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.