How does pKa affect equilibrium?

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Answer 1 · 2018-05-05T17:40:43

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

$K_"a" = ([H^+][A^-])/([HA])$

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

Consider a reaction,

puu.sh

We can look at this from a perspective of the magnitudes of $"pK"_"a"$ 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.