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

Question

1991 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-21T13:41:30

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

Low $[{OH}^{-}]$ $["OH"^"-"]$

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

Then

$pOH = - log [{OH}^{-}] = - {log 10}^{-10} = 10$ $"pOH" = -log["OH"^"-"] = -log10^"-10" = 10$

Thus, a low $[{OH}^{-}]$ $["OH"^"-"]$ means a high $pOH$ $"pOH"$ .

High $[{OH}^{-}]$ $["OH"^"-"]$

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

Then

$pOH = - log [{OH}^{-}] = - {log 10}^{10} = 1$ $"pOH" = -log["OH"^"-"] = -log10^"10" = 1$

Thus, a high $[{OH}^{-}]$ $["OH"^"-"]$ means a low $pOH$ $"pOH"$ .

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