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Answer 1 · 2017-08-12T02:51:08

$pH = 11.94$ $"pH" = 11.94$

We're asked to find the $pH$ $"pH"$ of a $0.064$ $0.064$ $M$ $M$ ${Ba(OH)}_{2}$ $"Ba(OH)"_2$ solution.

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