Question

A 0.0202M solution of chloroacetic acid is 24.1% ionized at 25°C calculate the Ka for this acid

Answer 1

The dissociation equilibrium reaction for chloroacetic acid is:

ClCH2COOH <------> ClCH2COO⁻ + H⁺

The acid-dissociation equilibrium constant is:

Ka = [ClCH2COO⁻][H⁺]/[ClCH2COOH]

We can find [H⁺] using the concentration and the percent dissociation:
0.241 x 0.0202 = 4.86 x 10^-3 M

To find [ClCH2COO^-], we use the 1:1 stoichiometry from the above equation:

[ClCH2COO^-] = [H⁺] = 4.86 x 10^-3 M

For the [HAc], we can subtract 4.86 x 10^-3 M from the given concentration of 0.02002M:
[HAc] = 0.02002 - 4.86 x 10^-3= 0.01533M

Now we can calculate Ka by inserting the concentrations found,
Ka = [(4.86 x 10^-3)( 4.86 x 10^-3)] /0.01533
Ka = 1.54 x 10^-3