Answer
(a) pH = 0.92
(b) pH = -0.38
(c) pH = 3.49
Work Step by Step
(a) Since $HCl$ is a strong acid: $[H_3O^+] = [HCl] = 0.12M$
-Find the pH.
$pH = -log[H^+]$
$pH = -log(0.12)$
$pH = 0.92$
(b) Since $HNO_3$ is a strong acid: $[H_3O^+] = [HNO_3] = 2.4M$
-Find the pH
$pH = -log(2.4)$
$pH = -0.38$
(c) Since $HClO_4$ is a strong acid: $[H_3O^+] = [HClO_4] = 3.2 \times 10^{-4}$
- Find the pH:
$pH = -log[H^+]$
$pH = -log( 3.2 \times 10^{- 4})$
$pH = 3.49$
