Question

The specific conductance of 0.01 M solution of acetic acid was found to be 0.0162 S m⁻¹ at 25°C. Calculate the percentage degree of dissociation of the acid. (Molar conductance of acetic acid at infinite dilution is 390.6 ×10⁻⁴ S m² mol⁻¹ at 25°C). (Nearest Integer)

Answer 1

4

Answer 2

To calculate the percentage degree of dissociation of acetic acid, we follow these steps:

1. Convert the concentration units:
   The given concentration is in M (mol L⁻¹), and specific conductance is in S m⁻¹, while molar conductance at infinite dilution is in S m² mol⁻¹. To ensure consistent units, we convert the concentration from mol L⁻¹ to mol m⁻³.
   $1 \text{ L} = 10^{-3} \text{ m}^3$
   So, $0.01 \text{ M} = 0.01 \text{ mol L}^{-1} = 0.01 \text{ mol} / (10^{-3} \text{ m}^3) = 10 \text{ mol m}^{-3}$

2. Calculate the molar conductance ($\Lambda_m$) at the given concentration:
   The formula for molar conductance is:
   $\Lambda_m = \frac{\kappa}{C}$
   where $\kappa$ is the specific conductance and $C$ is the concentration.
   Given $\kappa = 0.0162 \text{ S m}^{-1}$ and $C = 10 \text{ mol m}^{-3}$.
   $\Lambda_m = \frac{0.0162 \text{ S m}^{-1}}{10 \text{ mol m}^{-3}} = 0.00162 \text{ S m}^2 \text{ mol}^{-1}$

3. Calculate the degree of dissociation ($\alpha$):
   The degree of dissociation is given by the ratio of molar conductance at a given concentration to the molar conductance at infinite dilution:
   $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$
   Given $\Lambda_m^\circ = 390.6 \times 10^{-4} \text{ S m}^2 \text{ mol}^{-1} = 0.03906 \text{ S m}^2 \text{ mol}^{-1}$.
   $\alpha = \frac{0.00162 \text{ S m}^2 \text{ mol}^{-1}}{0.03906 \text{ S m}^2 \text{ mol}^{-1}} = 0.0414746...$

4. Calculate the percentage degree of dissociation:
   Percentage degree of dissociation $= \alpha \times 100\%$
   Percentage degree of dissociation $= 0.0414746... \times 100\% = 4.14746... \%$

5. Round to the nearest integer:
   Rounding 4.14746...% to the nearest integer gives 4%.

The final answer is 4.

Explanation of the solution:

1. Convert concentration from mol/L to mol/m³: $0.01 \text{ M} = 10 \text{ mol m}^{-3}$.
2. Calculate molar conductance at given concentration: $\Lambda_m = \kappa / C = 0.0162 \text{ S m}^{-1} / 10 \text{ mol m}^{-3} = 0.00162 \text{ S m}^2 \text{ mol}^{-1}$.
3. Calculate degree of dissociation: $\alpha = \Lambda_m / \Lambda_m^\circ = 0.00162 \text{ S m}^2 \text{ mol}^{-1} / (390.6 \times 10^{-4} \text{ S m}^2 \text{ mol}^{-1}) = 0.04147$.
4. Convert to percentage: Percentage $\alpha = 0.04147 \times 100\% = 4.147\%$.
5. Round to the nearest integer: $4\%$.