Answer
(a) pH = 2.34
(b) pH = 3.21
(c) pH = 11.83
(d) pH = 11.98
Work Step by Step
(a)
Strong acid: $[HCl] = [H^+]$
$pH = -log[H^+]$
$pH = -log( 4.5 \times 10^{- 3})$
$pH = 2.34$
(b)
Strong acid: $[HNO_3] = [H^+]$
$pH = -log[H^+]$
$pH = -log( 6.14 \times 10^{- 4})$
$pH = 3.21$
(c)
Strong base: $[NaOH] = [OH^-]$
$pOH = -log[OH^-]$
$pOH = -log( 6.83 \times 10^{- 3})$
$pOH = 2.16$
$pH + pOH = 14$
$pH + 2.16 = 14$
$pH = 11.83$
(d)
Strong base with 2 $OH^-$: $[OH^-] = [Ba(OH)_2] * 2$
$pOH = -log[OH^-]$
$pOH = -log( 9.6 \times 10^{- 3})$
$pOH = 2.01$
$pH + pOH = 14$
$pH + 2.01 = 14$
$pH = 11.98$
