Answer
$Ka = 2.6\times 10^{- 8}$
Work Step by Step
1. Calculate the hydronium concentration.
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 4.79}$
$[H_3O^+] = 1.622 \times 10^{- 5}$
2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is at the end of this answer.
-$[H_3O^+] = [Conj. Base] = x$
-$[Acid] = [Acid]_{initial} - x$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$
$Ka = \frac{x^2}{[InitialAcid] - x}$
$Ka = \frac{( 1.622\times 10^{- 5})^2}{ 0.01- 1.622\times 10^{- 5}}$
$Ka = \frac{ 2.63\times 10^{- 10}}{ 9.984\times 10^{- 3}}$
$Ka = 2.6\times 10^{- 8}$
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