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

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$\mathrm{p}K_\mathrm{a}$ 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.

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

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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.

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