Problem 56 - Chemistry: Principles and Reactions

Question

Ammonium bromide can be used as a wood preservative. Its solubility in water at \(20^{\circ} \mathrm{C}\) is \(75.5 \mathrm{~g} / 100 \mathrm{~g}\) water. At \(50^{\circ} \mathrm{C}\) its solubility is \(99.2 \mathrm{~g} / 100 \mathrm{~g}\) water. Calculate (a) the mass of ammonium bromide that dissolves in \(82.5 \mathrm{~g}\) of water at \(20^{\circ} \mathrm{C} .\) (b) the mass of water required to dissolve \(43.7 \mathrm{~g}\) of ammonium bromide at \(50^{\circ} \mathrm{C}\). (c) the mass of ammonium bromide that would not remain in solu- tion if a solution made up of \(29.0 \mathrm{~g}\) of ammonium bromide in \(35.0 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\) is cooled to \(20^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

Question: Calculate the mass of ammonium bromide that would not remain in the solution when a mixture of 29g of ammonium bromide and 35g of water is cooled from 50°C to 20°C. Answer: Approximately 2.575g of ammonium bromide would not remain in the solution.

Step 1: Write a solubility proportion

Using the solubility of ammonium bromide in water at 20°C, we can set up a proportion: \(\frac{75.5 \mathrm{~g\, of\, ammonium\, bromide}}{100 \mathrm{~g\, of\, water}}\). We want to find the mass of ammonium bromide that can dissolve in 82.5g of water at 20°C, so our proportion will be: \(\frac{m \mathrm{~g\, of\, ammonium\, bromide}}{82.5 \mathrm{~g\, of\, water}}\).

Step 2: Solve for the mass of ammonium bromide

Now we can set up an equation and solve for the mass of ammonium bromide: \(\frac{75.5 \mathrm{~g\, of\, ammonium\, bromide}}{100 \mathrm{~g\, of\, water}} = \frac{m \mathrm{~g\, of\, ammonium\, bromide}}{82.5 \mathrm{~g\, of\, water}}\) Cross multiply and solve for \(m\): \(75.5 \mathrm{~g\, of\, ammonium\, bromide} \times 82.5 \mathrm{~g\, of\, water} = 100 \mathrm{~g\, of\, water} \times m \mathrm{~g\, of\, ammonium\, bromide}\) Divide by \(100 \mathrm{~g\, of\, water}\) to find the mass of ammonium bromide that can dissolve in 82.5g of water: \(m = \frac{75.5 \mathrm{~g\, of\, ammonium\, bromide} \times 82.5 \mathrm{~g\, of\, water}}{100 \mathrm{~g\, of\, water}} \approx 62.2875 \mathrm{~g\, of\, ammonium\, bromide}\) #Part b: Mass of water required to dissolve 43.7g of ammonium bromide at 50°C#

Step 1: Write a solubility proportion

Using the solubility of ammonium bromide in water at 50°C, we can set up a proportion: \(\frac{99.2 \mathrm{~g\, of\, ammonium\, bromide}}{100 \mathrm{~g\, of\, water}}\). We want to find the mass of water needed to dissolve 43.7g of ammonium bromide at 50°C, so our proportion will be: \(\frac{43.7 \mathrm{~g\, of\, ammonium\, bromide}}{m \mathrm{~g\, of\, water}}\).

Step 2: Solve for the mass of water

Now we can set up an equation and solve for the mass of water: \(\frac{99.2 \mathrm{~g\, of\, ammonium\, bromide}}{100 \mathrm{~g\, of\, water}} = \frac{43.7 \mathrm{~g\, of\, ammonium\, bromide}}{m \mathrm{~g\, of\, water}}\) Cross multiply and solve for \(m\): \(99.2 \mathrm{~g\, of\, ammonium\, bromide} \times m \mathrm{~g\, of\, water} = 100 \mathrm{~g\, of\, water} \times 43.7 \mathrm{~g\, of\, ammonium\, bromide}\) Divide by \(99.2 \mathrm{~g\, of\, ammonium\, bromide}\) to find the mass of water required to dissolve 43.7g of ammonium bromide at 50°C: \(m = \frac{100 \mathrm{~g\, of\, water} \times 43.7 \mathrm{~g\, of\, ammonium\, bromide}}{99.2 \mathrm{~g\, of\, ammonium\, bromide}} \approx 44.0446 \mathrm{~g\, of\, water}\) #Part c: Mass of ammonium bromide not remaining in solution when cooled to 20°C#

Step 1: Calculate the mass of ammonium bromide that can dissolve at 20°C

We know the mass of ammonium bromide (29g) and the mass of water (35g) in the solution at 50°C. We will use the solubility of ammonium bromide at 20°C to set up a proportion to find the mass of ammonium bromide that would dissolve at 20°C: \(\frac{75.5 \mathrm{~g\, of\, ammonium\, bromide}}{100 \mathrm{~g\, of\, water}} = \frac{m \mathrm{~g\, of\, ammonium\, bromide}}{35 \mathrm{~g\, of\, water}}\) Solve for \(m\): \(m = \frac{75.5 \mathrm{~g\, of\, ammonium\, bromide} \times 35 \mathrm{~g\, of\, water}}{100 \mathrm{~g\, of\, water}} \approx 26.425 \mathrm{~g\, of\, ammonium\, bromide}\)

Step 2: Calculate the mass of ammonium bromide that would not remain in the solution

Now, subtract the mass of ammonium bromide that can dissolve at 20°C from the initial mass of ammonium bromide in the solution: \(29.0 \mathrm{~g\, of\, ammonium\, bromide} - 26.425 \mathrm{~g\, of\, ammonium\, bromide} \approx 2.575 \mathrm{~g\, of\, ammonium\, bromide}\) So, if the solution is cooled to 20°C, approximately 2.575g of ammonium bromide would not remain in the solution.

Key Concepts

Ammonium Bromide Solubility

Understanding the solubility of a compound is crucial for applications in various industries, including its use as a wood preservative where ammonium bromide plays a role. Solubility is generally expressed in terms of the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.

For instance, at a temperature of 20°C, 75.5 grams of ammonium bromide can be dissolved in 100 grams of water. This is a clear indication of how soluble the compound is at room temperature. When exercising a solubility calculation, it is essential to establish a solubility proportion. This proportion helps us make predictions about dissolving different amounts of a substance under varying conditions.

To put this into a practical context, if one has less water, say 82.5 grams instead of 100 grams, the amount of ammonium bromide that can dissolve at 20°C would also proportionally decrease. Thus, we find that approximately 62.2875 grams of ammonium bromide would be soluble in 82.5 grams of water.

Temperature-Dependent Solubility

Temperature profoundly affects the solubility of substances. For ammonium bromide, its solubility increases from 75.5 g/100 g water at 20°C to 99.2 g/100 g water at 50°C. This property of temperature-dependent solubility means that a solution can hold more of the solute at higher temperatures.

In addressing an example concern, to know the mass of water needed to dissolve a specific mass of ammonium bromide at a higher temperature, the altered solubility ratio is used. To dissolve 43.7 grams of ammonium bromide at 50°C, one would need approximately 44.0446 grams of water. This is because the increased kinetic energy at higher temperatures allows more solute particles to interact with solvent molecules, resulting in increased solubility.

Solubility Proportion

The concept of solubility proportion comes in handy when we are dealing with changes in solubility due to temperature variations. Taking the ammonium bromide example, if a solution is made up of 29.0 grams of the compound in 35.0 grams of water at 50°C and is then cooled to 20°C, some of the dissolved substance will precipitate out because the solubility at the lower temperature cannot accommodate all of the solute.

Through the setup of a new solubility proportion using the solubility data at 20°C, the calculation shows that only 26.425 grams of ammonium bromide would remain dissolved, henceforth, 2.575 grams would not remain in solution. This precipitation upon cooling is a direct result of reduced solubility at the lower temperature, and understanding these proportions are vital in predicting the outcome of such temperature changes on a solution's composition.