Question #b7542

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Answer 1 · 2017-08-12T02:51:08

#"pH" = 11.94#

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.