Questions: Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. Highest freezing point Lowest freezing point NH4Br K3PO4 K2CO3

Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points.

Highest freezing point

Lowest freezing point

NH4Br
K3PO4
K2CO3

Transcript text: Assuming equal concentrations and complete dissociation, arrange these aqueous solutions by their freezing points. Highest freezing point $\square$ Lowest freezing point $\square$ $\mathrm{NH}_{4} \mathrm{Br}$ $\mathrm{K}_{3} \mathrm{PO}_{4}$ $\mathrm{K}_{2} \mathrm{CO}_{3}$

Solution

Solution Steps

Step 1: Understanding Freezing Point Depression

Freezing point depression is a colligative property, which means it depends on the number of solute particles in a solution, not the identity of the solute. The formula for freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

where:

$\Delta T_f$ is the freezing point depression.
$i$ is the van't Hoff factor, representing the number of particles the solute dissociates into.
$K_f$ is the freezing point depression constant of the solvent.
$m$ is the molality of the solution.

Step 2: Determine the Van't Hoff Factor for Each Compound

To determine the freezing point depression, we need to calculate the van't Hoff factor ($i$) for each compound, assuming complete dissociation:

$\mathrm{NH}_4\mathrm{Br}$ dissociates into $\mathrm{NH}_4^+$ and $\mathrm{Br}^-$, so $i = 2$.
$\mathrm{K}_3\mathrm{PO}_4$ dissociates into 3 $\mathrm{K}^+$ and 1 $\mathrm{PO}_4^{3-}$, so $i = 4$.
$\mathrm{K}_2\mathrm{CO}_3$ dissociates into 2 $\mathrm{K}^+$ and 1 $\mathrm{CO}_3^{2-}$, so $i = 3$.

Step 3: Arrange Solutions by Freezing Point

The solution with the highest freezing point will have the smallest $\Delta T_f$, which corresponds to the smallest $i$. Conversely, the solution with the lowest freezing point will have the largest $\Delta T_f$, corresponding to the largest $i$.

$\mathrm{NH}_4\mathrm{Br}$ with $i = 2$ will have the highest freezing point.
$\mathrm{K}_2\mathrm{CO}_3$ with $i = 3$ will have a lower freezing point.
$\mathrm{K}_3\mathrm{PO}_4$ with $i = 4$ will have the lowest freezing point.

Final Answer

\[ \begin{align_} \text{Highest freezing point:} & \quad \boxed{\mathrm{NH}_4\mathrm{Br}} \\ \text{Lowest freezing point:} & \quad \boxed{\mathrm{K}_3\mathrm{PO}_4} \end{align_} \]

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