## Chemistry 10th Edition

$Ka = 1.332\times 10^{- 4}$
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: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Uric Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{( 6.761\times 10^{- 3})^2}{ 0.35- 6.761\times 10^{- 3}}$ $Ka = \frac{ 4.571\times 10^{- 5}}{ 0.3432}$ $Ka = 1.332\times 10^{- 4}$