Question

# 1.What volume of a 1.50 M NaOH solution should be added to 50.0 mL of 1.20...

1.What volume of a 1.50 M NaOH solution should be added to 50.0 mL of 1.20 M acetic acid (CH3CO2H; Ka = 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.

Ka(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. Ka(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 Ka = 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, Kb = 5.95×10-4.

a) 3.05

b) 3.40

c) 10.92

d) 9.89

e) 10.59

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

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

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

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