K_{sp}\) values indicate greater solubility, while lower ones signify less.

In solutions, ionic concentration dictates the properties of the resulting mixture and involves understanding how much of each type of ion is present. From the \(\mathrm{LiF}\) dissolution,

• \(\mathrm{Li^+}\) and \(\mathrm{F^-}\) concentrations are both 0.0501 M.

These arise because one mole of \(\mathrm{LiF}\) gives one mole each of its constituent ions. This simplicity is due to the 1:1 dissolution equation.
The significance of knowing ionic concentration includes:

• Predicting physical properties like conductivity.
• Calculating further reactions or predicting precipitate formation in more complex scenarios.

Calculating ionic concentrations is straightforward with clear equations and understanding of molarity, reflecting the balance in the chemistry of solutes in solutions.