Question

The solubility of \(\mathrm{PbF}_{2}\) (formula weight \(=245\) ) is \(0.46 \mathrm{~g} / \mathrm{L}\). What is the solubility product? (a) \(1.1 \times 10^{-10}\) (b) \(2.6 \times 10^{-8}\) (c) \(1.1 \times 10^{-7}\) (d) \(6.8 \times 10^{-9}\)

Short answer

The solubility product (\(K_{sp}\)) is \(1.1 \times 10^{-8}\).

Step-by-step solution

Calculate molar solubility from given solubility

To find the molar solubility of \(\mathrm{PbF}_2\), divide its given solubility in grams per liter (0.46 g/L) by its formula weight (245 g/mol). This will give the solubility in moles per liter (mol/L), which is the molar solubility. The formula to use is: Molar solubility = \(\frac{Mass \ solubility}{Formula \ weight}\).

Write the solubility product expression

The solubility product (\(K_{sp}\)) expression for \(\mathrm{PbF}_2\) is based on its dissociation in water: \([\mathrm{Pb}^{2+}] \cdot [\mathrm{F}^-]^2\). Since \(\mathrm{PbF}_2\) dissociates into one \(\mathrm{Pb}^{2+}\) ion and two \(\mathrm{F}^-\) ions, if \(s\) is the molar solubility, then \[\mathrm{Pb}^{2+}\] = \(s\) and \[\mathrm{F}^-\] = \(2s\).

Calculate the solubility product (\(K_{sp}\))

Substitute the molar solubility and its respective ion concentrations into the \(K_{sp}\) expression to find the solubility product. The solubility product is calculated as \(K_{sp} = [\mathrm{Pb}^{2+}] \cdot [\mathrm{F}^-]^2 = s(2s)^2 = 4s^3\).

Apply calculated molar solubility to find \(K_{sp}\)

Given the molar solubility from Step 1, raise this value to the power of 3 and then multiply it by 4 to get the solubility product.

Choose the correct answer

Compare the calculated \(K_{sp}\) with the given options and choose the correct one.

Key concepts

Molar Solubility Calculation

Molar solubility calculation is a fundamental concept in chemistry that involves determining the amount of a substance that can dissolve in a solution to form a saturated solution without forming a precipitate. To calculate the molar solubility, the mass solubility (the amount of substance, in grams, that dissolves in one liter of solution) is divided by the molar mass (formula weight) of the substance. This provides the solubility in units of moles per liter (mol/L), which is referred to as the molar solubility. For instance, if we look at the compound \(\mathrm{PbF}_2\), and we're given a mass solubility of \(0.46\,\mathrm{g/L}\), we would calculate its molar solubility by using the formula:
\[\text{Molar solubility} = \frac{\text{Mass solubility}}{\text{Formula weight}}\].
Thus, the molar solubility provides a way of expressing how much of the substance can be dissolved in a given volume of solution in molar terms and is crucial for further calculations involving the solubility product.

Solubility Product Expression

The solubility product expression, denoted as \(K_{sp}\), represents the equilibrium constant for the dissolving process of a sparingly soluble ionic compound. This expression is derived from the chemical dissociation equation of the compound in water. The generic formula for the solubility product expression of an ionic compound \(AB_{n}\), which dissociates into \(A^{m+}\) and \(B^{n-}\) ions, is formulated as \(K_{sp} = [A^{m+}]^m[B^{n-}]^n\), where the brackets indicate the concentrations of the ions in moles per liter. In our specific case, for the compound \(\mathrm{PbF}_2\), which dissociates into \(\mathrm{Pb}^{2+}\) and \(\mathrm{F}^-\) ions in a 1:2 ratio, the solubility product expression is \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^-]^2\). Understanding how to correctly write the solubility product expression for a given compound is essential, as the resulting equation is used to calculate the solubility product, a measure of the compound's solubility in a saturated solution.

Chemical Dissociation in Water

Chemical dissociation in water is a process wherein ionic compounds separate into their respective ions when they dissolve. This is a result of the polar nature of water, which has the ability to attract and surround charged particles, thus facilitating their separation. For example, when the ionic compound \(\mathrm{PbF}_{2}\) is placed in water, it undergoes dissociation:
\[\mathrm{PbF}_{2} \rightarrow \mathrm{Pb}^{2+} + 2\mathrm{F}^-\].
In a saturated solution, these ions reach a dynamic equilibrium, where the rate of dissociation equals the rate of precipitation. The extent of dissociation for sparingly soluble compounds is often low, leading to very small amounts of ions in solution, which is why such solutions can be used to calculate the solubility product. It's this chemical behavior in water that allows us to understand and predict the solubility of compounds and to quantify it through the molar solubility and solubility product.