Question

The Ka of a monoprotic weak acid is 2.35 × 10-3. What is the percent ionization of a 0.118 M solution of this acid?

Answer 1

HA ⇋ H(+) + A(-)
C                   0                0               at t=0

. C(1-x)         Cx               Cx           at eq

Now, [HA] initially is C(=0.118 M). If 'x' is the degree of dissociation of acid, at equilibrium, [HA] remaining will be C-Cx, i.e. C(1-x), and [H(+)] & [A(-)] formed would be Cx.

Now, Ka = [A(-)][H(+)]/[HA]
=> Ka = Cx²/(1-x)
Now,
Given that Ka = 2.5×10^-3
and for since solution is dilute, we can approximate (1-x) as 1.
Thus, inducing the above said results and value of C as 0.118M, we get

2.5×10^-3 = 0.118x²
=> x ≈ 0.1455

Thus, percent ionisation = 14.55% ≈ 15%