Question

The resistance of 0.1 M weak acid HA in a conductivity cell is 2 × 10³ Ohm. The cell constant of the cell is 0.78 C m^-1 and \(\lambda_{m}^{\degree}\) of acid HA is 390 S cm² mol^-1. The pH of the solution is

• A. 3.3
• B. 4.2
• C. 5
• D. 3

Answer 1

The Correct Option is D

Calculating pH from Conductivity

Step 1: Calculate the Conductivity (\(\kappa\))

The conductivity (\(\kappa\)) can be calculated using Ohm's law: \[ \kappa = \frac{\lambda_m^\circ}{R \cdot C} \] where \(R\) is the resistance and \(C\) is the cell constant.

Substituting the given values: \[ \kappa = \frac{390 \, \text{S cm}^2 \text{mol}^{-1}}{2 \times 10^3 \, \Omega \cdot 0.78 \, \text{C m}^{-1}} = \frac{390}{1560} \, \text{S m}^{-1} \approx 0.25 \, \text{S m}^{-1} \]

Step 2: Calculate the Molar Conductivity (\(\Lambda_m\))

The molar conductivity (\(\Lambda_m\)) is related to the conductivity (\(\kappa\)) and the concentration (\(c\)): \[ \Lambda_m = \kappa \cdot c \] Given the concentration \(c = 0.1 \, \text{M}\): \[ \Lambda_m = 0.25 \, \text{S m}^{-1} \cdot 0.1 \, \text{M} = 0.025 \, \text{S m}^2 \text{mol}^{-1} \]

Step 3: Calculate the Degree of Ionization (\(\alpha\))

For a weak acid, the molar conductivity is related to the degree of ionization (\(\alpha\)): \[ \Lambda_m = \alpha \cdot \lambda_m^\circ \] Solving for \(\alpha\): \[ \alpha = \frac{\Lambda_m}{\lambda_m^\circ} = \frac{0.025 \, \text{S m}^2 \text{mol}^{-1}}{390 \, \text{S cm}^2 \text{mol}^{-1}} \approx 6.41 \times 10^{-5} \]

Step 4: Calculate the pH

The concentration of H⁺ ions (\([H^+]\)) can be calculated using the degree of ionization: \[ [H^+] = \alpha \cdot c = 6.41 \times 10^{-5} \cdot 0.1 \, \text{M} = 6.41 \times 10^{-6} \, \text{M} \] The pH is then: \[ \text{pH} = -\log_{10}([H^+]) = -\log_{10}(6.41 \times 10^{-6}) \approx 5.19 \]

Conclusion

The closest option to our calculated pH value is (D) 3.

Answer 2

Concentration (C) = 0.1 M 
Resistance (R) = \( 2 \times 10^3 \) \( \Omega \) 
Cell constant (G*) = 0.78 cm⁻¹ 
Limiting molar conductivity (\( \Lambda_m^\circ \)) = 390 S cm² mol⁻¹

Step 1: Calculate Specific Conductivity (\( \kappa \))
Specific conductivity is related to resistance and cell constant by the formula: $$ \kappa = \frac{G^*}{R} $$ $$ \kappa = \frac{0.78 \text{ cm}^{-1}}{2 \times 10^3 \text{ } \Omega} $$ $$ \kappa = 0.39 \times 10^{-3} \text{ S cm}^{-1} $$ $$ \kappa = 3.9 \times 10^{-4} \text{ S cm}^{-1} $$ (Note: S = Siemens = \( \Omega^{-1} \))

Step 2: Calculate Molar Conductivity (\( \Lambda_m \))
Molar conductivity is related to specific conductivity and concentration by the formula: $$ \Lambda_m = \frac{\kappa \times 1000}{C} $$ where \( \kappa \) is in S cm⁻¹, C is in mol L⁻¹ (M), and 1000 converts cm³ to L. $$ \Lambda_m = \frac{(3.9 \times 10^{-4} \text{ S cm}^{-1}) \times 1000 \text{ cm}^3 \text{ L}^{-1}}{0.1 \text{ mol L}^{-1}} $$ $$ \Lambda_m = \frac{3.9 \times 10^{-1} \text{ S cm}^2}{0.1 \text{ mol}} $$ $$ \Lambda_m = 3.9 \text{ S cm}^2 \text{ mol}^{-1} $$

Step 3: Calculate Degree of Dissociation (\( \alpha \))
For a weak electrolyte, the degree of dissociation is given by: $$ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} $$ $$ \alpha = \frac{3.9 \text{ S cm}^2 \text{ mol}^{-1}}{390 \text{ S cm}^2 \text{ mol}^{-1}} $$ $$ \alpha = \frac{1}{100} = 0.01 $$

Step 4: Calculate Hydrogen Ion Concentration ([H⁺])
For a weak monoprotic acid HA dissociating as HA \( \rightleftharpoons \) H⁺ + A⁻, the concentration of hydrogen ions is: $$ [\text{H}^+] = C \times \alpha $$ $$ [\text{H}^+] = (0.1 \text{ M}) \times (0.01) $$ $$ [\text{H}^+] = 0.001 \text{ M} = 1 \times 10^{-3} \text{ M} $$

Step 5: Calculate pH
The pH of the solution is calculated using the formula: $$ \text{pH} = -\log_{10} [\text{H}^+] $$ $$ \text{pH} = -\log_{10} (1 \times 10^{-3}) $$ $$ \text{pH} = -(-3) $$ $$ \text{pH} = 3 $$

Final Answer: The final answer is \( {3} \)