## Chemistry 10th Edition

$[OBr^-] \approx [H_3O^+ \approx 3.354 \times 10^{- 5} M$ $[HOBr] \approx 0.45M$
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [OBr^-] = x$ -$[HOBr] = [HOBr]_{initial} - x = 0.45 - x$ For approximation, we consider: $[HOBr] = 0.45M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][OBr^-]}{ [HOBr]}$ $Ka = 2.5 \times 10^{- 9}= \frac{x * x}{ 0.45}$ $Ka = 2.5 \times 10^{- 9}= \frac{x^2}{ 0.45}$ $1.125 \times 10^{- 9} = x^2$ $x = 3.354 \times 10^{- 5}$ Percent ionization: $\frac{ 3.354 \times 10^{- 5}}{ 0.45} \times 100\% = 0.007454\%$ %ionization < 5% : Right approximation. Therefore: $[OBr^-] = [H_3O^+] = x = 3.354 \times 10^{- 5}$ $[HOBr] \approx 0.45M$