Define the following terms: Enthalpy, Lattice energy, Hydration energy, Entropy, Heat of combustion. Calculate the pH and pOH of a 2.3 x 10⁻² M HCl solution diluted to 100 mL.

Here are the solutions to your assignment questions.

1. Definitions:

• Enthalpy: A thermodynamic property representing the total heat content of a system at constant pressure. It is the sum of the system's internal energy and the product of its pressure and volume.
• Lattice energy: The energy required to completely separate one mole of a solid ionic compound into its gaseous ions. It is a measure of the strength of the electrostatic forces between ions in a crystal lattice.
• Hydration energy: The energy released when one mole of gaseous ions dissolves in water to form hydrated ions. It quantifies the attractive forces between ions and water molecules.
• Entropy: A thermodynamic property that measures the degree of randomness or disorder within a system. Systems tend to move towards states of higher entropy.
• Heat of combustion: The heat released when one mole of a substance undergoes complete combustion with oxygen under standard conditions. It is a specific type of enthalpy change.

2. Calculate the pH and pOH of each solution:

a) 25.0 mL of $2.3 \times 10^{-2}$ M HCl, diluted to 100 mL

Step 1: Calculate the initial moles of HCl.

$$\text{Moles of HCl} = M_1 V_1 = (2.3 \times 10^{-2} \text{ mol/L}) \times (0.0250 \text{ L}) = 5.75 \times 10^{-4} \text{ mol}$$

Step 2: Calculate the final concentration of HCl after dilution.

$$M_2 = \frac{\text{Moles of HCl}}{V_2} = \frac{5.75 \times 10^{-4} \text{ mol}}{0.100 \text{ L}} = 5.75 \times 10^{-3} \text{ M}$$

Since HCl is a strong acid, $[\text{H}^+] = 5.75 \times 10^{-3} \text{ M}$.

Step 3: Calculate pH.

$$\text{pH} = -\log[\text{H}^+] = -\log(5.75 \times 10^{-3}) = 2.24$$

Step 4: Calculate pOH.

$$\text{pOH} = 14 - \text{pH} = 14 - 2.24 = 11.76$$

The pH is $\boxed{\text{2.24}}$ and the pOH is $\boxed{\text{11.76}}$.

b) 5.0 mL of 1.87 M NaOH, diluted to 125 mL

Step 1: Calculate the initial moles of NaOH.

$$\text{Moles of NaOH} = M_1 V_1 = (1.87 \text{ mol/L}) \times (0.0050 \text{ L}) = 9.35 \times 10^{-3} \text{ mol}$$

Step 2: Calculate the final concentration of NaOH after dilution.

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