A 50.0 ml sample of 0.3 M acetic acid, HC_2H_3 O_2 is titrated with 0.1500 M NaO

Question

A 50.0 ml sample of 0.3 M acetic acid, HC_2H_3 O_2 is titrated with 0.1500 M NaOH. Ka (acetic acid) = 1.8 times 10^-5 a. Write a balanced net-ionic equation for the reaction of the above titration weak acid strong base) b. What is the pH before any NaOH is added? c. What is the pH half way to the equivalence point? d. How many mls of NaOH are required to reach the equivalence point? e. How many moles of acetate are formed at the equivalence point and what is the concentration of acetate C_2H_3O_2 at the equivalence point? f. What is the pH at the equivalence point?

Explanation / Answer

2)

a) net ionic equation :

CH3COOH (aq) + OH- (aq) ----------------> CH3COO- (aq) + H2O (l)

b)

pKa = -log Ka = -log (1.8 x 10^-5)

       = 4.74

pH = 1/2 (pKa - log C)

      = 1/2 (4.74 - log 0.3)

pH = 2.63

c)

pH = pKa

pH = 4.74

d)

At equivalence point :

millimoles of acid = millimoles of NaOH

50 x 0.3 = 0.15 x V

V = 100 mL

volume of NaOH = 100 mL

e)

moles of acetate formed = 15

concentration of acetate = 15 / (50 + 100) = 0.1 M

concentration of acetate = 0.100 M

f)

pH = 7 + 1/2 (4.74 + log 0.1)

   = 8.87

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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