# What is pH of a solution prepared from a 313*mg mass of barium hydroxide dissolved in a 1*L volume of water...?

Jul 1, 2017

$p H = 9.56$.......

#### Explanation:

By definition, $p H + p O H = 14$, and thus.......

$p O H = 14 - p H$, and $p H = 14 - p O H$......

But $p O H = - {\log}_{10} \left[H {O}^{-}\right]$

$= - {\log}_{10} \left\{\frac{\frac{2 \times 3.13 \times {10}^{-} 3 \cdot g}{171.34 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L}\right\}$

$= - {\log}_{10} \left\{3.65 \times {10}^{-} 5\right\} = - \left(- 4.44\right) = 4.44$

And so $p H = 14 - p O H = 14 - 4.44 = 9.56$

Why did I double the $\left[B a {\left(O H\right)}_{2}\right]$ to get $\left[H {O}^{-}\right]$?