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The dissociation constant of n-butyric a...

The dissociation constant of n-butyric acid is `1.6 xx 10^(-5)` and the molar conductivity at infinite dilution is `380 xx 10^(-4) Sm^(2) mol^(-1)`. The specific conductance of the `0.01M` acid solution is

A

`1.52 xx 10^(-5) Sm^(-1)`

B

`1.52 xx 10^(-2) Sm^(-1)`

C

`1.52 xx 10^(-3) Sm^(-1)`

D

none

Text Solution

AI Generated Solution

The correct Answer is:
To find the specific conductance (κ) of a 0.01 M n-butyric acid solution, we can follow these steps: ### Step 1: Write down the given data - Dissociation constant (Ka) of n-butyric acid = \(1.6 \times 10^{-5}\) - Molar conductivity at infinite dilution (\( \Lambda_m^\infty \)) = \(380 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}\) - Concentration (C) = \(0.01 \, \text{M}\) ### Step 2: Calculate the degree of dissociation (α) For a weak acid, the dissociation constant is given by: \[ K_a = C \alpha^2 \] Rearranging this gives: \[ \alpha = \sqrt{\frac{K_a}{C}} \] Substituting the values: \[ \alpha = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} = 0.04 \] ### Step 3: Calculate the molar conductivity at the given concentration (\( \Lambda_{mC} \)) The molar conductivity at the given concentration can be calculated using: \[ \Lambda_{mC} = \alpha \cdot \Lambda_m^\infty \] Substituting the values: \[ \Lambda_{mC} = 0.04 \times (380 \times 10^{-4}) = 15.2 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1} \] ### Step 4: Convert the molar conductivity to specific conductance (κ) The relationship between molar conductivity and specific conductance is given by: \[ \Lambda_{mC} = \frac{\kappa \cdot 1000}{C} \] Rearranging gives: \[ \kappa = \frac{\Lambda_{mC} \cdot C}{1000} \] Substituting the values: \[ \kappa = \frac{15.2 \times 10^{-4} \cdot 0.01}{1000} = 1.52 \times 10^{-4} \, \text{S cm}^{-1} \] ### Step 5: Convert to S m⁻¹ To convert the specific conductance from S cm⁻¹ to S m⁻¹: \[ \kappa = 1.52 \times 10^{-4} \, \text{S cm}^{-1} = 1.52 \times 10^{-2} \, \text{S m}^{-1} \] ### Final Answer The specific conductance of the 0.01 M n-butyric acid solution is: \[ \kappa = 1.52 \times 10^{-2} \, \text{S m}^{-1} \] ---

To find the specific conductance (κ) of a 0.01 M n-butyric acid solution, we can follow these steps: ### Step 1: Write down the given data - Dissociation constant (Ka) of n-butyric acid = \(1.6 \times 10^{-5}\) - Molar conductivity at infinite dilution (\( \Lambda_m^\infty \)) = \(380 \times 10^{-4} \, \text{S m}^2 \text{mol}^{-1}\) - Concentration (C) = \(0.01 \, \text{M}\) ### Step 2: Calculate the degree of dissociation (α) ...
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Knowledge Check

  • A weak monobasic acid is 5% dissociated in 0.01 mol dm^(-3) solution. The limiting molar conductivity at infinite dilution is 4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1) . Calculate the conductivity of a 0.05 mol dm^(-3) solution of the acid.

    A
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    B
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    C
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    D
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