Question

The molar conductivity of a 0.02 M weak acid HA is 3.2 mS m\(^2\) mol\(^{-1}\) at 298 K. The \( pK_a \) of HA is ....................

Hint:

To calculate the \( pK_a \) of a weak acid, use the molar conductivity formula and the degree of dissociation \( \alpha \) to find \( K_a \), then take the negative logarithm.

Answer 1

We are given the molar conductivity \( \lambda \) for 0.02 M weak acid HA, which is 3.2 mS m\(^2\) mol\(^{-1}\). We are also given the limiting molar conductivity \( \lambda_0 \) of HA, which is 39 mS m\(^2\) mol\(^{-1}\) at 298 K.
The formula for the molar conductivity \( \lambda \) of a weak acid is: \[ \lambda = \lambda_0 \left( 1 - \alpha \right) \] where \( \alpha \) is the degree of dissociation, given by: \[ \alpha = \frac{c - [HA]}{c} \] where \( c \) is the concentration of the acid and \( [HA] \) is the concentration of the undissociated acid.
We can rearrange the formula for \( \alpha \) as: \[ \alpha = 1 - \frac{\lambda}{\lambda_0} \] Substituting the given values: \[ \alpha = 1 - \frac{3.2}{39} = 1 - 0.0821 = 0.9179 \] The dissociation constant \( K_a \) is related to the degree of dissociation by the formula: \[ K_a = \frac{\alpha^2}{1 - \alpha} \] Substituting \( \alpha = 0.9179 \): \[ K_a = \frac{(0.9179)^2}{1 - 0.9179} = \frac{0.843}{0.0821} = 10.26 \] Now, \( pK_a \) is the negative logarithm of \( K_a \): \[ pK_a = -\log(10.26) = 1.01 \] Final Answer: \[ \boxed{1.0} \]