Answer
(a) pOH = 1.18, and pH = 12.82
(b) pOH = -0.734, and pH = 14.734
(c) pOH = -0.17 and pH = 14.17
Work Step by Step
(a)
KOH is a strong base, therefore: $[OH^-] = [KOH] = 0.066M$
$pOH = -log[OH^-]$
$pOH = -log( 6.6 \times 10^{- 2})$
$pOH = 1.18$
$pH + pOH = 14$
$pH + 1.18 = 14$
$pH = 12.82$
(b)
NaOH is a strong base, therefore: $[OH^-] = [NaOH] = 5.43M$
$pOH = -log[OH^-]$
$pOH = -log( 5.43)$
$pOH = -0.734$
$pH + pOH = 14$
$pH + (-0.734) = 14$
$pH = 14.734$
(c)
$[Ba(OH)_2]$ is a strong base with 2 $OH^-$ for each molecule, therefore:
$[OH^-] = [Ba(OH)_2]*2 = 0.74 * 2 = 1.48M$
$pOH = -log[OH^-]$
$pOH = -log( 1.48)$
$pOH = -0.17$
$pH + pOH = 14$
$pH + (-0.17) = 14$
$pH = 14.17$
