Question

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.

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

a) A 0.1500 M of AgNO3 solution was employed to
titrate a 25.00 mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that
Ksp, AgI(s) = 8.3*10-17, and Ksp, AgCl(s) =
1.8*10-10, please calculate the concentration of
Ag+ ion after 6.00 mL of AgNO3 was added.
b) please calculate the pAg after 100.00 mL of AgNO3
was added

a) A 0.1500 M of AgNO3 solution was employed to titrate a 25.00
mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that Ksp, AgI(s) =
8.3*10-17, and Ksp, AgCl(s) = 1.8*10-10, please calculate the
concentration of Ag+ ion after 6.00 mL of AgNO3 was added.
Answer: 1.2 x 10-15 M
b) please calculate the pAg after 100.00 mL of AgNO3 was
added
Answer: 1.35
I just want to see how they got that.

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.

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with
0.05012 M KI. ksp = AgI(s) = 8.3*10-17.
1) For the same titration as shown in Question 2,
please calculate the pAg value after 90.00 mL of KI is added.
2) For the same titration as shown in Question 2,
please calculate the volume of KI that is required to reach a pAg
of 2.00.

A 0.1000 M NaOH solution was employed to titrate a 25.00-mL
solution that contains 0.1000 M HCl and 0.0500 M HOAc. Please
determine the pH of the solution after 27.00 mL of NaOH is added.
Ka, HOAc = 1.75*10-5 .
Answer. 4.04
Just want to see how they got that.

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with
0.05012 M KI. Ksp, AgI(s) = 8.3*10-17. For the same titration as
shown in Question 2, please calculate the volume of KI that is
required to reach a pAg of 2.00.

Consider the titration of 25.00 mL of 0.1250 M AgNO3 with
0.05012 M KI.
Ksp, AgI(s) = 8.3*10-17.
please calculate the volume of KI that is required to reach a
pAg of 2.00.
Answer is 47.82 mL. How they got that?

Calculate the pH during the titration of 25.00 mL of 0.1000 M
LiOH(aq) with 0.1000 M HI(aq) after 24.1 mL of the acid have been
added.

Please show all steps.
If you titrate 25.00 mL of a 0.09797 M solution of HBrO, having
Ka = 2.8 x 10-9, with 0.05000 M Ba(OH)2
solution:
a) Calculate the number of mL of Ba(OH)2 solution
needed to reach the equivalence point.
b) Calculate the pH inititally, before any Ba(OH)2
solution is added.
c) Calculate the pH at halfway.
d) Calculate the pH at the equivalence point.

A group of students conducted a titration of 25.00-mL saturated
Ca(OH)2 solution with 0.0480M HCl. The students found
that 10.10-mL of acid was required to reach the equivalence point.
Calculate the molar concentration of OH- and
Ca2+, the molar solubility, and Ksp of the
analyte. Show detailed calculation for each.
[OH-] in the analyte:
[Ca2+] in the analyte:
molar solubility of Ca(OH)2:
Ksp of Ca(OH)2:

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