Finding the Van't Hoff Factor (i) of a Complex Salt
In the Solutions chapter, calculating colligative properties (like boiling point elevation or osmotic pressure) for electrolytes requires you to know the Van't Hoff Factor (i).
If a salt is completely dissociated, calculating i is easy. But what if the salt is only partially dissociated? In this scenario, we must use the Degree of Dissociation (α) formula. Let's solve a classic JEE/NEET problem involving a coordination complex.
Video Tutorial: The α Formula Method
Watch Abhishek Sengar sir from CHEMCA break down the ionization of Potassium Ferrocyanide and use the dissociation formula to find the exact Van't Hoff factor.
Step-by-Step Calculation
For dissociation, the relationship between the Degree of Dissociation (α) and the Van't Hoff Factor (i) is:
α = (i - 1) / (n - 1) Where n is the total number of ions produced by one formula unit of the salt.
Problem Statement:
If α = 0.6, find the Van't Hoff factor (i) for K4[Fe(CN)6].
-
Ionize the Complex (Find n):
The most common mistake students make is breaking the complex bracket apart. Remember, the coordination sphere [Fe(CN)6] stays completely intact in an aqueous solution! It acts as a single, large polyatomic ion.
K4[Fe(CN)6] → 4 K+ + [Fe(CN)6]4-We get 4 Potassium ions + 1 Complex ion. Therefore, n = 5.
Fig: Ionization of Potassium Ferrocyanide yields 5 total ions.
-
Apply the Formula:
Substitute n = 5 and α = 0.6 into the master equation.
0.6 = (i - 1) / (5 - 1)
0.6 = (i - 1) / 4 -
Solve for i:
Multiply 4 by 0.6:
2.4 = i - 1
i = 2.4 + 1
You can now use this value of i = 3.4 in formulas like ΔTf = i · Kf · m to find the precise freezing point depression!
Practice Questions for JEE & NEET
Test your speed and accuracy with these two common variations of the Van't Hoff calculation.
Question 1: What would the Van't Hoff factor (i) be for K4[Fe(CN)6] if the question stated that the salt undergoes complete dissociation (100% ionization)?
Answer: i = 5
Reasoning:
- Complete dissociation means the degree of dissociation α = 1 (or 100%).
- If we plug α = 1 into the formula: 1 = (i - 1) / (n - 1)
- This mathematically simplifies to: i = n.
- Since n = 5, the Van't Hoff factor is exactly 5. This is a great shortcut to remember: For strong, fully dissociated electrolytes, i simply equals the number of ions!
Question 2: Calculate the Van't Hoff factor for an aqueous solution of Potassium Sulfate (K2SO4) if it is 80% dissociated in water.
Answer: i = 2.6
Step-by-step Solution:
- Step 1 (Find n): Potassium sulfate ionizes as K2SO4 → 2 K+ + SO42-. Thus, n = 3.
- Step 2 (Identify α): 80% dissociation means α = 0.8.
- Step 3 (Formula): α = (i - 1) / (n - 1)
- 0.8 = (i - 1) / (3 - 1)
- 0.8 = (i - 1) / 2
- 1.6 = i - 1 → i = 2.6
No comments:
Post a Comment