Problem 81 - General Chemistry: Principles and Modern Applications

Question

Predict the approximate freezing points of \(0.10 \mathrm{m}\) solutions of the following solutes dissolved in water: (a) \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\text { urea }) ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{NO}_{3} ;(\mathrm{c}) \mathrm{HCl} ;(\mathrm{d}) \mathrm{CaCl}_{2}\) (e) \(\mathrm{MgSO}_{4} ;\) (f) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (ethanol); \((\mathrm{g}) \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (acetic acid).

Step-by-Step Solution

Answer

The freezing points of the solutions are approximately: (a) \(CO(NH_{2})_{2}\) (urea) : -0.186ºC, (b) \(NH_{4}NO_{3}\) : -0.372ºC, (c) \(HCl\) : -0.372ºC, (d) \(CaCl_{2}\) : -0.558ºC, (e) \(MgSO_{4}\) : -0.558ºC, (f) \(C_{2}H_{5}OH\) (ethanol) : -0.186ºC, (g) \(HC_{2}H_{3}O_{2}\) (acetic acid) : -0.186ºC.

Step 1: Identify the Type of Compound

The first step is to identify whether the compound is ionic or covalent. Ionic compounds typically consist of a metal and a non-metal and dissociate into ions when dissolved in water. In contrast, covalent compounds typically consist of two or more non-metals and do not usually disassociate into ions when dissolved in water. For example, \(CO(NH_{2})_{2}\) and \(C_{2}H_{5}OH\) are covalent, while \(NH_{4}NO_{3}\), \(HCl\), \(CaCl_{2}\), and \(MgSO_{4}\) are ionic.

Step 2: Calculate the Van 't Hoff Factor

The next step is to calculate the Van 't Hoff factor (i), which is the number of ions that the compound dissociates into when dissolved in water. For covalent compounds like urea and ethanol, i = 1. For ionic compounds, i depends on the number of ions formed. For \(NH_{4}NO_{3}\) and \(HCl\), i = 2. For \(CaCl_{2}\) and \(MgSO_{4}\), i = 3.

Step 3: Calculation of Freezing Point Depression

Using the freezing point depression equation \(ΔT_f = i * m * K_f\), where \(ΔT_f\) is the change in freezing point, m is the molality of the solute, and \(K_f\) is the cryoscopic constant of water (-1.86 ºC/m), you can calculate the freezing point depression. For example, for urea, \(ΔT_f = 1 * 0.10 * 1.86 = 0.186ºC\). Repeat calculation for each solute.

Step 4: Calculate the New Freezing Point

The final step is to subtract the freezing point depression from the normal freezing point of water (0ºC) to obtain the new, lower freezing point of the solution: \(T_f = T_{f, water} - ΔT_f\). For example, for urea, \(T_f = 0ºC - 0.186ºC = -0.186ºC\). Repeat calculation for each solute.

Key Concepts

Ionic and Covalent Compounds

When discussing solutions and solubility, it's crucial to grasp the difference between ionic and covalent compounds. Ionic compounds form when metals bond with non-metals, and they dissociate into ions in water. For instance, when table salt (\( \text{NaCl} \)) dissolves, it separates into \( \text{Na}^+ \) and \( \text{Cl}^- \) ions.

• Ionic compounds are usually solid at room temperature and have high melting and boiling points.
• They conduct electricity when dissolved in water because their ions are free to move.

In contrast, covalent compounds form when non-metals share electrons. Substances like sugar (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) dissolve but don't dissociate into ions. As such, they maintain their molecular integrity in solution.

• Covalent compounds often exist in gaseous or liquid states at room temperature and have lower melting and boiling points compared to ionic compounds.
• They do not conduct electricity in solution, as no charged particles are present to carry a current.

Van 't Hoff Factor

The Van 't Hoff factor, denoted as \( i \), is an important concept in chemistry when dealing with colligative properties. It refers to the number of particles a solute breaks into when it is dissolved in a solvent. This is particularly vital when predicting the freezing point depression of a solution.
For instance:

• Non-electrolytes, such as covalent compounds, do not dissociate into ions. Therefore, their Van 't Hoff factor is generally \( i = 1 \), like urea.
• Ionic compounds dissociate into ions, leading to a Van 't Hoff factor that equals the total number of ions formed. For example, \( \text{CaCl}_2 \) disassociates into three ions: one \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \), setting \( i = 3 \)

Understanding the Van 't Hoff factor helps in determining the effect of solutes on colligative properties like boiling point elevation, freezing point depression, and osmotic pressure.

Cryoscopic Constant

The cryoscopic constant, symbolized as \( K_f \), plays a crucial role in the calculation of freezing point depression. It is a property specific to each solvent and represents how much the freezing point of the solvent decreases per molal concentration of a non-volatile solute. For water, the cryoscopic constant is known to be \( -1.86^\circ\text{C/m} \).

• The higher the cryoscopic constant, the more significantly the freezing point will be lowered by a given quantity of solute.
• The negative sign indicates that as the solute concentration increases, the freezing point decreases.

When you calculate the freezing point depression using the formula \( \Delta T_f = i \cdot m \cdot K_f \), you realize that the cryoscopic constant directly influences how much the freezing point declines for a given system. This constant is intrinsic to the solvent and is used across various solvents to predict changes in physical properties due to solute presence.