Question

question 34 of 36
assuming 100% dissociation, calculate the freezing point ($t_f$) and boiling point ($t_b$) of $1.92\\ m\\ k_3po_4(aq)$.
colligative constants can be found in the chempendix.
$t_f =$ ____ $^\circ c$
$t_b =$ ____ $^\circ c$

Explanation:

Step1: Find van't Hoff factor

For $\text{K}_3\text{PO}_4$, 100% dissociation gives $3\text{K}^+ + \text{PO}_4^{3-}$, so $i=4$.

Step2: Get colligative constants

For water: $K_f=1.86\ ^\circ\text{C}/m$, $K_b=0.512\ ^\circ\text{C}/m$, pure $T_f=0^\circ\text{C}$, pure $T_b=100^\circ\text{C}$.

Step3: Calculate freezing point depression

$\Delta T_f = i \times K_f \times m = 4 \times 1.86 \times 1.92$
$\Delta T_f = 14.3616\ ^\circ\text{C}$

Step4: Find final freezing point

$T_f = T_f^\circ - \Delta T_f = 0 - 14.3616$

Step5: Calculate boiling point elevation

$\Delta T_b = i \times K_b \times m = 4 \times 0.512 \times 1.92$
$\Delta T_b = 3.93216\ ^\circ\text{C}$

Step6: Find final boiling point

$T_b = T_b^\circ + \Delta T_b = 100 + 3.93216$

Answer:

$T_f = -14.4^\circ\text{C}$
$T_b = 103.9^\circ\text{C}$