Question

What is the
pH of a solution of
0.28 M acid and
0.84 M of its conjugate base. If the ionization
constant of acid is
4 x 10-4​

Answer 1

Answer:

Let’s assume that the acid you are referring to is HA and therefore the conjugate base is A-. Then, the equilibrium we are considering is:

HA = H+ + A-

and therefore the ionization expression is:

Ka = [H+]{A-]/[HA]

Rearrange this and solving for [H+] gives

[H+] = Ka*[HA]/[A-]

Putting in the numbers given:

[H+] = 4*10^-4 * (0.28)/(0.84)

[H+] = 1.32 * 10^-4

Then, remembering that

pH = -log[H+] then

pH = -log (1.32 * 10^-4) = 3.88