100 ml of 0.02 M acetic acid (pKa= 4.76) is titrated with 0.02 N KOH. After adding

some base to the acid solution, the observed pH is 2.76. At this pH degree of protonation
is

To determine the degree of protonation at pH 2.76, we need to calculate the concentration of acetic acid and the concentration of its conjugate base (acetate ion) at this pH.

First, let's calculate the concentration of acetic acid (CH3COOH) at pH 2.76 using the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

Rearranging the equation, we have:

log ([A-]/[HA]) = pH - pKa

Substituting the values, we have:

log ([A-]/[HA]) = 2.76 - 4.76
log ([A-]/[HA]) = -2.00

Taking the antilog of both sides, we get:

[A-]/[HA] = 10^(-2.00)
[A-]/[HA] = 0.01

Now, we know that the concentration of acetic acid ([HA]) is 0.02 M. We can calculate the concentration of acetate ion ([A-]) as follows:

[A-] = 0.01 * [HA]
[A-] = 0.01 * 0.02
[A-] = 0.0002 M

Finally, to get the degree of protonation, we divide the concentration of acetic acid ([HA]) by the sum of the concentrations of acetic acid ([HA]) and acetate ion ([A-]):

Degree of Protonation = [HA] / ([HA] + [A-])
Degree of Protonation = 0.02 M / (0.02 M + 0.0002 M)
Degree of Protonation = 0.02 M / 0.0202 M
Degree of Protonation ≈ 0.9901

Therefore, at pH 2.76, the degree of protonation is approximately 0.9901, meaning 99.01% of acetic acid molecules are still protonated at this pH.

To determine the degree of protonation at a given pH, we can make use of the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

In this case, acetic acid (CH3COOH) is a weak acid and its conjugate base is acetate ion (CH3COO-). Let's denote [A-] as the concentration of acetate ion and [HA] as the concentration of acetic acid.

We know that the concentration of acetic acid is 0.02 M (because the initial concentration is given as 0.02 M) and the pH is 2.76. The pKa of acetic acid is 4.76.

Let's substitute these values into the Henderson-Hasselbalch equation:

2.76 = 4.76 + log ([A-]/0.02)

Now, let's solve for [A-]:

2.76 - 4.76 = log ([A-]/0.02)

-2 = log ([A-]/0.02)

Taking the antilog of both sides:

10^(-2) = [A-]/0.02

0.01 = [A-]/0.02

[A-] = 0.01 * 0.02

[A-] = 0.0002 M

Therefore, at a pH of 2.76, the concentration of acetate ion is 0.0002 M. The degree of protonation can be considered as 1 - [A-] since [A-] represents the concentration of the deprotonated form. Therefore, the degree of protonation at this pH is:

Degree of protonation = 1 - 0.0002

Degree of protonation = 0.9998

the answer is 99%

calculate the concentration of acid before addition of base and after addition of base and divide
you will get your answer