## Chemistry (4th Edition)

$Ka = 4.801 \times 10^{-9}$
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium: -$[H_3O^+] = [Conj. Base] = x$ -$[Acid] = [Acid]_{initial} - x$ 2. Convert the pH into hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.52}$ $[H_3O^+] = 3.01 \times 10^{- 5}M$ 3.. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][Conj. Base]}{ [Acid]}$ $Ka = \frac{x^2}{[Initial Acid] - x}$ $Ka = \frac{( 3.02\times 10^{- 5})^2}{ 0.19- 3.02\times 10^{- 5}}$ $Ka = \frac{ 9.12\times 10^{- 10}}{ 0.19}$ $Ka = 4.801\times 10^{- 9}$