The pKa of propanoic acid (propionic acid), CH3CH2COOH, is 4.87. Consider an equilibrium mixture of propanoic acid and its conjugate base with a pH of 4.87. Draw the structure of the form (acid or conjugate base) that predominates after a decrease in [H3O ]. Include all hydrogen atoms and any appropriate formal charges.

Question

Anonymous · Accepted Answer

What is the PH of 0.02 mol/L solution of propanoic acid, given that the pKa value is 4.87

DrBob222 · Answer

pH = pKa + log (base)/(acid)
pH = 4.87
pKa = 4.87
Therefore,
4.87 = 4.87 + log (base)/(acid)
and (base)/(acid) = 1 which means
(base) = (acid)

If you lower the (H3O^+), that raises the pH so let's substitute a higher number for pH, say 5.87 and see what happens.
5.87 = 4.87 + log (base)/(acid)
1 = log (base)/(acid) and
(base)/(acid) = 10 or
(base) = 10(acid)
which means base predominates by a factor of 10 over the acid.
The base is CH3CH2COO^-. You must draw the structures and any charges since we can't do that on this forum.