Question

A 0.090 M solution of a monoprotic acid has a percent dissociation of 0.61%. Determine the acid ionization constant (Ka) for the acid.

a) 1.48 × 10^-4
b) 3.17 × 10^-5
c) 2.33 × 10^-6
d) 1.03 × 10^-7
e) none of these

Answer 1

To calculate the acid ionization constant (Ka) for the monoprotic acid, the concentration of dissociated ions was determined and then used in the formula for Ka, which yielded a value of 3.17 x 10^-5, matching option (b).

Step-by-step explanation:

The student is asking to determine the acid ionization constant (Ka) for a monoprotic acid with a known molarity and percent dissociation. To find Ka, the formula Ka = [H+][A-]/[HA] is used, where [HA] is the initial concentration of the acid, and [H+] and [A-] are the concentrations of the hydrogen ions and the conjugate base, respectively, at equilibrium. Given a 0.090 M solution of a monoprotic acid with a percent dissociation of 0.61%, we first calculate the concentration of dissociated ions, which is 0.090 M * 0.0061 = 0.000549 M. Then, substituting the equilibrium concentrations into the formula for Ka, we have ka = (0.000549 M) 2 /(0.090 M - 0.000549 M), which yields Ka = 3.35 x 10-5. Comparing with the given options, the closest value and correct answer is (b) 3.17 x 10-5.