From what I understand, the concentration of hydronium and hydroxide molecules is constant in pure water (and equals $10^{-7}\ \text{M}$, which is measured experimentally).

What I don't understand is why this remains true for an aqueous solution with an acid or base in it. Shouldn't it affect these concentrations?

When you write the equilibrium reactions for a weak/weak acid/base pairs (one with its conjugate), and get that $K_\mathrm a\cdot K_\mathrm b = [\ce{H+}][\ce{OH- }]$, what's the explanation for it? Why does their multiplication (either as pure water or in a acidic/basic solution) remain constant?


1 Answer 1


The concentrations do change in the presence of acid or base and are no longer $10^{-7}$M.

Instead, [H3O+][OH-] $= 10^{-14}M^2$

This comes from:

$\ce{2H2O <=> H3O+ + OH-}$

$K = \frac{a(\ce{H3O+})a(\ce{OH-})}{(a(\ce{H2O}))^2}$

Where "a(x)" is activity of species "x".

Then, approximating activity of $\ce{H2O}$ as 1, and the activity of the ions as the concentration of the ions there is a constant $K_w$, such that:

[H3O+][OH-] $= 10^{-14}M^2 = K_w$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.