To find the value of the Van't Hoff factor (i) for acetic acid (CH3COOH) given that it dissociates 60% and dimerizes 40%, we need to consider the contributions from both processes. The Van't Hoff factor is a measure of the number of particles into which a solute dissociates in solution. In this case, acetic acid can dissociate into ions or dimerize into pairs of molecules. Let's break this down step by step.
Understanding the Processes
Acetic acid (CH3COOH) can behave in two ways in a solution:
- Dissociation: When it dissociates, it breaks down into ions: CH3COOH → CH3COO⁻ + H⁺. This reaction produces two particles from one molecule.
- Dimerization: In some cases, two acetic acid molecules can combine to form a dimer: 2 CH3COOH → (CH3COOH)2. This reaction decreases the number of particles since it combines two molecules into one.
Calculating the Van't Hoff Factor
Let's assume we start with 1 mole of acetic acid. According to the problem:
- 60% dissociates: This means 0.6 moles dissociate into ions.
- 40% dimerizes: This means 0.4 moles form dimers.
Now, let’s analyze the contributions:
Dissociation Contribution
For the 0.6 moles that dissociate:
- Each mole produces 2 particles (1 CH3COO⁻ and 1 H⁺).
- Total particles from dissociation = 0.6 moles × 2 = 1.2 particles.
Dimerization Contribution
For the 0.4 moles that dimerize:
- Each mole of dimer (formed from 2 molecules) counts as 1 particle.
- Total particles from dimerization = 0.4 moles × 1 = 0.4 particles.
Combining the Contributions
Now, we can sum the total number of particles:
- Total particles = Particles from dissociation + Particles from dimerization
- Total particles = 1.2 + 0.4 = 1.6.
Final Result
The Van't Hoff factor (i) is thus equal to 1.6. Therefore, the correct answer is:
This calculation shows how the dissociation and dimerization processes affect the overall particle concentration in solution, which is essential in understanding colligative properties and other related phenomena in chemistry.