Question

1.What volume of a 1.50 M NaOH solution should be added to 50.0 mL of 1.20 M acetic acid (CH₃CO₂H; K_a = 1.76×10^-5) to obtain a buffer with pH = 5.450?

a) 33.3 mL

b) 6.70 mL

c) 16.4 mL

d) 67.0 mL

e) 42.1 mL

2. Find the concentration of hydronium ion, H₃O⁺, in a 0.200 M solution of sodium hypochlorite, NaOCl.

K_a(HOCl) = 3.0×10^-8.

a) 7.8×10^-5 M

b) 1.3×10^-10 M

c) 1.0×10^-7 M

d) 3.9×10^-11 M

e) 2.6×10^-4 M

3. Consider a buffer solution made of 0.155 M acetic acid, CH₃COOH, and 0.205 M sodium acetate, NaCH₃COO. K_a(CH₃COOH) = 1.8 x 10^-5. After addition of 0.030 moles of NaOH to 1.0 L of this buffer, the pH becomes [X].  Fill in the blank. Show the number only. Report with 2 digits after the decimal point.

4. Nitrous acid, HNO₂, has the K_a = 7.1×10^-4. The pH of a 0.1750 M solution of nitrous acid is [X]. (Fill in the blank.  Report with correct number of significant figures).

5. A 1.25 M solution of chlorous acid, HClO₂, is 9.2% dissociated. What is the pH of this solution?

a) 1.13

b) 0.94

c) 2.16

d) 0.64

e) 0.097

6. What is the pH of a buffer solution that is 0.20 M in dimethylamine, (CH₃)₂NH, and 0.30 M in dimethylammoniumchloride, (CH₃)₂NH₂Cl? For (CH₃)₂NH, K_b = 5.95×10^-4.

a) 3.05

b) 3.40

c) 10.92

d) 9.89

e) 10.59

Answer 1

1. Using Hendersen-Haselbalck equation,

pH = pKa + log(base/acid)

5.45 = 4.76 + log(x/1.2 x 50 - x)

300 - 5x = x

moles of NaOH required = x = 300/6 = 50 mmol

Volume of NaOH = 50/1.5 = 33.3 ml

Answer: 1) 33.3 ml

2. OCl- + H2O <==> HOCL + OH-

Kb = 3.5 x 10^-8 = x^2/0.2

x = [OH-] = 8.37 x 10^-5 M

[H3O+] = 1 x 10^-14/8.37 x 10^-5 = 1.3 x 10^-10 M

Answer : B) 1.3 x 10^-10 M

3. ph after addition of NaOH

pH = 4.76 + log[(0.155 + 0.03)/(0.205 - 0.03)]

     = 4.72

4. HNO2

7.1 x 10^-4 = x^2/0.1750

x = [H3O+] = 1.11 x 10^-2 M

pH= -log[H3O+] = 2.00

5. pH = -log(0.092 x 1.25) = 0.94

Answer: B) 0.94

6. pH = 3.27 + log(0.2/0.3) = 3.05

Answer : a) 3.05