Question

The van't Hoff factor of Al$_2$(SO$_4$)$_3$ in water is 4.00. The percentage degree of dissociation of Al$_2$(SO$_4$)$_3$ is _______.

Answer 1

75

Answer 2

The dissociation of Al$_2$(SO$_4$)$_3$ in water can be represented by the equation:

Al$_2$(SO$_4$)$_3$ $\rightleftharpoons$ 2Al$^{3+}$ + 3SO$_4^{2-}$

From the dissociation equation, one formula unit of Al$_2$(SO$_4$)$_3$ produces 2 aluminum ions (Al$^{3+}$) and 3 sulfate ions (SO$_4^{2-}$) upon complete dissociation.
The total number of ions produced per formula unit upon complete dissociation is $n = 2 + 3 = 5$.

The van't Hoff factor (i) is related to the degree of dissociation ($\alpha$) by the formula:

$i = 1 + \alpha(n - 1)$

We are given the van't Hoff factor $i = 4.00$ and we have determined $n = 5$. Substituting these values into the formula:

$4.00 = 1 + \alpha(5 - 1)$
$4.00 = 1 + \alpha(4)$
$4.00 = 1 + 4\alpha$

Now, we solve for $\alpha$:

$4\alpha = 4.00 - 1$
$4\alpha = 3.00$
$\alpha = \frac{3.00}{4}$
$\alpha = 0.75$

The degree of dissociation ($\alpha$) is 0.75. The percentage degree of dissociation is $\alpha \times 100\%$.
Percentage degree of dissociation = $0.75 \times 100\% = 75\%$.

The percentage degree of dissociation of Al$_2$(SO$_4$)$_3$ is 75.