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.

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

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.

To determine the structure that predominates after a decrease in [H3O+], we need to consider the relationship between pH and pKa. The Henderson-Hasselbalch equation provides a convenient way to do this:

pH = pKa + log([conjugate base]/[acid])

Since the pH is equal to the pKa, we can assume that the concentration of the acid and the conjugate base are equal.

Now let's draw the structure of propanoic acid, CH3CH2COOH:

H3C – CH2 – COOH

To convert propanoic acid into its conjugate base (propanoate), we remove a hydrogen atom from the carboxylic acid functional group (-COOH). The hydrogen atom will be removed from the oxygen atom, resulting in a negative charge on the oxygen:

H3C – CH2 – COO-

Therefore, the structure of the form that predominates after a decrease in [H3O+] would be:

H3C – CH2 – COO-

To determine the structure that predominates after a decrease in [H3O+], we need to compare the pH value with the pKa value.

Given that the pH is equal to the pKa, it suggests that we are at the half-equivalence point, where the concentration of the acid and its conjugate base are equal. At this point, we have an equilibrium mixture.

Since the pKa of propanoic acid is 4.87 and the pH is also 4.87, this implies that the ratio of the concentrations of the protonated form (acid) and the deprotonated form (conjugate base) is 1:1.

To draw the structure, let's start with propanoic acid:

H
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CH3-CH2-COOH

Since the pH is equal to the pKa, it means the concentration of hydrogen ions (H3O+) is equal to the concentration of the conjugate base.

Therefore, after a decrease in [H3O+], the form that predominates is the deprotonated form (conjugate base). To represent this, we remove the hydrogen atom (H) from the carboxylic group:

|
CH3-CH2-COO-

The negative charge should be on the oxygen atom since oxygen is more electronegative than carbon.

Thus, the structure that predominates after a decrease in [H3O+] is:

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CH3-CH2-COO-