Answer
$Ka = 4.02\times 10^{- 3}$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 1.91}$
$[H_3O^+] = 1.23 \times 10^{- 2}M$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3COCOO^-] = x$
-$[CH_3COCOOH] = [CH_3COCOOH]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][CH_3COCOO^-]}{ [CH_3COCOOH]}$
$Ka = \frac{x^2}{[InitialCH_3COCOOH] - x}$
$Ka = \frac{( 0.0123)^2}{ 0.05- 0.0123}$
$Ka = \frac{ 1.514\times 10^{- 4}}{ 0.0377}$
$Ka = 4.02\times 10^{- 3}$
