The resistance of $0.1 \ M$ weak acid $HA$ in a conductivity cell is $2 \times 10^3 \ \Omega$. The cell constant of the cell is $0.78 \ cm^{-1}$ and $\lambda_{m}^{\circ}$ of acid $HA$ is $390 \ S \ cm^2 \ mol^{-1}$. The $pH$ of the solution is

  • A
    $3.3$
  • B
    $4.2$
  • C
    $5$
  • D
    $3$

Explore More

Similar Questions

The equivalent conductance of $H_2SO_4$ at infinite dilution is $384 \, \Omega^{-1} \, cm^2 \, eq^{-1}$. If the solution contains $49 \, g$ of $H_2SO_4$ per liter and the specific conductance is $18.4 \, \Omega^{-1} \, cm^{-1}$,calculate the degree of dissociation in percentage.

Difficult
View Solution

Given that $\Lambda_{m}^{\infty} = 133.4 \, S \, cm^{2} \, mol^{-1} (AgNO_{3})$; $\Lambda_{m}^{\infty} = 149.9 \, S \, cm^{2} \, mol^{-1} (KCl)$ and $\Lambda_{m}^{\infty} = 144.9 \, S \, cm^{2} \, mol^{-1} (KNO_{3})$,the molar conductivity at infinite dilution for $AgCl$ is $....... \, S \, cm^{2} \, mol^{-1}$.

Conductivity of a saturated solution of a sparingly soluble salt $AB$ at $298 \ K$ is $1.85 \times 10^{-5} \ S \ m^{-1}$. Solubility product of the salt $AB$ at $298 \ K$ is. Given $\Lambda_{m}^{\circ}(AB) = 140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$.

What is the molar conductivity of $CH_3COOH$ at infinite dilution if the molar conductivities of $H_2SO_4$, $K_2SO_4$, and $CH_3COOK$ at infinite dilution are $x$, $y$, and $z$ $\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ respectively?

Which of the following expressions represents the molar conductivity of an $AB_3$ type electrolyte?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo