Problem 68
Question
The solubility of ammonium formate, \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) in \(100 .\) g of water is 102 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
Verified Answer
888 g of NH4CHO2 precipitates.
1Step 1: Determine Maximum Solubility at 80°C
At 80°C, the solubility of \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) is 546 g per 100 g of water. Since 200 g of water is used, the maximum amount of \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) that can dissolve is \((546 \text{ g/100 g water}) \times 2 = 1092 \text{ g of } \mathrm{NH}_{4} \mathrm{CHO}_{2}\).
2Step 2: Determine Solubility at 0°C
At 0°C, the solubility of \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) is 102 g per 100 g of water. With 200 g of water, the soluble amount at 0°C is \((102 \text{ g/100 g water}) \times 2 = 204 \text{ g of } \mathrm{NH}_{4} \mathrm{CHO}_{2}\).
3Step 3: Calculate the Mass of Precipitate
Subtract the solubility at 0°C from the amount dissolved at 80°C to find the mass of \(\mathrm{NH}_{4} \mathrm{CHO}_{2}\) that precipitates: \(1092 \text{ g} - 204 \text{ g} = 888 \text{ g}\).
Key Concepts
Ammonium FormateTemperature Effect on SolubilityMass of PrecipitateChemical SolutionsChemistry Problem-Solving
Ammonium Formate
Ammonium formate is a chemical compound with the formula \( \mathrm{NH}_{4} \mathrm{CHO}_{2} \). It acts as the ammonium salt of formic acid. This compound is quite versatile and has several applications. It's used as a buffering agent and has roles in the production of various applications like leather tanning and preparation of buffered aqueous solutions.
It decomposes readily upon heating, producing gases like formamide and water.
Due to its solubility properties, it's an interesting substance for solubility studies and experiments in chemistry, highlighting the behavior of salts in solution at varying temperatures.
It decomposes readily upon heating, producing gases like formamide and water.
Due to its solubility properties, it's an interesting substance for solubility studies and experiments in chemistry, highlighting the behavior of salts in solution at varying temperatures.
Temperature Effect on Solubility
The solubility of substances changes with temperature. Generally, for many solids dissolved in liquids, solubility increases with temperature. Ammonium formate is an example of a compound whose solubility is highly dependent on temperature changes.
At \( 0^{\circ} \mathrm{C} \), its solubility in water is \( 102 \text{ g per 100 g} \) of water, whereas at \( 80^{\circ} \mathrm{C} \), it skyrockets to \( 546 \text{ g per 100 g} \).
This dramatic change emphasizes the significant impact temperature has on dissolving capabilities of substances.
At \( 0^{\circ} \mathrm{C} \), its solubility in water is \( 102 \text{ g per 100 g} \) of water, whereas at \( 80^{\circ} \mathrm{C} \), it skyrockets to \( 546 \text{ g per 100 g} \).
This dramatic change emphasizes the significant impact temperature has on dissolving capabilities of substances.
- At lower temperatures, molecules have less kinetic energy, resulting in lower solubility.
- As temperature rises, molecular movement increases, enhancing solubility.
Mass of Precipitate
Precipitation refers to the process where solutes become insoluble and form solid particles. When a solution containing a solute, like ammonium formate, is cooled, it might not be able to hold as much solute.
In the presented problem, the solution initially holds \( 1092 \text{ g of } \mathrm{NH}_{4} \mathrm{CHO}_{2} \) at \( 80^{\circ} \mathrm{C} \). Cooling to \( 0^{\circ} \mathrm{C} \) reduces solubility to \( 204 \text{ g} \), prompting \( 888 \text{ g} \) to precipitate out.
This concept illustrates saturation limits and how exceeding them results in precipitation. Important aspects include:
In the presented problem, the solution initially holds \( 1092 \text{ g of } \mathrm{NH}_{4} \mathrm{CHO}_{2} \) at \( 80^{\circ} \mathrm{C} \). Cooling to \( 0^{\circ} \mathrm{C} \) reduces solubility to \( 204 \text{ g} \), prompting \( 888 \text{ g} \) to precipitate out.
This concept illustrates saturation limits and how exceeding them results in precipitation. Important aspects include:
- Initial solubility vs. new solubility—subtraction provides precipitate mass.
- Understanding that temperature shifts significantly affect solute solubility.
Chemical Solutions
Chemical solutions are homogeneous mixtures of substances. In this context, water acts as the solvent, while ammonium formate is the solute.
Several factors influence the properties of a solution, including temperature, pressure, and the nature of solute and solvent. In the case of ammonium formate, solubility varies dramatically with temperature, indicating a strong dependency on the thermal state of the system.
Chemical solutions are characterized by their concentration, which in this exercise is represented by the maximum amount of solute that can dissolve in a solvent at a given condition.
Several factors influence the properties of a solution, including temperature, pressure, and the nature of solute and solvent. In the case of ammonium formate, solubility varies dramatically with temperature, indicating a strong dependency on the thermal state of the system.
Chemical solutions are characterized by their concentration, which in this exercise is represented by the maximum amount of solute that can dissolve in a solvent at a given condition.
- Composition of solutions is pivotal in reactions and product formation.
- Manipulating these factors allows chemists to design specific reactions or processes.
Chemistry Problem-Solving
Chemistry problem-solving involves using known principles to find unknowns. The exercise illustrates how to apply solubility data to solve a practical problem.
Solving such problems requires:
Practicing such problems helps students solidify their grasp on theoretical concepts and strengthens their problem-solving strategies, fundamental in both academic and real-world chemistry scenarios.
Solving such problems requires:
- Identifying the key information: solubility data at different temperatures.
- Using calculations: determining total solubility, and consequently precipitation mass.
- Understanding underlying concepts: the temperature impact on solubility and precipitation mechanisms.
Practicing such problems helps students solidify their grasp on theoretical concepts and strengthens their problem-solving strategies, fundamental in both academic and real-world chemistry scenarios.
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