The Inverse Relation between [H₃O⁺] and [OH⁻]

Since hydronium and hydroxide are the two sole products of an equilibrium reaction, in order to maintain the equilibrium state of $[H^+][OH^-]=K_w$, if the concentration of one ion increases, the other must decrease. In other words, we know that if a solution has “extra” H3O+, it must have less OH than there would be in pure water.

A solution of an acid in water has a hydronium ion concentration of $2.0×10^{-6}\;M$. What is the concentration of hydroxide ion at 25 °C?

Use the value of the ion-product constant for water at 25 °C $$2H_2O\;(l)⇌H_3O^+\;(aq)+OH^-\;(aq)\qquad K_w=[H_3O^+][OH^-]=1.0×10^{-14}$$

to calculate the missing equilibrium concentration.

Rearrangement of the $K_w$ expression shows that [OH] is inversely proportional to [H3O+]:


Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Châtelier’s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.

Substituting the ion concentrations into the $K_w$ expression confirms this calculation, resulting in the expected value:


Check Your Learning

What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C?


[H3O+] = $1×10^{−11}\;M$