Problem 66 - Chemistry and Chemical Reactivity

Question

The solubility of ammonium formate, \(\mathrm{NH}_{4} \mathrm{CHO}_{2},\) in 100. 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 \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) precipitates? (Assume that no water evaporates and that the solution is not supersaturated.)

Step-by-Step Solution

Answer

888 g of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) precipitates at \( 0^{\circ} \mathrm{C} \).

Step 1: Calculate Initial Solubility at 80°C

Given that the solubility of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) is \( 546 \text{ g} \) per \( 100 \text{ g} \) of water at \( 80^{\circ} \text{C} \), calculate the solubility in \( 200 \text{ g} \) of water. Multiply \( 546 \text{ g} \) by \( 2 \): \( 546 \times 2 = 1092 \text{ g} \). This is the total amount of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) that can dissolve in \( 200 \text{ g} \) of water at \( 80^{\circ} \text{C} \).

Step 2: Calculate Solubility at 0°C

The solubility of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) at \( 0^{\circ} \text{C} \) is \( 102 \text{ g} \) per \( 100 \text{ g} \) of water. For \( 200 \text{ g} \) of water, the solubility is \( 102 \times 2 = 204 \text{ g} \). This means \( 204 \text{ g} \) of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) can remain dissolved at \( 0^{\circ} \text{C} \).

Step 3: Determine Mass of Precipitated Compound

At \( 80^{\circ} \text{C} \), \( 1092 \text{ g} \) of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) is dissolved. At \( 0^{\circ} \text{C} \), only \( 204 \text{ g} \) can stay dissolved. The mass of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) that precipitates is \( 1092 - 204 = 888 \text{ g} \). Thus, \( 888 \text{ g} \) of \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \) will precipitate.

Key Concepts

Ammonium Formate

Ammonium formate, represented by the chemical formula \( \mathrm{NH}_4 \mathrm{CHO}_2 \), is a white crystalline solid that readily dissolves in water. This compound plays significant roles in various applications, such as being a useful reagent in chemical synthesis and a preserving agent in tissues. The solubility of ammonium formate is a key property to understand, especially when dealing with temperature variations in solutions.

In general, ammonium formate forms ions in water: ammonium ions \( (\mathrm{NH}_4^+) \) and formate ions \( (\mathrm{CHO}_2^-) \). These ions move freely in solution, making the compound an ionic solute. Understanding how these ions interact with water molecules helps explain the solubility behavior of ammonium formate.

Temperature Dependence of Solubility

The solubility of many solids, including ammonium formate, varies significantly with temperature. In the given exercise, the solubility of ammonium formate increases from 102 g per 100 g of water at 0°C to 546 g per 100 g of water at 80°C.

This relationship between temperature and solubility can be generally explained by the absorption or release of energy during the dissolving process.

• At higher temperatures, more ammonium formate can dissolve because increased kinetic energy allows for better interaction between the water molecules and the ions of ammonium formate.
• As temperature increases, the entropy or disorder of the system also increases, making dissolution more favorable.

Understanding this dependence is crucial for predicting the behavior of solutions when cooled or heated.

Precipitation Process

The precipitation process involves the formation of solids within a solution that were previously dissolved. This occurs when the dissolved solute exceeds the solubility limit at a given temperature, as seen when cooling a saturated solution.

For ammonium formate, cooling a solution from 80°C to 0°C reduces its solubility.

• At 80°C, 1092 g of ammonium formate can dissolve in 200 g of water.
• As the temperature drops to 0°C, solubility decreases to 204 g in the same amount of water.

Consequently, the excess ammonium formate precipitates out of the solution. In this exercise, 888 g precipitates after cooling, because 888 g is the difference between the initial dissolved amount and the new solubility level.

Chemical Equilibrium

Chemical equilibrium in a solution is the state where the rate of dissolution of a solute equals the rate of precipitation. This dynamic balance ensures that the concentration of solute in the solution remains constant over time. However, changes in temperature can disrupt this equilibrium.

In the case of ammonium formate, at 80°C, the system is in equilibrium with a higher concentration of ammonium formate dissolved.

• When cooled to 0°C, the solubility limit decreases, shifting equilibrium towards precipitation.
• The newly formed solid represents the shifted equilibrium, with less dissolved ammonium formate in the solution.

Understanding these principles of chemical equilibrium is essential in the study of chemistry, as it applies to many reversible reactions and solubility processes.