Question

A solution is saturated with magnesium palmitate \(\left[\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2}, \text { a component of bathtub ring }\right] \mathrm{at}\) \(50^{\circ} \mathrm{C} .\) How many milligrams of magnesium palmitate will precipitate from \(965 \mathrm{mL}\) of this solution when it is cooled to \(25^{\circ} \mathrm{C} ?\) For \(\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2},\) \(K_{\mathrm{sp}}=4.8 \times 10^{-12}\) at \(50^{\circ} \mathrm{C}\) and \(3.3 \times 10^{-12}\) at \(25^{\circ} \mathrm{C}\).

Short answer

To find the amount of magnesium palmitate that will precipitate, subtract the molar solubility at \(25^{\circ}C\) from the molar solubility at \(50^{\circ}C\) and multiply by the volume of the solution in liters. Then multiply by the molar mass of magnesium palmitate and convert to milligrams. Please plug in the values into the formula mentioned in Step 3 for obtaining the exact figure.

Step-by-step solution

Find the Solubility at given Temperatures

Using the formula for solubility product \(K_{sp} = [Mg^{2+}][A^{2-}]^2\), where A is the palmitate ion, we can find the molar solubility of the salt at the given temperatures by taking the square root of each \(K_{sp}\). Let's denote the molar solubility at \(50^{\circ}C\) as \(x\) and the molar solubility at \(25^{\circ}C\) as \(y\). So, \(x = \sqrt{4.8 \times 10^{-12}}\) and \(y = \sqrt{3.3 \times 10^{-12}}\).

Calculate the Difference in Solubility

The difference in molar solubility tells us how much of the solute will precipitate out when the solution is cooled from \(50^{\circ}C\) to \(25^{\circ}C\). This difference, \(d\), is found by subtraction: \(d = x - y\).

Determine the Mass in milligrams

A solution's concentration is its molarity, which is the number of moles per liter of solution. Because we have 965 mL of solution, we need to convert this to liters to find the mass of the precipitate. The number of moles of precipitate is \(d \times 0.965\) L. Then, using the molecular weight of magnesium palmitate (i.e., 591.27 g/mol), we calculate the mass of the precipitate in grams and then convert it to milligrams: \(Precipitate (mg) = d \times 0.965 \times 591.27 \times 10^3\).

Key concepts

Saturated Solutions

A saturated solution is a type of solution where no additional solute can dissolve at a given temperature and pressure. In essence, it has reached its maximum concentration. Any extra solute will remain undissolved and settle at the bottom. This concept is crucial when thinking about solutions and their capacities for holding solutes.

For example, if you keep adding sugar to a cup of tea, eventually the sugar will stop dissolving. The tea has reached saturation, and at that point, any extra sugar just sits at the bottom. Similarly, magnesium palmitate forms a saturated solution when it no longer dissolves in the solvent at a specific temperature.

This condition is significant when predicting solubility changes. As the temperature drops, the solubility can change. In our scenario, cooling the solution from 50°C to 25°C affects the amount of solute that can remain dissolved, leading to precipitation.

Solubility Product (Ksp)

The solubility product, denoted as Ksp, is a constant that provides insight into the solubility of ionic compounds in water. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. Specifically, it represents the product of the concentrations of the ions, each raised to the power of their coefficient in the balanced equation.

For magnesium palmitate, the reaction can be written as:\[ \mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2} \rightarrow \mathrm{Mg}^{2+} + 2 \mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}^{-} \]The Ksp expression is:\[ K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}^{-}]^2 \]
By plugging in the given Ksp values (like 4.8 × 10^-12 at 50°C), we can calculate the concentration of ions at different temperatures. Understanding Ksp allows us to predict whether a precipitation reaction will occur when conditions change.

Chemical Precipitation

Chemical precipitation refers to the process of converting a dissolved substance into a solid that separates from the solution. This often occurs when a solution is saturated and conditions change, like temperature drops, causing the solutes to exceed their solubility limits and form precipitates.

The exercise provides a perfect example of how temperature influences solubility. At 50°C, the magnesium palmitate is fully dissolved. However, when cooled to 25°C, its solubility decreases according to the Ksp at that temperature. The difference in solubility results in the formation of solid magnesium palmitate from the solution.

In practical terms, the precipitate is the excess compound that can no longer remain dissolved under new conditions. In calculations, using the difference in molar solubility and considering the volume of the solution in liters allows us to determine the mass of the precipitate.