What #["OH"^(-)]# and #"pOH"# give a basic solution?

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Answer 1 · 2017-06-21T13:41:30

A low #["OH"^"-"]# and a high #"pOH"# best state the basicity of a solution.

Low #["OH"^"-"]#

Let's say that #["OH"^"-"] = 10^"-10"color(white)(l) "mol/L"# (low #["OH"^"-"]#.

Then

#"pOH" = -log["OH"^"-"] = -log10^"-10" = 10#

Thus, a low #["OH"^"-"]# means a high #"pOH"#.

High #["OH"^"-"]#

Let's say that #["OH"^"-"] = 10color(white)(l) "mol/L" # (high #["OH"^"-"])#.

Then

#"pOH" = -log["OH"^"-"] = -log10^"10" = 1#

Thus, a high #["OH"^"-"]# means a low #"pOH"#.