Why Do Different Electrolytes Exhibit Varying Molar Conductivities?

[physicist888]

strong and weak electrolytes-need help pleasez

hi all
i need help to solve the following exercise

Given: those are a strong electrolyt limiting , and we give the molar conductivity for each ion in ( S.Cm2/mol)
HCl 426.1
CH3COONA 91.05
HF 405
NaCl 126.5
1- why did thos electrolyts has a different molar conductivity?
2- calculate the degree of dissociation alpha and the constant of dissociation K for a weak electrolyt CH3COOH dilute at 0.6. given the molar conductivity equal 195.3 Cm2/mol. (show the calculation in details)

 

[chemisttree]

The conductivities here are ionic in nature. That is, the conduction requires that the ion be mobile in the solvent. Ions with smaller solvent shells will be more mobile than ions with larger solvent shells. Ions that form better hydrogen bonds will conduct better than those that form weaker hydrogen bonds.

Show me the formula for ionic conductivity that has the K term for dissociation contained within and we can talk about the second question.

 

[Blueshift5]

I can provide some explanations and guidance for this exercise.

Firstly, electrolytes are substances that can conduct electricity when dissolved in water. They are divided into two categories: strong and weak electrolytes. Strong electrolytes are substances that completely dissociate into ions when dissolved in water, while weak electrolytes only partially dissociate.

Now, looking at the given molar conductivity values, we can see that HCl, HF, and NaCl have significantly higher values compared to CH3COONA. This is because HCl, HF, and NaCl are strong electrolytes, meaning they have a higher degree of dissociation and produce more ions in solution, resulting in a higher conductivity. On the other hand, CH3COONA is a weak electrolyte, so it has a lower degree of dissociation and thus a lower molar conductivity.

To answer the first question, the different molar conductivity values for these electrolytes can be attributed to the difference in their degree of dissociation. Strong electrolytes have a higher degree of dissociation, while weak electrolytes have a lower degree of dissociation, resulting in different molar conductivity values.

Moving on to the second question, to calculate the degree of dissociation alpha and the constant of dissociation K for a weak electrolyte like CH3COOH, we can use the following equations:

Molar conductivity (Λ) = k * (α * c) (1)

Where:
k = constant of dissociation
α = degree of dissociation
c = concentration in mol/L

We are given the molar conductivity (Λ) and concentration (c) in the exercise, so we can rearrange equation (1) to solve for α:

α = Λ / (k * c) (2)

To solve for k, we can use the formula:

K = (α^2 * C) / (1 - α) (3)

Where:
α = degree of dissociation
C = concentration in mol/L

Now, substituting the given values in equation (2), we get:

α = 195.3 / (k * 0.6) (4)

To solve for k, we need to know the value of α. So, we can use equation (3) and substitute the value of α from equation (4):

K = (0.6 * 0.6 * 0.6) / (1