The pKa of CH3COOH is 4.76. What statement describes a solution of CH3COOH at pH of 6.4?

There are no statements.

To determine how a solution of CH3COOH behaves at a pH of 6.4, we need to consider the pKa value of CH3COOH. The pKa is a measure of the acid's strength and indicates the pH at which the acid is half-dissociated.

The pKa of CH3COOH is 4.76, which means that at a pH below 4.76, CH3COOH will predominantly exist in its undissociated form, while at a pH above 4.76, it will predominantly exist in its dissociated form as CH3COO- and H+ ions.

In this case, the pH is 6.4, which is higher than the pKa of CH3COOH. As a result, CH3COOH will be mostly dissociated into CH3COO- and H+ ions. Therefore, a solution of CH3COOH at a pH of 6.4 would be considered as a weak acid solution in which most of the CH3COOH molecules have dissociated into their respective ions.

To determine the statement that describes a solution of CH3COOH at a pH of 6.4, we need to compare the pH of the solution to the pKa of CH3COOH.

First, let's understand the concept of pKa. The pKa is the measure of the acidity of a substance, specifically, the negative logarithm of the acid dissociation constant (Ka). It tells us the tendency of a compound to donate a proton (H+) in a solution.

The pKa value of CH3COOH is 4.76, which means that CH3COOH is a weak acid. It partially dissociates in water to release H+ ions.

When the pH of a solution is lower than the pKa of an acid, there are more H+ ions in solution than the acid form. Conversely, when the pH is higher than the pKa, there are more acid molecules than H+ ions.

In this case, the pH of the solution (6.4) is higher than the pKa (4.76) of CH3COOH. Therefore, the statement that describes a solution of CH3COOH at a pH of 6.4 is:

"At pH 6.4, the concentration of the CH3COOH acid form will be higher than the concentration of H+ ions."