## Chemistry 9th Edition

The $K_a$ value for $HOBr$ is equal to $1.9 \times 10^{-9}$
1. Calculate the $[H_3O^+]$ of that solution. $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 4.95}$ $[H_3O^+] = 1.1 \times 10^{- 5}$ Therefore : $x = 1.1\times 10^{- 5}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $HOBr(aq) + H_2O(l) \lt -- \gt OBr^-(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $OBr^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [OBr^-] = 0 + x = x$ -$[HOBr] = [HOBr]_{initial} - x$ For approximation, we are going to consider $[HOBr]_{initial} = [HOBr]$ 3. Write the Ka equation, and find its value: $Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$ $Ka = \frac{x^2}{[InitialHOBr] - x}$ $Ka = \frac{( 1.1\times 10^{- 5})^2}{ 0.063- 1.1\times 10^{- 5}}$ $Ka = 1.9 \times 10^{- 9}$