Answer
$[H_3O^+] = 0.072M$
$[OH^-] = 1.389 \times 10^{- 13}M$
$pH = 1.143$
Work Step by Step
1. To calculate the $[H_3O^+]$ of a mixture of 2 strong acids, all we need to do is to sum the concentrations:
$HBr$ is a strong acid, therefore: $[H_3O^+]$ from $ HBr = [HBr] = 0.052M$
$HNO_3$ is a strong acid, therefore: $[H_3O^+]$ from $HNO_3 = [HNO_3] = 0.020M$
$0.052M + 0.020M = 0.072M = Total [H_3O^+]$
2. Calculate the pH value
$pH = -log[H_3O^+]$
$pH = -log( 0.072)$
$pH = 1.143$
3. Use $K_w$ to calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 7.2 \times 10^{- 2} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 7.2 \times 10^{- 2}}$
$[OH^-] = 1.389 \times 10^{- 13}$
