Problem 72 - Chemistry and Chemical Reactivity

Question

The solubility of ammonium formate, \(\mathrm{NH}_{4} \mathrm{CHO}_{2},\) in \(100 \mathrm{g}\) of water is \(102 \mathrm{g}\) at \(0^{\circ} \mathrm{C}\) and \(546 \mathrm{g}\) at \(80^{\circ} \mathrm{C} .\) A solution is prepared by dissolving \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) in \(200 \mathrm{g}\) of water until no more will dissolve at \(80^{\circ} \mathrm{C}\). The solution is then cooled to \(0^{\circ} \mathrm{C} .\) What mass of \(\mathrm{NH}_{4} \mathrm{CH} \mathrm{O}_{2}\) precipitates? (Assume that no water evaporates and that the solution is not supersaturated.)

Step-by-Step Solution

Answer

888 g of ammonium formate precipitates.

Step 1: Understanding the Problem

The problem gives us the solubility of ammonium formate at two different temperatures in 100 g of water and asks us how much ammonium formate will precipitate when a saturated solution at 80°C is cooled to 0°C.

Step 2: Calculate Mass of Solute Dissolved at 80°C

Since 546 g of ammonium formate can dissolve in 100 g of water at 80°C, double the mass can dissolve in 200 g of water. Therefore, \( 546 \times 2 = 1092 \) g of ammonium formate can dissolve in 200 g of water at 80°C.

Step 3: Calculate Mass of Solute Required at 0°C

The solubility of ammonium formate is 102 g per 100 g of water at 0°C. Thus, in 200 g of water, \( 102 \times 2 = 204 \) g can be dissolved. This is the amount that will remain dissolved after cooling.

Step 4: Calculate Mass of Precipitated Solute

Subtract the amount that remains dissolved at 0°C from the initial amount dissolved at 80°C: \( 1092 - 204 = 888 \) g. This is the mass of ammonium formate that precipitates when the solution is cooled from 80°C to 0°C.

Key Concepts

Temperature Effect on Solubility

The solubility of a substance refers to the maximum amount that can be dissolved in a particular solvent at a specified temperature. How much a solute dissolves in a solvent often changes with temperature. For many solids, solubility increases as temperature rises. This is because higher temperatures provide more energy to overcome intermolecular forces, allowing more solute to break apart and mix into the solvent.

In the case of ammonium formate in water, its solubility dramatically increases from 102 g in 100 g of water at 0°C to 546 g at 80°C. This significant increase highlights how temperature can play a crucial role in determining how much solute can be dissolved.

Understanding this concept is important for anticipating changes in a solution when the temperature is altered, as it predicts how much of the solute can remain in solution and how much might come out as precipitation.

Precipitation

Precipitation occurs when a solute comes out of solution and forms a solid or "precipitate." This happens when the solution becomes oversaturated under new conditions, such as a lower temperature than when it was initially saturated.

In this exercise, starting with a saturated solution of ammonium formate at 80°C, the solution can hold a maximum of 1092 g of solute per 200 g of water. When the solution is cooled to 0°C, the solubility reduces to 204 g per 200 g of water. The excess solute that cannot remain dissolved at the lower temperature will precipitate.

This process is practical in various applications, including in the purification of compounds and in industries where crystal formation is needed.

Saturated Solution

A saturated solution is one where no more solute can be dissolved in the solvent at a particular temperature. It represents a state of equilibrium between the dissolved ions and the undissolved solid.

In the exercise, a saturated solution of ammonium formate is prepared at 80°C, dissolving 1092 g in 200 g of water. This means that if more ammonium formate was added, it would remain undissolved, settling at the bottom.

Saturating solutions is key in many chemical processes where controlling solute quantities is essential, such as in recrystallization methods or in creating solutions for product formulation.