Answer
$Ka = 1.4\times 10^{- 5}$
Work Step by Step
1. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.78}$
$[H_3O^+] = 1.66 \times 10^{- 3}M$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [CH_3(CH_2)_4COO^-] = x$
-$[CH_3(CH_2)_4COOH] = [CH_3(CH_2)_4COOH]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][CH_3(CH_2)_4COO^-]}{ [CH_3(CH_2)_4COOH]}$
$Ka = \frac{x^2}{[InitialCH_3(CH_2)_4COOH] - x}$
$Ka = \frac{( 1.66\times 10^{- 3})^2}{ 0.2- 1.66\times 10^{- 3}}$
$Ka = \frac{ 2.754\times 10^{- 6}}{ 0.1983}$
$Ka = 1.4\times 10^{- 5}$
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