How does pKa affect equilibrium?

Question
Answer 1

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,

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.

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Answer 2
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