# Question b7542

Aug 12, 2017

$\text{pH} = 11.94$

#### Explanation:

We're asked to find the $\text{pH}$ of a $0.064$ $M$ ${\text{Ba(OH)}}_{2}$ solution.

Since one mole of barium hydroxide contains two moles of ${\text{OH}}^{-}$ ions, the molar concentration of ${\text{OH}}^{-}$ is

$2 \times 0.064 \textcolor{w h i t e}{l} \text{mol/L" = color(red)(ul(0.128color(white)(l)"mol/L}$

Now that we know the $\left[{\text{OH}}^{-}\right]$ of the solution, we can calculate the $\text{pOH}$ of the solution by the equation

$\underline{{\text{pOH" = -log["OH}}^{-}}$

So

"pOH" = -log(color(red)(0.128color(white)(l)M)) = color(green)(ul(2.056

At $25$ $\text{^"o""C}$, the $\text{pH}$ is given by

ul("pH" = 14 -"pOH"

So

color(blue)("pH") = 14- color(green)(2.056) = color(blue)(ulbar(|stackrel(" ")(" "11.94" ")|)#

rounded to two decimal places, because the problem gave us the concentration to two significant figures.