Answer
$Ka = 1.434\times 10^{- 4}$
$pKa = 3.843$
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.
-$[H_3O^+] = [C_3H_5O_3^-] = x$
-$[C_3H_6O_3] = [C_3H_6O_3]_{initial} - x$
2. Calculate the $[H_3O^+]$
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.43}$
$[H_3O^+] = 3.715 \times 10^{- 3}$
Therefore : $x = 3.715\times 10^{- 3}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][C_3H_5O_3^-]}{ [C_3H_6O_3]}$
$Ka = \frac{x^2}{[InitialC_3H_6O_3] - x}$
$Ka = \frac{( 3.715\times 10^{- 3})^2}{ 0.1- 3.715\times 10^{- 3}}$
$Ka = \frac{ 1.38\times 10^{- 5}}{ 0.09628}$
$Ka = 1.434\times 10^{- 4}$
4. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.434 \times 10^{- 4})$
$pKa = 3.843$
