Answer
(a) pH = 10.75
(b) pH = 3.28
Work Step by Step
(a)
1. Find the concentration of $[OH^-]$
$Ba(OH)_2$ is a strong base, and has 2 $OH$, therefore, the concentration of $OH^-$ will be the double of the base concentration.
$[OH^-] = [Ba(OH)_2] * 2 = 2.8 \times 10^{-4} * 2 = 5.6 \times 10^{-4}M$
2. Calculate the pOH and then the pH.
$pOH = -log[OH^-] = -log(5.6 \times 10^{-4}) = 3.25$
$pOH + pH = 14$
$3.25 + pH = 14$
$pH = 14 - 3.25 = 10.75$
(b)
1. Find the concentration of $[H^+]$:
$HNO_3$ is a strong acid.
$[HNO_3] = [H^+] = 5.2 \times 10^{-4}M$
2. Calculate the pH.
$pH = -log[H^+] = -log(5.2 \times 10^{-4}) = 3.28$
