# What is the pH of a mass of 4.66*mg of barium hydroxide dissolved in a 1*L volume of water?

Oct 26, 2016

$p H = 9.74$
We need to find (i), $\left[H {O}^{-}\right]$ $=$ $\frac{2 \times 4.66 \times {10}^{-} 3 g}{171.34 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{L}$ $=$ $5.44 \times {10}^{-} 5 m o l \cdot {L}^{-} 1$.
And (ii), $p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$ $=$ $- {\log}_{10} \left(5.44 \times {10}^{-} 5\right)$ $=$ $4.26 .$
And since (iii) $p H + p O H = 14 , p H = 14 - 4.26 = 9.74$.