# What is the difference between pH and pKa?

Is there any difference between the two, when talking of an acid? I am a bit confused. Please help me out.

• – jerepierre Jun 19 '16 at 22:24

## 3 Answers

$\mathrm{p}K_\mathrm{a}$ is the intrinsic property of an acid. It does not depend on how much acid there is in a solution. For example, acetic acid has a $\mathrm{p}K_\mathrm{a}$ of $4.76$. This $\mathrm{p}x$ means $-\log x,$ we can calculate $K_\mathrm{a}$ to be $10^{-4.76}$ is $1.7\times 10^{-5}$, which means that at equilibrium, $\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = 1.7\times 10^{-5}$. On the other hand, $\mathrm{pH}$ is related to the actual amount of $\ce{H+}$ ions, so if you had more acetic acid, the solution would be more acidic ($\mathrm{pH}$ is lower). This is why on Wikipedia, you see $\mathrm{p}K_\mathrm{a}$ given for acids, rather than a $\mathrm{pH}$, because $\mathrm{pH}$ is amount dependent. Similarly, in experiments, we are more interested in $\mathrm{pH}$ because that tells us the actual acidity of a solution.

Also know that though $K_\mathrm{a}$ may always be constant, percent ionization isn't; as there is more weak acid, they dissociate less.

‡ Just like the ideal gas law, aqueous equilibrium constants based on concentrations are for ideal solutions, but they is very accurate for most work.

pKa is the pH for which the protonated ($\ce{HA}$) and unprotonated species ($\ce{A^{-}}$) of the acid are equal. So $$\ce{10^{pKa} = \dfrac{[H+][A^{-}]}{[HA]} = \dfrac{[H+]^{2}}{[HA]}}$$

The Ka, like all other equilibrium constants, gives the concentrations of the substances at equilibrium. From that, you can find the concentration of H+ at equilibrium, which in turn can be used to find the pH.

Note that Ka can only be used for equilibrium. pH can be used any time you have a concentration for H+ and just need to convert it to a scale so you can measure it/compare it to others.