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

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."