#### Answer

$[H^+] = 4.05 \times 10^{-14}M$
$pH = 13.39$

#### Work Step by Step

1. Find the molar mass of $Ba(OH)_2 {\circ} 8 H_2O$
Ba: 137.3 * 1 = 137.3
O: 16.0 * 10 = 160.0
H: 1.0 * 18 = 18.0
137.3 + 160 + 18 = 315.3
2. Find the number of moles.
$mm(g/mol) = \frac{mass(g)}{n(moles)}$
$ 315.3= \frac{ 3.9}{n(moles)}$
$n(moles) = \frac{ 3.9}{ 315.3}$
$n(moles) = 0.01236$
3. Find the concentration.
$Concentration(M) = \frac{n(moles)}{Volume(L)}$
$Concentration(M) = \frac{ 0.01236}{ 0.1}$
$Concentration(M) = 0.1236M$
4. Calculate $[OH^-]$
** This base has 2 "OH"s.
$[OH^-] = 2* [Base] $
$[OH^-] = 0.247$
5. Calculate $[H^+]$
$[H^+] * [OH^-] = 10^{-14}$
$[H^+] = \frac{10^{-14}}{[OH^-]}$
$[H^+] = \frac{10^{-14}}{ 0.247}$
$[H^+] = 4.05 \times 10^{-14}$
6. Find the pH.
$pH = -log[H^+] = -log( 4.05 \times 10^{-14})$
$pH = 13.39$