The molar conductances of NaCl, HCl and CH₃COOHa at infinite dilution are 126.45, 426.16 and 91 Ω⁻¹-cm²-mol⁻¹ respectively. The molar conductance of CH₃COOH at infinite dilution i s

To determine the molar conductance of acetic acid (CH₃COOH) at infinite dilution, we can use the information provided about the molar conductances of its ions and the relationship between the substances involved. The key here is to recognize that acetic acid is a weak acid and partially dissociates in solution.

Understanding Molar Conductance

Molar conductance is a measure of how well a solution can conduct electricity, and it is influenced by the concentration of ions in the solution. At infinite dilution, each electrolyte is fully dissociated into its ions. For the substances you mentioned:

• NaCl: Dissociates into Na⁺ and Cl⁻.
• HCl: Dissociates into H⁺ and Cl⁻.
• CH₃COOH: Partially dissociates into CH₃COO⁻ and H⁺.

Using Known Molar Conductances

The molar conductance at infinite dilution for a weak acid like acetic acid can be understood by considering its dissociation. The molar conductance for the ions produced during the dissociation of CH₃COOH can be expressed as:

Λ(CH₃COOH) = Λ(CH₃COO⁻) + Λ(H⁺)

From the given data, we know:

• Λ(NaCl) = 126.45 Ω⁻¹-cm²-mol⁻¹
• Λ(HCl) = 426.16 Ω⁻¹-cm²-mol⁻¹
• Λ(CH₃COO⁻) = 91 Ω⁻¹-cm²-mol⁻¹ (This is typically the value for acetate ions, derived from other sources.)

Finding Λ(H⁺)

We can find the molar conductance of H⁺ from HCl, where we know it dissociates completely. The molar conductance of HCl can be expressed as:

Λ(HCl) = Λ(H⁺) + Λ(Cl⁻)

We know that HCl dissociates completely, so we can rearrange this to find Λ(H⁺):

Λ(H⁺) = Λ(HCl) - Λ(Cl⁻)

From NaCl, we can deduce that:

Λ(Cl⁻) = 126.45 Ω⁻¹-cm²-mol⁻¹

Substituting values:

Λ(H⁺) = 426.16 Ω⁻¹-cm²-mol⁻¹ - 126.45 Ω⁻¹-cm²-mol⁻¹ = 299.71 Ω⁻¹-cm²-mol⁻¹

Calculating Molar Conductance of CH₃COOH

Now, we can substitute the values back into our original equation for the molar conductance of acetic acid:

Λ(CH₃COOH) = Λ(CH₃COO⁻) + Λ(H⁺)

Λ(CH₃COOH) = 91 Ω⁻¹-cm²-mol⁻¹ + 299.71 Ω⁻¹-cm²-mol⁻¹ = 390.71 Ω⁻¹-cm²-mol⁻¹

Thus, the molar conductance of acetic acid (CH₃COOH) at infinite dilution is approximately 390.71 Ω⁻¹-cm²-mol⁻¹.

Summary

The molar conductance of acetic acid at infinite dilution is derived from the molar conductances of its dissociated ions. By breaking down the components and using the known conductances of other substances, we can accurately calculate the conductance for CH₃COOH. This approach not only highlights the connections between different acids and their respective ions but also enhances our understanding of electrolyte behavior in solution.