Question

What is the pH of a soft drink in which the major buffer components are 8.50 g of NaH2PO4 and 9.23 g of Na2HPO4 per 355 mLs of solution?

Answer 1

Data.

8.50 g NaH₂PO₄; m.wt = 119.98 g/mol

9.23 g Na₂HPO₄; m.wt = 118.97 g/mol

Volume = 355 ml = 0.355 L

---

Phosphate buffer: NaH₂PO₄ (the weak acid) + Na₂HPO₄ (the salt, conjugated base)

NaH₂PO₄ <==> Na⁺ + H₂PO₄^-
Na₂HPO₄ <==> 2Na⁺ + HPO₄^2-

From here, we have the 3 stages of phosphoric acid (H₃PO₄), it is a weak poliprotic acid:

H₃PO₄ <==> H⁺ + H₂PO₄^-

H₂PO₄^- <==> H⁺ + HPO4^2-

HPO₄^2- <==> H⁺ + PO₄^3-

The condition for a effective buffer is that the pKa of the acid component is close to the desired pH. As phosforic acid is a triprotic one, we have to analyze the 3 stages of ionization as is shown above:

We are using NaH₂PO₄ as the weak acid, so we are going to use the Ka of this component from tables

Ka = 6.2 x 10^-8

pKa = -log(Ka) = -log(6.2 x 10^-8)

pKa = 7.21

---

Molarity of each component:

moles [H₂PO₄^-] = 8.50 gr/(119.98 g/mol) = 0.0708 mol/0.355L

= 0.1995 M

moles [HPO₄^2-] = 9.23 g/(118.97 g/mol) = 0.0775 mol/0.355L

= 0.2185 M

[HPO₄^2-]/[H₂PO₄^-] = 0.2185 M/0.1995 M = 1.0954

---

From Henderson-Hasselbalch equation:

pH = pKa + log [congujated base]/[weak acid]

pH = pKa + log [HPO₄^2-]/[H₂PO₄^-]

pH = 7.21 + log (1.0954)

pH = 7.21 + 0.039 = 7.245

pH = 7.25