# Is it true that a pH higher than the pKa of an acid make the H automatically disassociate?

Will the hydrogen atom always disassociate if the pH of the solution is higher than the pKa of the acid?

• It's an equilibrium, so no. – canadianer Aug 22 '14 at 15:05
• Could you elaborate? – FinlandRules Aug 22 '14 at 15:07

The $\text{p}K_{\text{a}}$ of an acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ mainly exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists mainly as undissociated $\ce{HA}$. This doesn't mean that at $\text{pH}$s higher than the $\text{p}K_{\text{a}}$ suddenly every molecule of the acid dissociates, it means that a greater portion of the initial concentration of $\ce{HA}$ will be dissociated as you go higher in $\text{pH}$. There will always be some (potentially tiny) amount of the undissociated acid left even at very high $\text{pH}$ because it is an equilibrium. For more information on $\text{p}K_{\text{a}}$ see here.

• "This doesn't mean that at pHs lower (?) than the pKa suddenly every molecule of the acid dissociates..." – don't you mean "higher"? – FinlandRules Aug 22 '14 at 16:15
• @FinlandRules Oh yes, sorry for that. Confused the pH-directions there. – Philipp Aug 22 '14 at 18:25

$$\ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$ The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.
$$\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}$$ From this equation, we can deduce that when $\ce{pH}>\ce{pK}_A$, $[A^-]>[\ce{H}A]$, i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.