# How does pKa affect equilibrium?

May 5, 2018

Consider what $\text{pK"_"a}$ represents: the negative logarithm of the equilibrium expression for acid dissociation,

${K}_{\text{a}} = \frac{\left[{H}^{+}\right] \left[{A}^{-}\right]}{\left[H A\right]}$

To be sure, its magnitude is inversely proportional to the amount of dissociation the acid undergoes in solution.

Consider a reaction,

We can look at this from a perspective of the magnitudes of $\text{pK"_"a}$ values.

Alkanes: ~55

Phenol: ~10

The latter is far more likely to dissociate in this solution, and hence, the equilibrium will probably favor the left.

In non-scientific language (which helped me more during organic chemistry): the alkane is happy with its proton, but the phenol isn't.