Question

The molar conductivity of $0.007 \,M$ acetic acid is $20\, S\, cm ^{2} mol ^{-1}$. What is the dissociation constant of acetic acid? Choose the correct option. $\left[\Lambda_{ H ^{+}}^{\circ}=350 \,S \,cm ^{2} mol ^{-1}, \Lambda_{ CH _{3} COO ^{-}}^{\circ}=50\, S\, cm ^{2} mol ^{-1}\right]$

• A. $1.75 \times 10^{-4}\, mol\, L ^{-1}$
• B. $2.50 \times 10^{-4}\, mol\, L ^{-1}$
• C. $1.75 \times 10^{-5}\, mol\, L ^{-1}$
• D. $2.50 \times 10^{-5} \, mol\, L ^{-1}$

Answer 1

The Correct Option is C

To find the dissociation constant (\(K_a\)) of acetic acid, we need to use the relationship between molar conductivity and dissociation. Here is the step-by-step solution to find the dissociation constant: 

Determine the limiting molar conductivity (\(\Lambda^\circ_m\)) of acetic acid:

The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. For acetic acid, \(CH_3COOH\), it dissociates into \(H^+\) and \(CH_3COO^-\).

\[\Lambda^\circ_m = \Lambda_{H^+}^\circ + \Lambda_{CH_3COO^-}^\circ\]

Given, \(\Lambda_{H^+}^\circ = 350 \, S \, cm^2 \, mol^{-1}\) and \(\Lambda_{CH_3COO^-}^\circ = 50 \, S \, cm^2 \, mol^{-1}\).

\[\Lambda^\circ_m = 350 + 50 = 400 \, S \, cm^2 \, mol^{-1}\]

Calculate the degree of dissociation (\(\alpha\)) using the formula:

\[\alpha = \frac{\Lambda_m}{\Lambda^\circ_m} = \frac{20}{400} = 0.05\]

Use the degree of dissociation to find the dissociation constant \(K_a\):

\[K_a = c \alpha^2 = 0.007 \times (0.05)^2\]

1.  

\[K_a = 0.007 \times 0.0025 = 0.0000175 = 1.75 \times 10^{-5} \, mol \, L^{-1}\]

Therefore, the dissociation constant of acetic acid is \(1.75 \times 10^{-5} \, mol \, L^{-1}\). The correct option is:

\(1.75 \times 10^{-5} \, mol \, L^{-1}\)