Question

\( \text{K}_2\text{HgI}_4 \) is \( 40\% \) ionised in aqueous solution. The value of its van't Hoff factor (i) is:

• A. 1.6
• B. 1.8
• C. 2.2
• D. 2.0

Hint:

Remember that the complex ion \( [\text{HgI}_4]^{2-} \) remains intact as a single entity in aqueous solution; it does not dissociate further into mercury and iodine ions.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the van't Hoff factor \( i \) for the complex salt \( \text{K}_2\text{HgI}_4 \) when it undergoes \( 40\% \) ionization in water.

Step 2: Key Formula or Approach:
For association or dissociation, the van't Hoff factor \( i \) is related to the degree of ionization \( \alpha \) by:
\[ i = 1 + (n - 1)\alpha \] where \( n \) is the number of ions produced per formula unit of solute.

Step 3: Detailed Explanation:
The ionization of the complex salt \( \text{K}_2[\text{HgI}_4] \) in aqueous solution occurs as follows:
\[ \text{K}_2[\text{HgI}_4] \rightleftharpoons 2\text{K}^+ + [\text{HgI}_4]^{2-} \] From this equation, 1 molecule of the salt dissociates to give a total of 3 ions (2 potassium ions and 1 complex tetraiodomercurate ion).
Thus, the number of particles \( n = 2 + 1 = 3 \).
Given:
- Degree of ionization, \( \alpha = 40\% = 0.4 \)
Using the relation:
\[ i = 1 + (3 - 1)\alpha \] \[ i = 1 + 2(0.4) \] \[ i = 1 + 0.8 = 1.8 \] Thus, the van't Hoff factor is \( 1.8 \).

Step 4: Final Answer:
(B) 1.8