Conductance

To find the dissociation constant ( $K_a$ ) of acetic acid, we need to use the relationship between molar conductivity and dissociation. Here is the step-by-step solution to find the dissociation constant:

Determine the limiting molar conductivity ( $%5CLambda%5E%5Ccirc_m$ ) of acetic acid:

The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. For acetic acid, $CH_3COOH$ , it dissociates into $H%5E%2B$ and $CH_3COO%5E-$ .

%5CLambda%5E%5Ccirc_m%20%3D%20%5CLambda_%7BH%5E%2B%7D%5E%5Ccirc%20%2B%20%5CLambda_%7BCH_3COO%5E-%7D%5E%5Ccirc

Given, $%5CLambda_%7BH%5E%2B%7D%5E%5Ccirc%20%3D%20350%20%5C%2C%20S%20%5C%2C%20cm%5E2%20%5C%2C%20mol%5E%7B-1%7D$ and $%5CLambda_%7BCH_3COO%5E-%7D%5E%5Ccirc%20%3D%2050%20%5C%2C%20S%20%5C%2C%20cm%5E2%20%5C%2C%20mol%5E%7B-1%7D$ .

%5CLambda%5E%5Ccirc_m%20%3D%20350%20%2B%2050%20%3D%20400%20%5C%2C%20S%20%5C%2C%20cm%5E2%20%5C%2C%20mol%5E%7B-1%7D

Calculate the degree of dissociation ( $%5Calpha$ ) using the formula:

%5Calpha%20%3D%20%5Cfrac%7B%5CLambda_m%7D%7B%5CLambda%5E%5Ccirc_m%7D%20%3D%20%5Cfrac%7B20%7D%7B400%7D%20%3D%200.05

Use the degree of dissociation to find the dissociation constant $K_a$ :

K_a%20%3D%20c%20%5Calpha%5E2%20%3D%200.007%20%5Ctimes%20(0.05)%5E2

K_a%20%3D%200.007%20%5Ctimes%200.0025%20%3D%200.0000175%20%3D%201.75%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20mol%20%5C%2C%20L%5E%7B-1%7D

Therefore, the dissociation constant of acetic acid is $1.75%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20mol%20%5C%2C%20L%5E%7B-1%7D$ . The correct option is:

$1.75%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%20mol%20%5C%2C%20L%5E%7B-1%7D$

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

Conductance

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

Chapter Questions
1 MCQs