Problem 115 - A Complete Resource Book in Chemistry for JEE Main

Question

A \(0.2\) molal aqueous solution of a weak acid (HX) is \(20 \%\) ionized. The freezing point of this solution is (Given \(\mathrm{K}_{\mathrm{r}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) for water \()\) (a) \(-0.45^{\circ} \mathrm{C}\) (b) \(-0.90^{\circ} \mathrm{C}\) (c) \(-0.21^{\circ} \mathrm{C}\) (d) \(-0.43^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The freezing point of the solution is approximately \(-0.45\, °C\), so the answer is (a) \(-0.45\, °C\).

Step 1: Determine the Van't Hoff factor (i)

The weak acid HX is 20% ionized. Therefore, out of 1 mole of HX, 0.2 moles ionize to give 0.2 moles of H⁺ and 0.2 moles of X⁻. This leaves 0.8 moles as non-ionized HX, resulting in a total of \(1 + 0.2 = 1.2\) moles of distinct particles in the solution. Thus, \(i = 1.2\).

Step 2: Use the freezing point depression formula

The formula for freezing point depression is \(\Delta T_f = i \cdot K_f \cdot m\), where \(\Delta T_f\) is the change in freezing point, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution. Given \(m = 0.2\, \text{mol/kg}\) and \(K_f = 1.86\, °C \cdot \text{kg/mol}\), substitute the known values: \(\Delta T_f = 1.2 \cdot 1.86 \cdot 0.2\).

Step 3: Calculate the change in freezing point

Calculate \(\Delta T_f = 1.2 \times 1.86 \times 0.2 = 0.4464\, °C\).

Step 4: Determine the new freezing point of the solution

The freezing point of pure water is \(0\, °C\). The freezing point of the solution is lowered by \(0.4464\, °C\) due to the presence of the solute. Thus, the new freezing point is \(-0.4464\, °C\).

Step 5: Choose the closest answer

The calculated freezing point is \(-0.4464\, °C\). The closest option provided is \(-0.45\, °C\). Therefore, the answer is option (a) \(-0.45\, °C\).

Key Concepts

Van't Hoff Factor

The Van't Hoff factor, often symbolized as "i," plays a significant role in determining how solutions behave in terms of colligative properties like freezing point depression and boiling point elevation. It reflects the number of distinct particles a solute contributes to a solution when dissolved.

• For non-electrolytes like sugar, which do not ionize, the Van’t Hoff factor is generally 1.
• For strong electrolytes like NaCl, which completely dissociate in solution, "i" would equal the number of ions formed, which is 2 in the case of NaCl (1 Na⁺ and 1 Cl^-).
• For weak electrolytes like weak acids or bases that only partially ionize, "i" is typically between 1 and the number of ions that can be formed. In our problem, a weak acid ionizes to a certain extent, resulting in a Van’t Hoff factor of 1.2 because 20% of HX ionizes to form H⁺ and X^-.

Understanding the Van't Hoff factor is crucial as it directly affects how significantly a solute will affect the freezing and boiling points of a solution.

Cryoscopic Constant

The cryoscopic constant, denoted as "K_f," is a property unique to each solvent. It is crucial in the equation for freezing point depression, \[ \Delta T_f = i \cdot K_f \cdot m\] where \(\Delta T_f\) is the change in freezing point, "i" is the Van't Hoff factor, and "m" is the molality of the solution.

• Nature-Dependent: The value of the cryoscopic constant varies depending on the solvent. For instance, water has a cryoscopic constant K_f of 1.86 °C kg/mol, which is used in our problem.
• Role in Calculation: It quantifies the magnitude of freezing point depression a solute can cause per molal concentration. Hence, a higher K_f means a greater depression for the same solute concentration.

Cryoscopic constant provides a handy measure when calculating how the arrival of a solute changes the thermal properties of a particular solvent.

Ionization of Weak Acid

The ionization of a weak acid is a fundamental chemical process where a portion of the acid molecules dissociate into ions in a solution. Unlike strong acids, weak acids do not fully dissociate, meaning not all original acid molecules break into ions.

• Partial Ionization: A weak acid dissociates to a limited extent in water. For instance, if the acid HX is 20% ionized, this means 20% of HX molecules break into H⁺ and X^- ions, and 80% remain as non-ionized HX molecules.
• Calculation Implications: In freezing point depression, the incomplete ionization affects the Van’t Hoff factor. For example, with 20% ionization, we end up with an "i" value of 1.2 instead of 2.

Understanding weak acid ionization helps identify the degree to which a solution’s properties change compared to scenarios involving strong acids or non-electrolytes. It's essential in solving problems related to colligative properties where the nature of the solute affects physical properties of the solution.