Calculate the ph at the end of an enzyme-catalyzed reaction if it were carried out in a 0.1 m phosphate buffer, ph 6.86 and 0.005 m of acid was produced during the reaction.

6.77

Explanation:

A buffer is a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid. For polyprotic acids (with more than one acidic hydrogen), like phosphoric acid , it'll be more than one equilibrium, each one with a different pKa, the value of -logKa, where Ka is the equilibrium constant. For H₃PO₄:

H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ pKa₁ = 2.14

H₂PO₄⁻ + H₂O ⇄ HPO₄²⁻ + H₃O⁺ pKa₂ = 6.88

HPO₄²⁻ + H₂O ⇄ PO₄³⁻ + H₃O⁺ pKa₃ = 12.4

So, the pH of the solution is close to pKa₂, and its the principal equilibrium. The initial concentration of H₂PO₄⁻ and HPO₄²⁻ are 0.05 M (the total was 0.1 M). When acid is produced during the reaction, it reacts with the conjugate base and forms the acid, so:

[H₂PO₄⁻] = 0.05 + 0.005 = 0.055M

[HPO₄²⁻] = 0.05 - 0.005 = 0.045M

By Henderson-Hasselbalch equation:

pH = pKa + log ([A⁻]/[HA]), where [A⁻] is the conjugate concentration, and [HA] the acid concentration, s:

pH = 6.86 + log(0.045/0.055)

pH = 686 - 0.087

pH = 6.77

i do not get it   at all