Question

A student is given a weak monoprotic acid "HA" that has a K_a = 0.00009897. What is the % ionization of a 0.11 molar aqueous solution of HA? (Enter your number with two significant figures and DO NOT include the percent sign in your answer)

Answer 1

Answer 2

To find the percent ionization of the weak monoprotic acid "HA," we can use the formula for percent ionization:

% Ionization = (ionized concentration / initial concentration) * 100

Given information: Ka (acid dissociation constant) = 0.00009897 Initial concentration of HA (c) = 0.11 M

The ionized concentration of HA is equal to the concentration of H+ ions formed when the acid dissociates. For a monoprotic acid, the concentration of H+ ions formed is the same as the concentration of ionized HA (because 1 mole of acid dissociates to form 1 mole of H+ ions).

Let x be the concentration of H+ ions formed (which is also the concentration of ionized HA).

Now, we can set up the expression for the equilibrium constant (Ka) and solve for x:

Ka = [H+][A-] / [HA] 0.00009897 = x * x / (0.11 - x)

Since x is much smaller than 0.11 (as the acid is weak), we can approximate 0.11 - x as approximately 0.11:

0.00009897 = x^2 / 0.11

Now, solve for x:

x^2 = 0.00009897 * 0.11 x^2 = 0.0000108867 x ≈ √0.0000108867 x ≈ 0.003298

Now, calculate the percent ionization:

% Ionization = (0.003298 M / 0.11 M) * 100 ≈ 2.99

The percent ionization of the 0.11 M aqueous solution of HA is approximately 2.99% (rounded to two significant figures).