Answer
$Ka = 1.352\times 10^{- 4}$
$pKa = 3.869$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Lactic acid] = [Lactic acid]_{initial} - x$
2. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.66}$
$[H_3O^+] = 2.188 \times 10^{- 3}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Lactic acid]}$
$Ka = \frac{x^2}{[InitialLactic acid] - x}$
$Ka = \frac{( 2.188\times 10^{- 3})^2}{ 0.0376- 2.188\times 10^{- 3}}$
$Ka = \frac{ 4.786\times 10^{- 6}}{ 0.03541}$
$Ka = 1.352\times 10^{- 4}$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.352 \times 10^{- 4})$
$pKa = 3.869$
