Answer
a) $pH = 0.32$
b) $pH = 11.40$
c) $pH = 12.88$
Work Step by Step
a) 1. Since $HCl$ is a strong acid: $[H_3O^+] = [HCl] = 0.48$
2. Calculate the pH value:
$pH = -log[H_3O^+]$
$pH = -log( 0.48)$
$pH = 0.3188 = 0.32$
b) 1. Since $Ba(OH)_2$ is a strong base, and it has 2 $OH^-$ in each molecules: $[OH^-] = 2 * [Ba(OH)_2] = 2 * 2.5\times 10^{- 3} = 5\times 10^{- 3}M$
2. Calculate the pOH, and then the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 2.5 \times 10^{- 3})$
$pOH = 2.602$
$pH + pOH = 14$
$pH + 2.602 = 14$
$pH = 11.398 = 11.40$
c) 1. Since $NaOH$ is a strong base: $[OH^-] = [NaOH] = 0.075$
2. Calculate the pOH, and then the pH value:
$pOH = -log[OH^-]$
$pOH = -log( 0.075)$
$pOH = 1.125$
$pH + pOH = 14$
$pH + 1.125 = 14$
$pH = 12.875 = 12.88$
