What is the expected van't Hoff factor for NaC2H3O2 in an aqueous solution, assuming 100% dissociation.
The van't Hoff factor (i) represents the number of particles a solute dissociates into when dissolved in a solvent. For NaC2H3O2, also known as sodium acetate, it will dissociate into two ions in an aqueous solution:
NaC2H3O2 (aq) → Na+ (aq) + C2H3O2- (aq)
Since there are two ions produced from one solute particle, the expected van't Hoff factor for NaC2H3O2 in an aqueous solution, assuming 100% dissociation, is i = 2.
The van't Hoff factor (i) represents the number of particles into which one formula unit of a compound dissociates in a solution. In this case, we are assuming 100% dissociation of NaC2H3O2 in an aqueous solution.
NaC2H3O2 is sodium acetate, which is an ionic compound that dissociates into Na+ and C2H3O2- ions in solution.
When sodium acetate (NaC2H3O2) fully dissociates, it breaks down into one Na+ ion and one C2H3O2- ion. Therefore, the expected van't Hoff factor (i) for NaC2H3O2 in an aqueous solution is 2.
So, the van't Hoff factor (i) for NaC2H3O2 is 2 when assuming 100% dissociation.
The van't Hoff factor, denoted as "i", represents the number of particles a solute dissociates into in a solution. It is used to adjust colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure.
In this case, we are considering the compound NaC2H3O2, which is sodium acetate. When sodium acetate dissolves in water, it dissociates into sodium ions (Na+) and acetate ions (C2H3O2-). The balanced chemical equation for this dissociation is:
NaC2H3O2(s) ⟶ Na+(aq) + C2H3O2-(aq)
Since we assume 100% dissociation in the aqueous solution, it means that every molecule of sodium acetate breaks apart into one sodium ion and one acetate ion. Therefore, the van't Hoff factor (i) for NaC2H3O2 in this case is 2.
To summarize, the expected van't Hoff factor for NaC2H3O2 in an aqueous solution, assuming 100% dissociation, is 2.