Answer
$Ka = 1.332\times 10^{- 4}$
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.
-$[H_3O^+] = [Conj. Base] = x$
-$[Uric Acid] = [Uric Acid]_{initial} - x$
2. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.17}$
$[H_3O^+] = 6.761 \times 10^{- 3}$
$x = 6.761 \times 10^{- 3}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][Conj. Base]}{ [Uric Acid]}$
$Ka = \frac{x^2}{[Initial Acid] - x}$
$Ka = \frac{( 6.761\times 10^{- 3})^2}{ 0.35- 6.761\times 10^{- 3}}$
$Ka = \frac{ 4.571\times 10^{- 5}}{ 0.3432}$
$Ka = 1.332\times 10^{- 4}$
