Question

A buffer that is a mixture of acetic acid and potassium acetate has a pH = 5.22. The molar ratio of the conjugate base to weak acid in this buffer is? (acetic acid Ka = 1.8 x 10-5 )

A. 1:3

B. 1:5

C. 3:1

D. 1:1

E. 5:1

Answer 1

P^{​​​​​H} of buffer solution will be given as follows.

P^{​​​​​​H} = P^{​​​​​​Ka} + log{[conjugate base]/[ACID]}

or

P^{​​​​​​H} = -logK_​​a + log{[conjugate base ]/[Acid]}

Using the both given values

5.22 = -log(1.8×10^-5) +log{[conjugate base]/[Acid]}

5.22 = 4.74 + log{[conjugate base]/[Acid]}

log{[conjugate base]/[Acid]} = 5.22 - 4.74

log{[conjugate base]/[Acid] = 0.48

Taking antilog on both sides

[Conjugation base]/[Acid] = 3

Or

[Conjugation base]/[Acid] = 3/1

Or

[Conjugation base] : [Acid] = 3 : 1

Answer = C) 3 : 1

Answer 2

The Henderson-Hasselbalch equation for a buffer solution is given as:

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

where pKa is the dissociation constant of the weak acid, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, the pH of the buffer is given as 5.22, and the dissociation constant of acetic acid is 1.8 x 10^-5. We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A^-]/[HA]:

[A^-]/[HA] = 10^(pH - pKa)

Substituting the values, we get:

[A^-]/[HA] = 10^(5.22 - (-log10(1.8 x 10^-5))) [A^-]/[HA] = 10^(5.22 + 4.74) [A^-]/[HA] = 2.39 x 10^10

This means that the ratio of [A^-] to [HA] is 2.39 x 10^10 : 1.

To simplify this ratio, we can assume that the total concentration of the buffer is 1 M (i.e., [HA] + [A^-] = 1). Then, we can write:

[A^-]/[HA] = x/1-x

where x is the molar fraction of A^- in the buffer. Substituting the ratio we found earlier, we get:

2.39 x 10^10 = x/(1-x)

Solving for x, we get:

x = 1.000000000239 x 10^-10

Therefore, the molar ratio of [A^-] to [HA] in the buffer is:

[A^-]:[HA] = x:(1-x) = 1: 9.94 x 10^9

This ratio is closest to option E, 5:1.