Question

To the nearest hundredths, what is the pH at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 M buffer initially at pH 6.65, and 0.006 M of acid (H+) was produced during the reaction? (The buffer used was a polyprotic acid of the general form: H3A; the three pKas of this polyprotic acid are 2.11, 6.65, and 11.33.)

Answer 1

The given pH = 6.65

The given pKa values of polyprotic acid (H₃A): pKa₁ = 2.11, pKa₂ = 6.65 and pKa₃ = 11.33

Here, pH = pKa2

i.e. The buffer mixture can be expressed as H₂A^-/HA^2-

According to Henderson-Hasselbalch equation, we can write as shown below.

pH = pKa₂ + Log[HA^2-]/[H₂A^-]

Log[HA^2-]/[H₂A^-] = 0 (since pH = pKa₂)

i.e. [HA^2-]/[H₂A^-] = 1

i.e. [HA^2-] = [H₂A^-]

Given that the concentration of buffer = 0.1 M

i.e. [HA^2-] + [H₂A^-] = 0.1

Therefore, [HA^2-] = [H₂A^-] = 0.1/2, i.e. 0.05 M

Given that 0.006 M of acid (H+) was produced during the reaction.

i.e. [HA^2-] = 0.05 - 0.006, i.e. 0.044 M (since HA^2- is a base, which reacts with produced H⁺, as a result its concentration decreases)

and [H₂A^-] = 0.05 + 0.006, i.e. 0.056 M (since H₂A^- is an acid, which adds up to the produced H⁺, as a result its concentration increases)

Rewrite the Henderson-Hasselbalch equation: pH = pKa₂ + Log[HA^2-]/[H₂A^-]

i.e. pH = 6.65 + Log(0.044/0.056)

i.e. pH = 6.65 + (-0.105)

i.e. The pH at the end of an enzyme-catalyzed reaction​ = 6.545