Answer to Question #109373 in General Chemistry for mya 18

Question

The Ka of a monoprotic weak acid is 0.00719. What is the percent ionization of a 0.181 M
solution of this acid?

Answer 1

Solution:

HA - monoprotic weak acid.

Initial concentration of the acid = C_o(HA) = 0.181 M

The dissociation of a monoprotic weak acid:

HA + H₂O = H₃O⁺ + A⁻

For which K_a = 0.00719.

Now make an ICE (Initial, Change, Equlibrium) table, considering the ionization of a monoprotic weak acid in water. Because only solutes and gases are incorporated into the equilibrium expression, you can ignore the concentration of water (a pure liquid) in the calculations.

0.00719 = x² / (0.181 - x)

so

x² = (0.00719*0.181) - 0.00719*x;

x² + 0.00719x - 0.0013 = 0

then use the quadratic formula to calculate

x = 0.0326

% ionization = (Concentration of the ionized portion of the acid) / (Initial concentration of the acid).

Since x is equal to [% ionization] * C_o(HA), then:

% ionization = x / C_o(HA) = 0.0326 / 0.181 = 0.18, or 18%

% ionization = 18%.

Answer: The percent ionization of a solution is 18%.