What is the value of Van't Hoff factor (i) for $K_{2}SO_{4}$?

Question

Accepted Answer

The correct option is (1):

The Van't Hoff factor (i) is a measure of the effect of a solute on colligative properties such as osmotic pressure, relative lowering of vapor pressure, elevation of boiling point, and depression of freezing point. It represents the number of particles (ions or molecules) that a solute dissociates into when dissolved in a solvent. For $K_{2}SO_{4}$ (potassium sulfate), we need to consider its dissociation in an aqueous solution.

Potassium sulfate is an ionic compound, and it dissociates completely in water as follows:
$\mathrm{K\_{2}SO\_{4} (aq) \rightarrow 2K^{+} (aq) + SO\_{4}^{2-} (aq)}$

From this dissociation equation, we can see that one molecule of $K_{2}SO_{4}$ produces two potassium ions ($\mathrm{K}^{+}$) and one sulfate ion ($\mathrm{SO_{4}^{2-}}$). Therefore, the total number of ions produced from one formula unit of $K_{2}SO_{4}$ is $2 + 1 = 3$.

Thus, for an ideal dilute solution where complete dissociation occurs, the Van't Hoff factor (i) for $K_{2}SO_{4}$ is 3.

The incorrect option is (2): If the Van't Hoff factor were 2, it would imply that $K\_{2}SO\_{4}$ dissociates into only two particles, which is incorrect as it forms three ions.
The incorrect option is (3): If the Van't Hoff factor were 1, it would mean that $K\_{2}SO\_{4}$ does not dissociate at all or exists as a single particle, which is typical for non-electrolytes. However, $K\_{2}SO\_{4}$ is a strong electrolyte.
The incorrect option is (4): A Van't Hoff factor of 4 would mean that $K\_{2}SO\_{4}$ dissociates into four particles, which is not the case based on its chemical formula and dissociation.