how would i set up a balanced equation? For #5-#9: consider the dipeptide (Pep)

Question

how would i set up a balanced equation?

For #5-#9: consider the dipeptide (Pep) consisting of an Histidine residue and a Lysine residue (the neutral form is pictured on right) Note: The most positive, low pH, form of this dispeptide has a charge +3 _OH - 37 Ring-N (CH2)4NH2 The two side groups ionize according to: -Ring-NHO ? H+ +-Ring-N and-(CH2)4NHp-> H+ +-(CH2)4NH2 The four pK, s are: pKa, (?-CO2H)-22, pKa"(Ring-NH+) and pKa""(-NH3) 10.6 6.0, pKa,"(?-NH3+)-9.8, What is the pH after 1.5 equivalents of NaOH are added to a solution containing the most acidic form of Pep? (A) 2.2 5. (B) 4.1 (C) 6.0 (D) 7.9

Explanation / Answer

The pKa values are given. If 1 equivalent OH would have added that would neutralize the most acidic group with least pKa value. Since it is 1.5 equivalent, therefore, it would be at the midpoint of neutralizing the 2nd least pKa containing group. At the midpoint of titration, pH = pKa, therefore, the pH is 6.0 at that point.

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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