# 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. Aug 22, 2014 at 15:05
• Could you elaborate? Aug 22, 2014 at 15:07

Let's consider the following equilibrium:

$$\ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$ 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: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.

we rewrite this equation as follows:

$$\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}$$ 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.

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.