The value of Ka in water at 25 °C for propionic acid (C2H5CO2H) is 3.4 x...

The value of Ka in water at 25 °C for propionic acid (C2H5CO2H) is 3.4 x 10-5 M. What is the pH of a 0.020 M aqueous solution of propionic acid?

Expert Solution

proponoic acid is a week acid

construct the ICE table

C2H5CO2H + H2O <------> C2H5CO2^- + H3O+

I 0.02 0 0

C -x +x +x

E 0.02-x +x +x

now dissociation constant

Ka = [C2H5CO2-][H3O+] / [C2H5CO2H]

3.4 x 10^-5 = [x][x] / [0.02-x]

x² + x 3.4 * 10-5 - 6.8 x 10^-7 = 0

solve the quadratic equation to get x value

x = 0.0008078 M = [H3O+]

now

pH = -log[H3O+] = -log[0.0008078] = 3.09

queen_honey_blossom answered 1 year ago

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