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 6.73, and 0.004 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 2.17, 6.73, and 11.27.)

Answer 1

Consider the dissociation of the triprotic acid, H₃A.

H₃A (aq) <=====> H⁺ (aq) + H₂A^- (aq); pK_a1 = 2.17

H₂A^- (aq) <=====> H⁺ (aq) + HA^2- (aq); pK_a2 = 6.73

HA^2- (aq) <=====> H⁺ (aq) + A^3- (aq); pK_a3 = 11.27

The initial pH of the enzyme-catalyzed reaction is 6.73. Since, the pH is close to the pK_a2 of the triprotic acid H₃A, hence the species present in the buffer solution are H₂A^-/HA^2-. Use the Henderson-Hasslebach equation to find the ratio of the molar concentrations of H₂A^-/HA^2- at the start of the reaction.

pH = pK_a2 + log [HA^2-]/[H₂A^-]

===> 6.73 = 6.73 + log [HA^2-]/[H₂A^-]

===> 0.00 = log [HA^2-]/[H₂A^-]

===> [HA^2-]/[H₂A^-] = 1 (since log 1 = 0)

===> [HA^2-] = [H₂A^-]

Therefore, the concentrations of the weak acid H₂A^- and the conjugate base HA^2- are equal at the start of the reaction. Again,

[H₂A^-] + [HA^2-] = 0.1 M

===> [H₂A^-] + [H₂A^-] = 0.1 M (since [H₂A^-] = [HA^2-])

===> 2*[H₂A^-] = 0.1 M

===> [H₂A^-] = (0.1 M)/2 = 0.05 M.

Therefore, [H₂A^-] = [HA^2-] = 0.05 M.

The reaction produces 0.004 M acid. H₂A^- is the acidic species in the buffer while HA^2- is the conjugate base. Possibly, the reaction involves neutralization of H₂A^- to give H⁺ and HA^2-.

As per the equations above, the molar ratio of H⁺ and HA^2- is 1:1; hence, the concentration of HA^2- produced by the reaction is 0.004 M. Therefore, the equilibrium concentration of H₂A^- is (0.05 – 0.004) M = 0.046 M and the equilibrium concentration of HA^2- is (0.05 + 0.004) M = 0.054 M.

Use the Henderson-Hasslebach equation again.

pH = pK_a + log [HA^2-]/[H₂A^-] = 6.73 + log (0.054 M)/(0.046 M) = 6.73 + log (1.1739) = 6.73 + 0.06963 = 6.79963 ≈ 6.80 (ans).