Question

assuming 100% dissociation, calculate the freezing point (t_f) and boiling point (t_b) of 3.11 m k_3po_4(aq). colligative constants can be found in the chempendix.

Explanation:

Step1: Determine the van't Hoff factor (i)

Since $K_3PO_4$ dissociates into $3K^+$ and $PO_4^{3 - }$, $i = 4$ (3 potassium ions and 1 phosphate ion).

Step2: Calculate the freezing - point depression ($\Delta T_f$)

The formula for freezing - point depression is $\Delta T_f=i\times K_f\times m$. For water, $K_f = 1.86^{\circ}C/m$. Given $m = 3.11m$ and $i = 4$. So, $\Delta T_f=4\times1.86^{\circ}C/m\times3.11m = 23.1^{\circ}C$. The normal freezing point of water is $T_{f,water}=0^{\circ}C$. Then $T_f=T_{f,water}-\Delta T_f=0 - 23.1=- 23.1^{\circ}C$.

Step3: Calculate the boiling - point elevation ($\Delta T_b$)

The formula for boiling - point elevation is $\Delta T_b=i\times K_b\times m$. For water, $K_b = 0.512^{\circ}C/m$. Given $m = 3.11m$ and $i = 4$. So, $\Delta T_b=4\times0.512^{\circ}C/m\times3.11m = 6.38^{\circ}C$. The normal boiling point of water is $T_{b,water}=100^{\circ}C$. Then $T_b=T_{b,water}+\Delta T_b=100 + 6.38 = 106.38^{\circ}C$.

Answer:

$T_f=-23.1^{\circ}C$
$T_b = 106.38^{\circ}C$