In a freezing point depression experiment, it is found that acetic acid is 4.2 %

Question

ionized in 0.010 M solution. Calculate its ionization constant from this
information

Answer 1

To calculate the ionization constant (Ka) of acetic acid, we need to use the equation for the degree of ionization (α):

α = (concentration of ionized species) / (initial concentration of acid)

In this case, we are given that the acetic acid is 4.2% ionized, which means that α = 0.042. We are also given that the initial concentration of acetic acid (CH3COOH) is 0.010 M.

Therefore, we can rewrite the equation as:

0.042 = (concentration of ionized species) / 0.010

Solving for the concentration of ionized species gives:

(concentration of ionized species) = 0.042 * 0.010 = 0.00042 M

Since acetic acid ionizes to form one hydronium ion (H3O+) and one acetate ion (CH3COO-), the concentration of ionized species is the sum of the concentrations of H3O+ and CH3COO-. Therefore, we can divide the concentration of ionized species by 2 to get the concentration of each ion:

(concentration of H3O+) = (concentration of CH3COO-) = 0.00042 M / 2 = 0.00021 M

Now, we can substitute the concentration of H3O+ and CH3COO- into the equation for the ionization constant (Ka):

Ka = (concentration of H3O+) * (concentration of CH3COO-) / (concentration of undissociated acid)

Plugging in the values:

Ka = (0.00021 M) * (0.00021 M) / 0.010 M = 0.00000441 M

Therefore, the ionization constant (Ka) of acetic acid is 0.00000441 M.