A student used the half-neutralization procedure described in this experiment an

Question

A student used the half-neutralization procedure described in this experiment and obtained a pH reading for an acid of 3.27. Assuming that 100.0 ml of a 1.00M solution of the acid was used: What is the ionization constant for the acid? What is the percent ionization for the acid? What fraction of the acid remains un-ionized? If the enthalpy of neutralization for the reaction of the acid with NaOH was found to be -52.3 kJ/mole, what is the enthalpy of ionization for the acid (assuming that none of the acid was ion­ized prior to reaction)? What is the enthalpy of ionization for the acid based upon the calculated fraction of un-ionized acid?

Explanation / Answer

a) At half neutralization point(or) half equivalence point ,

pH = pKa

3.27 = -logKa

Ka = 10-3.27

Ka = 0.00053

Therefore,

Ionization constant of the acid = 0.00053

b)

Ka = ca2 a = percent ionization

a = [Ka/C]1/2

= [0.00053/1.0]1/2

= 0.023

      = 2.3 %

Therefore,

percent ionization = 2.3 %

c)

Fraction of acid remains un-inized = 100 - 2.3 % = 97.7 % = 0.977

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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