Chemistry - Defining general acidic strength

Solution 1:

According to the Brønsted-Lowry acid base theory, an acid is a compound that releases a $\ce{H+}$ ion to give a conjugate base. For most acids, this reaction exists in a dynamic equilibrium. The equilibrium constant of this reaction is what defines $K_\mathrm{a}$. Generally more stable the product, the more of the forward reaction takes place.

Using this last statement, when comparing two compounds, we can say that if the conjugate base of one compound is more stable, it would mean that the value of its $K_\mathrm{a}$ would be greater. This implies its $\mathrm{p}K_\mathrm{a}$ would be less, meaning that the compound would be more acidic than the other.

For example, taking the case of two compounds, $\ce{HCOOH}$ and $\ce{C6H5-OH}$: Here formate ion would have resonance via oxygen whereas phenoxide ion would have resonance via carbon. This makes phenol less acidic than formic acid.

$\mathrm{p}K_\mathrm{a} \ {\text{ of phenol} = 10.0}$

$\mathrm{p}K_\mathrm{a} \ {\text{ of formic acid} = 3.75}$

Now, if we consider polyprotic acids, we can only consider one hydrogen at a time. Successive $K_\mathrm{a}$s decrease in magnitudes of very large orders.

For example, $\ce{H3PO4}$ has $\mathrm{p}K_\mathrm{a1} = 2.12$ whereas $\mathrm{p}K_\mathrm{a2} = 7.21$ and $\mathrm{p}K_\mathrm{a3} = 12.68$.

Hope this answered your question.

Solution 2:

$K_\mathrm{a}$ is the measurement of equilibrium constant when an acid dissociates as: $$\ce{HA <=> H+ + A-}$$ $$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$$ The term $\mathrm{p}K_\mathrm{a}$ is defined as $\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a})$. So lower the $\mathrm{p}K_\mathrm{a}$, the better is the acidity.

In the example given, if you wish to compare the acidity, you have to mention which proton of $\ce{H2CO3}$ you are interested to abstract. In other words, $\ce{H2CO3}$ is more acidic then phenol but $\ce{HCO3-}$ is less acidic then phenol.