# Assuming 100% dissociation, what is the freezing point and boiling point of 3.39 m K_3PO_4(aq)?

Jul 16, 2017

$\text{f.p.} = - 25.2$ $\text{^"o""C}$

$\text{b.p.} = 106.94$ $\text{^"o""C}$

#### Explanation:

We're asked to find the new freezing and boiling points of a $3.39 m$ ${\text{K"_3"PO}}_{4}$ solution (given it ionizes completely).

To do this, we can use the equations

$\Delta {T}_{f} = i m {K}_{f}$ (freezing point depression)

$\Delta {T}_{b} = i m {K}_{b}$ (boiling point elevation)

where

• $\Delta {T}_{f}$ and $\Delta {T}_{b}$ are the changes in freezing and boiling point temperatures, respectively.

• $i$ is the van't Hoff factor, which is essentially the number of dissolved ions per unit of solute (equal to $4$ here; there are $4$ ions per unit of ${\text{K"_3"PO}}_{4}$).

• $m$ is the molality of the solution, given as $3.39 m$

• ${K}_{f}$ is the molal freezing point constant for the solvent (water), which (although not given) is $1.86$ $\text{^"o""C/} m$

• ${K}_{b}$ is the molal boiling point constant for the solvent (water), equal to $0.512$ $\text{^"o""C/} m$

Plugging in known values, we have

$\Delta {T}_{f} = \left(4\right) \left(3.39 \cancel{m}\right) \left(1.86 \frac{\text{^"o""C}}{\cancel{m}}\right) = 25.2$ $\text{^"o""C}$

$\Delta {T}_{b} = \left(4\right) \left(3.39 \cancel{m}\right) \left(0.512 \frac{\text{^"o""C}}{\cancel{m}}\right) = 6.94$ $\text{^"o""C}$

These are by how much the freezing and boiling point temperatures decrease (freezing point) and increase (boiling point).

To find the new freezing point, we simply subtract this value from the normal freezing point of water, $0.0$ $\text{^"o""C}$:

color(red)("new f.p.") = 0^"o""C" - 25.2^"o""C" = color(red)(-25.2 color(red)(""^"o""C"

The new boiling point is found nearly the same way, but by adding the temperature change to the normal boiling point of water, $100.00$ $\text{^"o""C}$:

color(blue)("new b.p.") = 100.00^"o""C" + 6.94^"o""C" = color(blue)(106.94 color(blue)(""^"o""C"