The Question is from neet-chemistry , and it was created by the author Mr. S M HUSSAIN

Question

The observed freezing point depression of a 0.01 m aqueous solution of a weak acid HA is 0.0205 °C.  If the $K_f$ of water is 1.86 K kg/mol, which expression correctly calculates the percent ionization of the weak acid?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Freezing point depression and percent ionization of weak acids.
 **Related Topics**: Colligative properties, equilibrium constant.
 **Prerequisites**: Understanding of freezing point depression ($\Delta T_f = i K_f m$), where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality.
## SOLUTION PATHWAY
**Given Information**:
  $ \Delta T_f = 0.0205 ^\circ C $ 
  $ m = 0.01 m $
  $ K_f = 1.86 K kg/mol $
**Method & Steps**:
 1. Calculate the van't Hoff factor (i).
  Reasoning: The freezing point depression is given by $ \Delta T_f = i K_f m $.  We can solve for i to find the effective number of particles in solution.
  Formula/Rule: $ i = \frac{\Delta T_f}{K_f m} $
 2. Calculate the percent ionization.
  Working: Substitute the given values into the formula: $ i = \frac{0.0205}{1.86 	imes 0.01} = 1.102 $.  Since the acid is weak, the van't Hoff factor represents the degree of dissociation. For a weak monoprotic acid, $i = 1 + \alpha$, where $\alpha$ is the degree of ionization. Therefore, $\alpha = i - 1 = 1.102 - 1 = 0.102$. The percent ionization is $ \alpha 	imes 100\% $. 
  Key Point: The van't Hoff factor (i) reflects the number of particles the solute dissociates into in solution. For a completely undissociated weak acid, i=1. For complete dissociation, i=2 (for a monoprotic acid).
 3. Express the percent ionization.
  Result: Percent ionization $= 0.102 	imes 100\% = 10.2\%$. The expression that yields this result is option A.
  Verification: Option A correctly incorporates the calculation of i and then subtracts 1 to find α, finally multiplying by 100 for percentage. 
 ## OPTION ANALYSIS
A ✓: $ \frac{(0.0205/1.86/0.01) - 1}{(2-1)} 	imes 100$ correctly calculates i, subtracts 1 to account for the undissociated acid, and multiplies by 100 to get the percentage.
B :x: This expression incorrectly inverts the relationship between $\Delta T_f$ and i.
C :x: This expression calculates i but doesn't subtract 1 to find the degree of ionization.
D :x: This expression uses an incorrect combination of the given values.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly using the formula for freezing point depression: Ensure you use the correct formula and understand the meaning of each variable.
  2.Forgetting to subtract 1 from the van't Hoff factor to find the degree of ionization: Remember that i represents the total number of particles, and we need the fraction that ionized.
 **Quick Tips**:
  1.Always check the units to ensure consistency throughout the calculation.
  2.Remember that the van't Hoff factor for a weak acid is always between 1 and 2 (for a monoprotic acid).

Mr. S M HUSSAIN · Answer

The correct Option is "$ \frac{(0.0205/1.86/0.01) - 1}{(2-1)} 	imes 100$"

Mr. S M HUSSAIN · Answer

$ \frac{(1.86 	imes 0.01/0.0205) - 1}{(2-1)} 	imes 100$

Mr. S M HUSSAIN · Answer

$ \frac{(0.0205/1.86/0.01)}{(2-1)} 	imes 100$

Mr. S M HUSSAIN · Answer

$ \frac{(1.86 	imes 0.01 	imes 0.0205)}{(2-1)} 	imes 100$

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