Answer
$K_a$(Butyric acid) $\approx 1.6\times 10^{- 5}$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.21}$
$[H_3O^+] = 6.166 \times 10^{- 4}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x = 6.166 \times 10^{-4}$
-$[Butyric Acid] = [Butyric Acid]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Butyric Acid]}$
$Ka = \frac{x^2}{[InitialButyric Acid] - x}$
$Ka = \frac{( 6.166\times 10^{- 4})^2}{ 0.025- 6.166\times 10^{- 4}}$
$Ka = \frac{ 3.802\times 10^{- 7}}{ 0.02438}$
$Ka = 1.559\times 10^{- 5}$
