Question

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium...

1. a. 0.0500 M of AgNO3 is used to titrate a 25.00-mL containing 0.1000 M sodium chloride (NaCl) and 0.05000 M potassium iodide (KI), what is the pAg of the solution after 15.00 mL of AgNO3 is added to the solution? Ksp, AgCl (s) = 1.82 x 10-10; Ksp, AgI(s) = 8.3*10-17.

b. Same titration as in (a), what is the pAg of the solution after 25.00 mL of AgNO3 is added to the above solution?

c. Same titration as in (a), what is the concentration of the Cl- ion after 85.00 mL of AgNO3 is added to the solution?

2. a. What is the Ksp for the precipitation of aluminum hydroxide [Al(OH)3], given that its molar solubility is 1.83 x 10-9 M and its formular weight is 77.99 grams/mole?

b. What is the molar solubility of Fe(OH)3 in an aqueous solution buffered at pH = 10.50? Ksp is 2 x 10-39 for the reaction: Fe(OH)3 (s) ÍÎ Fe3+ + 3 OH- . Formular weight of Fe(OH)3 is 106.86 grams/mole.

3. Determine the pH of the following solutions:

a. A 25.00 mL aqueous solution containing 0.05000 M sodium acetate (CH3COONa), given that the Ka of acetic acid (CH3COOH) is 1.75 x 10-5. Kw = 1.00*10-14.

b. The solution in (a) after 5.50 mL of 0.1000 M HCl is added.

Homework Answers

Answer #1

you would have 0.025 L x 0.1000 mol/L = 0.0025 mol of Nacl ions
and 0.025 L * 0.0500 mol/L = 0.00125 mol KI ions
then total volume is 0.050 L after mixing
so
[KI] = 0.00125 mol / 0.050 L = 0.025 M
[Nacl] = 0.0025 mol / 0.050 L = 0.05 M


Ksp = [Agl][Nacl]
AgI(s) = 8.3*10-17, and AgCl(s) = 1.8*10-10
  8.3*10-17 x 0.05 = 4.15e-18

When 15 mL have been added, Volume = 25 mL
Moles Ag+ added = 0.0015 L X 0.100 mol/L = 0.00015 moles
Assume that initially, all of the added Ag+ precipitates. You will be left with a Cl- concentration of 3.5 X 10-3 M
some of the precipitated AgCl will dissolve in that solution to give:
Ksp = 1.82 X 10-10 = [Ag+] [3.5X10^-3]
[Ag+] = 5.2 X 10^-8 M
pAg = 7.28

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