Answer
(a) pOH = -0.0934, pH = 14.0934
(b) pOH = 0.356, pH = 13.644
(c) pOH = 1.07, pH = 12.93
Work Step by Step
(a)
LiOH is a strong base, therefore: $[OH^-] = [LiOH] = 1.24M$
$pOH = -log[OH^-]$
$pOH = -log( 1.24)$
$pOH = -0.0934$
$pH + pOH = 14$
$pH + (-0.0934) = 14$
$pH = 14.0934$
(b)
$Ba(OH)_2$ is a strong base with 2 $OH^-$, therefore:
$[OH^-] = 2 * [Ba(OH)_2] = 2 * 0.22 = 0.44M$
$pOH = -log[OH^-]$
$pOH = -log(0.44)$
$pOH = 0.356$
$pH + pOH = 14$
$pH + 0.356 = 14$
$pH = 13.644$
(c)
NaOH is a strong base, therefore: $[OH^-] = [NaOH] = 0.085$
$pOH = -log[OH^-]$
$pOH = -log( 8.5 \times 10^{- 2})$
$pOH = 1.07$
$pH + pOH = 14$
$pH + 1.07 = 14$
$pH = 12.93$
