Answer
$pH \approx 11.231$
Work Step by Step
1. Write the Ka equation:
$*[OH^-] = [HBrO] = x$
$4 \times 10^{-6} = \frac{x^2}{0.724 - x}$
* Because x is very small, we can disconsider the value of x in the subtraction.
$4 \times 10^{-6} = \frac{x^2}{0.724 }$
$2.896 \times 10^{-6} = x ^2$
$ x = 1.702 \times 10^{-3}$
2. Now, with $[OH^-]$, find the pOH, and them, the pH.
$[OH^-] = 1.702 \times 10^{-3}$
$pOH = -log[OH^-] $
$pOH = -log(1.702 \times 10^{-3})$
$pOH = 2.769$
$pH = 14 - 2.769$
$pH = 11.231$
