The pH and conductance of a weak acid (HX) was found to be 5 and 4 times 10^-5 S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1text cm^2 were at a distance of 15text cm apart. The value of the limiting molar conductivity is ..... textS m^2text mol^-1. (nearest integer) (Given: degree of dissociation of the weak acid (alpha) ll 1)

Numerical Answer Type:
Enter a numerical value Answer: 6 to 6 +4 marks

Solution & Explanation

### Related Formula kappa = G times fraclA Lambda_m = frackappa times 1000C alpha = fracLambda_mLambda_m^infty [H^+] = Calpha ### Core Logic Given: pH = 5, so [H^+] = 10^-5text M. Since [H^+] = C cdot alpha = C cdot fracLambda_mLambda_m^infty, we have 10^-5 = C cdot fracLambda_mLambda_m^infty. First, calculate conductivity (kappa): Conductance G = 4 times 10^-5text S. Cell constant G^* = fraclA = frac15text cm1text cm^2 = 15text cm^-1. kappa = G cdot G^* = (4 times 10^-5) times 15 = 6 times 10^-4text S cm^-1 Molar conductivity (Lambda_m): Lambda_m = frackappa times 1000C = frac6 times 10^-4 times 1000C = frac0.6C Substitute Lambda_m into the proton concentration formula: [H^+] = 10^-5 = C cdot frac0.6 / CLambda_m^infty 10^-5 = frac0.6Lambda_m^infty Lambda_m^infty = frac0.610^-5 = 60000text S cm^2text mol^-1 Convert units to textS m^2text mol^-1: Since 1text m^2 = 10^4text cm^2, Lambda_m^infty = 60000 times 10^-4text S m^2text mol^-1 = 6text S m^2text mol^-1 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium

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