Question

Question #b7542

Answer 1

`"pH" = 11.94`

Explanation:

We're asked to find the `"pH"` of a `0.064` `M` `"Ba(OH)"_2` solution.

Since one mole of barium hydroxide contains two moles of `"OH"^-` ions, the molar concentration of `"OH"^-` is

`2xx0.064color(white)(l)"mol/L" = color(red)(ul(0.128color(white)(l)"mol/L"`

Now that we know the `["OH"^-]` of the solution, we can calculate the `"pOH"` of the solution by the equation

`ul("pOH" = -log["OH"^-]`

So

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

At `25` `""^"o""C"`, the `"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.