The following are Weak Acid/Strong Base Titration Curve Questions based on the e

Question

The following are Weak Acid/Strong Base Titration Curve Questions based on the experiment described below: In a particular titration, 1.5 liters of 0.020 M acetic acid (a weak acid) was titrated with 1.0 M NaOH (a strong base). The Ka of acetic acid is 1.76 x 10 M. To answer the questions below, assume that the volume change during the titration is negligible, i.e., the total volume remains constant at 1.5 liters throughout the titration. The titration curve and chemical reaction for this experiment is shown below. Molecular eqn CHsCOOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(1) Net ionic eqn. CH3COOH(aq) + 0H(aq) CH3COO-(aq) + H2O(l) At Equivalence Point, 0.020 M Sodium Acetate Equivalence Point pH Half-Equivalence Point Initial Solution 0.020 M Acetic Acid Volume added

Explanation / Answer

1) The volume of acid = 1.5L

molarity of acid = 0.02

molarity of base = 1.0M

Volume of base = ?

At equivalence moles of acid = moles of base

moles = molarity x Volume in L

thus

1.5L x 0.020 moles of acid = V L x 1.0 moles of base

Thus V olume of base at equivalence = 0.03 L = 30 mL

2) Sodium acetate is a salt of weak acid and strong base. Thus the anion acetate undergoes hydrolysis and the solution is basic with pH > 7

The pH of a salt of weak acid and strong base is given by

pH = 1/2[pKw + pKa + log C]

given ka = 1.76x10-5 and pH = 4.754

and C = 0.02

Thus pH = 1/2 [14 + 4.754 + log 0.02]

= 8.5275

The pH at equivalence = 8.5275

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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