Question #702e2

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Answer 1 · 2017-08-03T14:49:07

#["OH"^- ] = 5.6xx10^-12# #M#

We're asked to find the hydroxide ion concentration #["OH"^- ]# of a solution, given its #"pOH"#.

To do this, we can use the equation

#"pOH" = -log["OH"^- ]#

But rearrange it to solve for #["OH"^- ]#:

#ul(["OH"^- ] = 10^(-"pOH")#

We're given that #"pOH"# #= 11.25#, so we have

#["OH"^- ] = 10^(-11.25) = color(red)(ulbar(|stackrel(" ")(" "5.6xx10^-12color(white)(l)M" ")|)#

Since the #"pOH"# has two decimal places, the hydroxide ion concentration has two significant figures.