Question

1. A 0.735 M solution of a weak acid is 12.5% dissociated. (a)Calculate the [H3O+], pH, [OH -], and pOH of the solution. (b) calculate the ka of the acid

Answer 1

the reaction is

HA + H20 ---> H30+ + A-

given

12.5 % dissociated

so

[H30+] = 12.5 x 0.735 / 100

[H30+] =0.091875

now

pH = -log [H30+]

so

pH = -log [0.091875]

pH = 1.037

now

we know that

pOH = 14 - pH

so

pOH = 14 - 1.037

pOH = 12.963

now

-log [OH-] = 12.963

[OH-] = 1.088 x 10-13

2)

now

HA + H20 ---> H30+ + A-

Ka = [H30+] [A-] / [HA]

[HA] = 0.735 - 0.091875 = 0.643125

now

[H30+] = [A-] = 0.091875

so

Ka = [0.091875] [0.091875] / [0.643125]

Ka = 0.013125

so

the Ka of the acid is 0.013125