Question

The Ka of a monoprotic weak acid is 3.16 × 10-3. What is the percent ionization of a 0.188 M solution of this acid?

Answer 1

HA <-----> H^+ + A^-

0.188        0         0   initial

-x             +x       +x change

0.188-x      x        x     equilibrium

Ka = x^2/(0.188-x) let assume 0.188-x = 0.188

3.16*10^-3 = x^2/0.188

x = 0.0244

percent ionization = 0.0244/0.188*100 =12.98%