## Chemistry: Atoms First (2nd Edition)

$Ka \approx 2\times 10^{- 9}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [CCl_3C{O_2}^-] = x$ -$[CCl_3CO_2H] = [CCl_3CO_2H]_{initial} - x$ 2. Using the pH, calculate the hydronium concentration: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.95}$ $[H_3O^+] = 1.122 \times 10^{- 5}$ So: $x = 1.122 \times 10^{-5}$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$ $Ka = \frac{x^2}{[InitialHOBr] - x}$ $Ka = \frac{( 1.122\times 10^{- 5})^2}{ 0.063- 1.122\times 10^{- 5}}$ $Ka = \frac{ 1.259\times 10^{- 10}}{ 0.06299}$ $Ka = 1.999\times 10^{- 9}$