the equivalent conductance of 0.0140 n solution of fluoroacetic acid is 109.0 at 25°c, if coundunce at infinite dilution ^° = 389.5.what is ionisation constant of fluoroacetic acid
To find the ionization constant (Ka) of fluoroacetic acid, we need to use the concept of equivalent conductance and the concept of degree of ionization. The equivalent conductance is given by the formula:
Λeq = (λ0 + Δλ) × 1000 / c
Where:
Λeq = Equivalent conductance in Siemens per meter (S m^2 mol^-1)
λ0 = Limiting molar conductance at infinite dilution in Siemens per meter (S m^2 mol^-1)
Δλ = Conductance due to the presence of ions in the solution that are not fully ionized in Siemens per meter (S m^2 mol^-1)
c = Concentration of the solution in moles per liter (mol L^-1)
In this case, the equivalent conductance (Λeq) of the fluoroacetic acid solution is given as 109.0 S m^2 mol^-1, the concentration (c) is given as 0.0140 n (where n represents normality), and the limiting molar conductance at infinite dilution (λ0) is given as 389.5 S m^2 mol^-1.
We need to rearrange the formula to solve for Δλ:
Δλ = (Λeq × c) / 1000 - λ0
Substituting the given values:
Δλ = (109.0 × 0.0140) / 1000 - 389.5
Δλ = 1.526 - 389.5
Δλ = -388.0 (Note: The negative sign indicates that there is a decrease in conductance due to incomplete ionization)
Now that we have Δλ, we can calculate the degree of ionization (α):
α = Δλ / λ0
Substituting the values:
α = -388.0 / 389.5
α ≈ -0.995 (Note: The negative sign indicates a decrease in conductance due to incomplete ionization)
The degree of ionization cannot be negative, so we take the absolute value:
α ≈ 0.995
The ionization constant (Ka) is related to the degree of ionization (α) using the formula:
Ka = α^2 × c
Substituting the values:
Ka = 0.995^2 × 0.0140
Ka ≈ 0.988 × 0.0140
Ka ≈ 0.0138
Therefore, the ionization constant (Ka) of fluoroacetic acid is approximately 0.0138.