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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.

User Icon for Explain Bot Explain Bot answered
1 year ago

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.