Consider the following buffer scenarios: A solution is made by mixing 50.0 mL of

Question

Consider the following buffer scenarios: A solution is made by mixing 50.0 mL of 2.0 M K_3PO_4 and 25.0 mL of 2.0 M Na_2HPO_4. The solution is diluted to a final volume of 250.0 mL. What is the pH of the final solution? Over what pH range would the two species that were mixed together in part (a) make an effective buffer? What percentage of arginine's side chain would be in its ionized form at this pH? If an enzymatic reaction was carried out in 25.0 mL of this buffer and the enzyme produced 0.0050 moles of protons (i.e. H^+ ions), what would the new phosphate ion concentrations be? What would be the new pH?

Explanation / Answer

[salt] i.e K3PO4 = 50.0*0.200 / 250.0 = 0.0400 M

[Acid] i.e NaHPO4 = 25.0 * 0.200 / 250.0 = 0.02 M

pKa of HPO42- = 12.35

Applying Henderson's equation for buffers,

pH = pKa + Log[salt]/[acid]

pH = 12.35 + Log(0.04 / 0.02)

p H = 12.65

(b) it acts as an effective buffer between 11.65 and 13.65 pH range.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.