Question

Care must be taken in preparing solutions of solutes that liberate heat on dissolving. The heat of solution of \(\mathrm{NaOH}\) is \(-44.5 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) To what maximum temperature may a sample of water, originally at \(21^{\circ} \mathrm{C},\) be raised in the preparation of \(500 \mathrm{mL}\) of \(7.0 \mathrm{M}\) NaOH? Assume the solution has a density of \(1.08 \mathrm{g} / \mathrm{mL}\) and specific heat of \(4.00 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Short answer

The maximum temperature that can be reached in the preparation of 500 mL of 7.0 M NaOH from a water sample originally at 21 degrees Celsius is \(21 + 72.20 = 93.20\) degrees Celsius, assuming the density and specific heat remain constant. The sample of water can reach a maximum final temperature of approximately 93.20 degrees Celsius.

Step-by-step solution

Calculation of moles of NaOH

Firstly, calculate the moles of NaOH in the solution using the formula Molarity (M) = moles/volume(L). Given, Molarity (M) = 7 M and Volume (V) = 500 mL = 0.5 L. Hence, moles of NaOH = 7 * 0.5 = 3.5 mol

Calculation of heat released

Next, calculate the amount of heat released by the NaOH using the formula Heat = moles of solute * heat of solution. Given, heat of solution for NaOH = -44.5 kJ/mol. Hence, Heat released = -44.5 kJ/mol * 3.5 mol = -155.75 kJ = -155750 J. The negative sign indicates that the heat is being released into the solution.

Calculation of temperature increase

Now, use this proper value for the heat released in the next calculation. We apply the formula for energy change, \(q = mc\Delta T \) where q is the heat transferred, m is the mass, c is the specific heat, and \(\Delta T \) is the change in temperature. We know that the specific heat of the solution is 4.00 J/g*degree Celsius, the density is 1.08 g/mL and we have 500 mL solution. The mass of the solution is the density times the volume. Hence Mass (m) = 1.08 g/mL * 500 mL = 540 g. We rearrange the formula to solve for \(\Delta T \), so \(\Delta T = q/(mc) \) . Hence \(\Delta T = -155750 J/(540 g * 4.00 J/g*degree Celsius) = -72.20 degree Celsius\)

Calculation of final temperature

Finally, calculate the final temperature by adding the temperature change to the initial temperature. Given, the initial temperature is 21 degree Celsius. Hence Final Temperature = Initial Temperature + Temperature Change = 21 degree Celsius - 72.20 degree Celsius = -51.20 degree Celsius which is not physically possible. Since \(\Delta T \) is the absolute temperature difference, the actual temperature rise would be 72.20 degree Celsius.

Key concepts

Heat of Solution

When certain solutes dissolve in a solvent, they can release or absorb heat, a process called the heat of solution. In this problem, sodium hydroxide (\(\mathrm{NaOH}\)) releases heat when it dissolves. The heat given off is measured in kilojoules per mole (kJ/mol).
The heat of solution here for \(\mathrm{NaOH}\) is \(-44.5 \mathrm{kJ/mol}.\) This negative value shows that the process releases energy, making it exothermic.
It's crucial to understand:

• Whether the heat of solution is positive or negative can influence the nature (endothermic or exothermic) of the reaction.
• Exothermic heats of solution indicate heat release during dissolution, while endothermic values indicate heat absorption.

In thermochemistry, keeping track of heat exchanges is vital for managing reactions, ensuring safety, and predicting changes in temperature during solution preparations.

Molarity Calculation

Molarity is a critical concept in chemistry that refers to the concentration of a solution. It's defined as the number of moles of a solute divided by the volume of the solution in liters. For this exercise, you need to prepare a \(7.0 \mathrm{M}\) solution of \(\mathrm{NaOH}.\)
Given that the molarity (M) is \(7.0 \mathrm{mol/L}\) and the volume is \(500\, \mathrm{mL} \,(0.5 \, \mathrm{L}),\) we use the formula:\[\text{Moles of } \mathrm{NaOH} = \text{Molarity} \times \text{Volume} = 7.0 \times 0.5 = 3.5 \, \text{moles}\]Calculating molarity is important for accurate solution preparation, which directly affects reaction efficiency and safety.
There are key points to remember:

• The volume must always be in liters when calculating molarity.
• Moles can be derived by rearranging the molarity formula, which is essential for determining how much solute is needed.

By understanding these principles, students can solve a wide range of similar problems.

Temperature Change Calculation

Calculating the temperature change when a solution is formed involves understanding how much heat is released or absorbed and how it affects the temperature of the solution. We use the formula \(q = mc\Delta T \) where \(q\) is the heat exchanged, \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the temperature change.
For our scenario:

• The heat \(q\) released is calculated to be \(-155,750 \mathrm{J}.\)
• The mass \(m\) of the solution is \(540 \mathrm{g},\) given the density of \(1.08 \mathrm{g/mL.}\)
• The specific heat \(c\) is \(4.00 \mathrm{J/g}^\circ\mathrm{C}.\)

The formula rearranges to:\[\Delta T = \frac{q}{mc} = \frac{-155,750}{540 \times 4.00} = -72.20^\circ\mathrm{C}\]The negative sign indicates the direction of heat release, but physically, the temperature rise is considered positive, so the increase is \(72.20^\circ\mathrm{C}.\) Understanding this calculation helps in predicting temperature outcomes of reactions and ensuring safe chemical processes.