Find out the ionisation constant of a weak ac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a weak acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$. The ionisation constant $K_a$ is given by $K_a = \frac{C \

Vedclass · Accepted Answer

$K_a = \frac{C (\Lambda_m^c)^2}{\Lambda_m^o(\Lambda_m^o - \Lambda_m^c)}$

Vedclass · Answer

(B) For a weak acid $HA$,the dissociation equilibrium is $HA ightleftharpoons H^+ + A^-$. The ionisation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{(1 - \alpha)}$,where $\alpha$ is the degree of dissociation. The degree of dissociation $\alpha$ is related to molar conductivity $\Lambda_m^c$ and limiting molar conductivity $\Lambda_m^o$ as $\alpha = \frac{\Lambda_m^c}{\Lambda_m^o}$. Substituting $\alpha$ into the expression for $K_a$: $K_a = \frac{C 	imes (\frac{\Lambda_m^c}{\Lambda_m^o})^2}{1 - \frac{\Lambda_m^c}{\Lambda_m^o}} = \frac{C 	imes \frac{(\Lambda_m^c)^2}{(\Lambda_m^o)^2}}{\frac{\Lambda_m^o - \Lambda_m^c}{\Lambda_m^o}} = \frac{C (\Lambda_m^c)^2}{\Lambda_m^o(\Lambda_m^o - \Lambda_m^c)}$.

Find out the ionisation constant of a weak acid $(HA)$ in terms of $\Lambda_m^o$ and $\Lambda_m^c$. (Given $\alpha$ cannot be ignored with respect to $1$)

Similar Questions

Discuss the construction and uses of a conductivity cell.

The resistance of a $0.2 \ M$ solution of an electrolyte is $30 \ \Omega$ and conductivity is $1.2 \ S \ m^{-1}$. What is the value of the cell constant (in $cm^{-1}$)?

The increase in equivalent conductance of an electrolyte solution with dilution is primarily due to the increase in:

$0.0129 \ S \ cm^{-1}$ is equivalent to how many $S \ m^{-1}$ (in $S \ m^{-1}$)?

The limiting molar conductivities of $X^{+}$ and $Y^{-2}$ ions are $45$ and $110 \ S \ cm^2 \ mol^{-1}$ respectively. The $\Lambda_{m}^{\infty}$ of $X_2Y$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module