A 0.15 M solution of butanoic acid, C3H7COOH, has a pH of 2.8210. Find the Ka of butanoic acid.
Call buanoic acid HB.
Then the ionization is
HB ==> H^+ + B^-
To start (HB) = 0.15
(H^+) = 0 and (B^-) = 0
After ionization, (H^+) = x but that is found by converting pH to (H^+)
(B^-) = x = same as (H^+)
(HB) = 0.15-x = 0.15 - (H^+) from above.
Set up Ka expression, plug in x and 0.15-x in the appropriate places and solve for Ka. Post your work if you get stuck.
To find the Ka of butanoic acid, we can start by converting the pH to the concentration of H^+ ions. The pH is given as 2.8210.
The pH is defined as the negative logarithm of the concentration of H^+ ions in a solution. Mathematically, it can be written as:
pH = -log[H+]
We can rearrange this equation to solve for [H+]:
[H+] = 10^(-pH)
Plugging in the given pH value of 2.8210, we find:
[H+] = 10^(-2.8210) = 0.005699 M
Since the concentration of H^+ ions is equal to the concentration of B^- ions (both are x), we can write:
[H+] = [B^-] = x
The initial concentration of butanoic acid (HB) is given as 0.15 M.
Therefore, the concentration of HB after ionization is:
[HB] = 0.15 - x
The Ka expression for the ionization of butanoic acid can be written as:
Ka = [H+][B^-] / [HB]
Substituting the values we found earlier, we have:
Ka = (x)(x) / (0.15 - x)
Simplifying further:
Ka = x^2 / (0.15 - x)
We can now solve for Ka using this expression.