What is the pH at the stoichiometric point for the titration of 0.100 M CH_3COOH

Question

What is the pH at the stoichiometric point for the titration of 0.100 M CH_3COOH(aq) with 0.100 M KOH(aq)? The value of K_c for acetic acid is 1.8 Times 10^-5. a) 8.72 b) 5.28 c) 9.26 d) 8.89 e) 7.00 38. Consider the dimerization reaction below: 2A rightarrow A_2 rate = k[A]^2 When the initial concentration of A is 2.0 M. it requires 30 mm for 60% of A to react. Calculate the rate constant. a. 1.1 times 10^-3 M^-1 s^-1 b. 1.9 times 10^-4 M^-1 s^-1 c. 3.2 times 10^-4 M^-1 s^-1 d. 4.2 times 10^-5 M^-1 s^-1 e. 5.0 times 10^-6 M^-1 s^-1 39. Consider the following reactiton: 2N_2O_5(g) rightarrow 4NO_2(g) + O_2(g) rate = k[N_2O_5] Calculate the time for the concentration of N_2O_5 to fall to one-eighth its initial value if the rate constant for the reaction is 520 Times 10^-3 s^-4. a. 533 s b. 400 s c. 16.7 s d. 33.3 s e. 1070 s

Explanation / Answer

37. At stoichiometric point, 0.1 M CH3COONa is present

Kb = 5.55 x 10^-10 = x^2/0.1

x = [OH-] = 7.45 x 10^-6 M

pOH= -log[OH-] = 5.12

pH = 14 - pOH = 8.89

So pH at stoichiometric point is,

d) 8.89

11. For second order reaction,

1/[A] = 1/[Ao] + kt

[Ao] = 2.0 M

[A] = 0.8 M

t = 30 x 60 = 180 s

we get,

1/0.8 - 1/2 = k x 180

rate constant k = 4.2 x 10^-4 M-1.s-1

d. 4.2 x 10^-4 M-1.s-1

12. For first order reaction,

ln([A]/[Ao]) = -kt

k = 5.2 x 10^-3 s-1

ln(1/8) = -5.2 x 10^-3 x t

t = 400 s

So time taken to reduce the concentration to 1/8 of initial would be,

b. 400 s

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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