Answer
$Ka = 6.847\times 10^{- 6}$
Work Step by Step
1. Caculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.95}$
$[H_3O^+] = 1.122 \times 10^{- 3}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[InitialAcid] - x}$
$Ka = \frac{( 1.122\times 10^{- 3})^2}{ 0.185- 1.122\times 10^{- 3}}$
$Ka = \frac{ 1.259\times 10^{- 6}}{ 0.1839}$
$Ka = 6.847\times 10^{- 6}$
