Answer
$Ka = 1.434\times 10^{- 5}$
$pKa = 4.843$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
** The image is at the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Caproic acid] = [Caproic acid]_{initial} - x$
2. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 3.43}$
$[H_3O^+] = 3.715 \times 10^{- 4}$
Therefore: $x = 3.715 \times 10^{- 4}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Caproic acid]}$
$Ka = \frac{x^2}{[InitialCaproic acid] - x}$
$Ka = \frac{( 3.715\times 10^{- 4})^2}{ 0.01- 3.715\times 10^{- 4}}$
$Ka = \frac{ 1.38\times 10^{- 7}}{ 9.628\times 10^{- 3}}$
$Ka = 1.434\times 10^{- 5}$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.434 \times 10^{- 5})$
$pKa = 4.843$
