Answer
(a) pH = 3
(b) pH = 13.88
Work Step by Step
(a)
1. HCl is a strong acid, therefore: $[H_3O^+] = [HCl] = 0.0010M$
$pH = -log[H_3O^+]$
$pH = -log( 1 \times 10^{- 3})$
$pH = 3$
(b)
1.KOH is a strong base, therefore: $[OH^-] = [KOH] = 0.76M$
$pOH = -log[OH^-]$
$pOH = -log( 0.76)$
$pOH = 0.119$
$pH + pOH = 14$
$pH + 0.119 = 14$
$pH = 13.88$
