The pH of a solution is 5. What is the concentration of OH- ions at this pH?

Answer 1

A simple way to approach this problem is to start with the assumption that pH + pOH =14

If the pH= 5 then the pOH =8

If we use the derivation of the pOH formula (pOH=-log [OH-] then [OH] =10^(-8) where -8 is -pOH

So your answer is 10^-8 or 1.0 X 10^8

Answer 2

Hi Jane J.,

This problem is getting you to recognize the inverse relationship between [H+] ion concentration and [OH-] ion concentration. This relationship, in pure water at ~25C (i.e. room temperature) is [H+] * [OH-] = 10^-14.

So for [H+] vs. pH, the relationship is [H+] = 10^(-pH) M and for [OH-], [OH-] = 10^(-14+pH)

You should be able to solve it from there?

By the way, because the self-dissociation of water is an equilibrium process, with an positive enthalpy change associated (I'm not sure if the entropy change is significant), that product does vary with temperature. (See reference table in https://www.chemguide.co.uk/physical/acidbaseeqia/kw.html , about 6 pp down.), so the value of [H+] in pure water does vary with temperature, somewhat. However, that change doesn't have much importance per se even when you're doing a reaction that requires [H+] as a reactant, because temperature has a direct effect on the speed of reactions -- it promotes all reactants into more energetic molecules which can access the "transition state" of the reaction more easily.

You will study all this presently, if you haven't already --

-- Cheers, -- Mr. d.