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 of acetic acid, we need to use the concept of molar conductivity and the relationship between the degree of dissociation (\(\alpha\)) and the dissociation constant (\(K_a\)). The relevant formulas are as follows:

1. The molar conductivity at any concentration is given by: \(\Lambda_m = \alpha \Lambda_m^{\circ}\), where \(\Lambda_m^{\circ}\) is the limiting molar conductivity.
2. The degree of dissociation (\(\alpha\)) is calculated as: \(\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}\).
3. The limiting molar conductivity of acetic acid is given by: \(\Lambda_m^{\circ} = \Lambda_{H^+}^{\circ} + \Lambda_{CH_3COO^-}^{\circ} = 350 \, S \, cm^2 \, mol^{-1} + 50 \, S \, cm^2 \, mol^{-1} = 400 \, S \, cm^2 \, mol^{-1}\).
4. Using the given molar conductivity of acetic acid: \(\Lambda_m = 20 \, S \, cm^2 \, mol^{-1}\), \[ \alpha = \frac{20}{400} = 0.05 \]
5. For weak electrolytes, the dissociation constant is calculated using: \(K_a = c\alpha^2\), where \(c\) is the concentration of acetic acid, \(c = 0.007 \, M\). \[ K_a = 0.007 \times (0.05)^2 = 0.007 \times 0.0025 = 0.0000175 \, mol \, L^{-1} \]

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