Question

What is the fraction of association (?) for the following potassium propionate solutions? Ignore activities. The Ka of propanoic acid is 1.34 × 10-5.
(a) 5.00 × 10-1 M K(C2H5CO2)
(b) 5.00 × 10-2 M K(C2H5CO2)
(c) 5.00 × 10-12 M K(C2H5CO2)

Answer 1

C₂H₅CO₂ + H₂O --------> HC₂H₅CO₂ + OH^- -----> Kb

With this overall reaction, tthe first thing is calculate Kb:

Kb = Kw/Ka = 1x10^-14 / 1.34x10^-5 = 7.46x10^-10

Now, let's calculate the concentration of the propanoic acid:

C₂H₅CO₂ + H₂O --------> HC₂H₅CO₂ + OH^-

i) 5x10^-1 -----------------------0 0

eq) 5x10^-1 - x -------------------x x

7.46x10^-10 = x² / 0.5 - x ----> however Kb is a low value, so we can assume that 0.5 - x = 0.5 so:

x = (7.46x10^-10 x 0.5)^1/2 = 1.93x10^-5 M

Doing the same thing with the other concentration:

x₂ = (7.46x10^-10 x 0.05)^1/2 = 6.11x10^-6 M

The α can be calculated like this:

α = [HC₂H₅CO₂] / [C₂H₅CO₂]

α1 = 1.93x10^-5 / 0.5 = 3.86x10^-5 = α1

α2 = 6.11x10^-6 / 0.05 = 1.222x10^-4 = α2

Answer 2

C₂H₅CO₂ + H₂O --------> HC₂H₅CO₂ + OH^- -----> Kb

With this overall reaction, tthe first thing is calculate Kb:

Kb = Kw/Ka = 1x10^-14 / 1.34x10^-5 = 7.46x10^-10

Now, let's calculate the concentration of the propanoic acid:

C₂H₅CO₂ + H₂O --------> HC₂H₅CO₂ + OH^-

i) 5x10^-1 -----------------------0 0

eq) 5x10^-1 - x -------------------x x

7.46x10^-10 = x² / 0.5 - x ----> however Kb is a low value, so we can assume that 0.5 - x = 0.5 so:

x = (7.46x10^-10 x 0.5)^1/2 = 1.93x10^-5 M

Doing the same thing with the other concentration:

x₂ = (7.46x10^-10 x 0.05)^1/2 = 6.11x10^-6 M

The α can be calculated like this:

α = [HC₂H₅CO₂] / [C₂H₅CO₂]

α1 = 1.93x10^-5 / 0.5 = 3.86x10^-5 = α1

α2 = 6.11x10^-6 / 0.05 = 1.222x10^-4 = α2