An aqueous solution of a salt $MX_2$ at certain temperature has a van?t Hoff factor of 2. The degree of dissociation for this solution of the salt is :

Question

Consider the dissociation equilibrium of the following weak acid: \[ \mathrm{HA \rightleftharpoons H^+(aq) + A^-(aq)} \] If the $pK_a$ of the acid is $4$, then the pH of a $10\ \text{mM}$ HA solution is ________ (Nearest integer). (Given: The degree of dissociation can be neglected with respect to unity)

Answer 1

To solve the problem of finding the degree of dissociation of the salt MX_2 at a certain temperature, given that the van't Hoff factor is 2, we need to understand the dissociation process and the relationship between the van't Hoff factor and degree of dissociation.

The van't Hoff factor (i) is given by:

i = 1 + \alpha(n - 1)

where:

\alpha is the degree of dissociation.
n is the number of particles the compound dissociates into.

For MX_2, the dissociation in solution can be represented as:

MX_2 \rightarrow M^{2+} + 2X^−

This indicates that MX_2 dissociates into 1 M^{2+} ion and 2 X^− ions, which implies n = 3.

Given that the van't Hoff factor is 2, we use the formula:

2 = 1 + \alpha(3 - 1)

Simplifying this:

2 = 1 + 2\alpha

Subtract 1 from both sides:

1 = 2\alpha

Solving for \alpha:

\alpha = \frac{1}{2} = 0.5

Thus, the degree of dissociation for this solution of the salt MX_2 is 0.5, which corresponds to the correct option.

Answer 2