# What is the pH of a 5.00 x 10^-2 M aqueous solution of Ba(OH)_2?

$p H = 13$
We know that in water under standard conditions, $p H + p O H = 14$.
Now for $5.00 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$ $B a {\left(O H\right)}_{2}$, $\left[H {O}^{-}\right] = 0.100 \cdot m o l \cdot {L}^{-} 1$; and thus $p O H = - {\log}_{10} \left[H {O}^{-}\right]$ $= - {\log}_{10} \left({10}^{-} 1\right) = 1$. And thus $p H = 13$.