what is the Ag+]init (M) [C2H3O2-]init (M for flask d e and f ?  What is the  Moles SCN- used in titration (mol) &  Moles of Ag+ in 20.0 mL titrated (mol) for flask d e and f

Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.11PAE
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what is the

Ag+]init (M)

[C2H3O2-]init (M

for flask d e and f ? 


What is the

 Moles SCN- used in titration (mol) & 
Moles of Ag+ in 20.0 mL titrated (mol)

for flask d e and f 

This will be done as a dry-lab. There will not be sufficient time to do these procedures, but the
steps below explain the procedure that generated the data you will find on your data sheet
1. Repeat part A. using three flasks labeled d, e and f. Notice that the data you will be given for
this part of the experiment is obtained by approaching equilibrium from the right of the
chemical equation, not the left.
Solution 0.300 M NaC₂H3O2
d
25.0 mL
25.0 mL
25.0 mL
e
f
AgC₂H3O₂ (s)
0.250 M AgNO3
30.0 mL
30.0 mL
30.0 mL
Temp. °C
ice bath
room temp
50 - 60°C
Ag+ (aq) + C₂H3O₂ (aq)
<--- Equilibrium approached from right.
2. During Step A.5 when the system comes to equilibrium, the three flasks are kept at different
temperatures for 1/2 hour. Flask d is placed in an ice water bath. Flask e is kept at room
temperature. Flask f is placed in a warm water bath between 50 and 60°C. All three of the
flasks have identical initial quantities of chemicals present. In fact, they are all identical to
flask b in part A, except for the temperature.
3. Samples d, e and f are filtered and labeled, just as in step A.6.
4. Titration procedures same as steps A.7 through A. 12.
Transcribed Image Text:This will be done as a dry-lab. There will not be sufficient time to do these procedures, but the steps below explain the procedure that generated the data you will find on your data sheet 1. Repeat part A. using three flasks labeled d, e and f. Notice that the data you will be given for this part of the experiment is obtained by approaching equilibrium from the right of the chemical equation, not the left. Solution 0.300 M NaC₂H3O2 d 25.0 mL 25.0 mL 25.0 mL e f AgC₂H3O₂ (s) 0.250 M AgNO3 30.0 mL 30.0 mL 30.0 mL Temp. °C ice bath room temp 50 - 60°C Ag+ (aq) + C₂H3O₂ (aq) <--- Equilibrium approached from right. 2. During Step A.5 when the system comes to equilibrium, the three flasks are kept at different temperatures for 1/2 hour. Flask d is placed in an ice water bath. Flask e is kept at room temperature. Flask f is placed in a warm water bath between 50 and 60°C. All three of the flasks have identical initial quantities of chemicals present. In fact, they are all identical to flask b in part A, except for the temperature. 3. Samples d, e and f are filtered and labeled, just as in step A.6. 4. Titration procedures same as steps A.7 through A. 12.
C. Calculations
1. Assume that the total volume of each saturated solution is 55.00 mL.
2. When you prepare this equilibrium by mixing solutions, you are approaching this equilibrium
from the right. Use an 'ICE Table' approach to getting the equilibrium concentrations
[Ag leq and [C₂H302 leq in each of your six flasks.
AgC₂H3O2 (s)
Initial
Change
Equilibrium
3. The Initial [Aglinit and [C₂H3O₂ linit concentrations can be calculated from the molarities
and volumes of solutions used. The initial concentration refers to the system after mixing but
before precipitation. You will need to refer to the tables on pages 2 and 3 to get this
information.
example a: [Ag+]init
Ag+ (aq) + C₂H3O₂ (aq)
[C₂H3O2 linit
[Ag+] init
-X
([Ag+]init - X)
total moles of Ag+ put into flask a
total liters solution in flask a.
in sample a.
4. The Change referred to in the' ICE table' occurs when precipitate forms after mixing. The
equilibrium shifts to the left, reducing the silver and the acetate ion concentrations by X
mol/L.
=
=
0.0350 L solutions x 0.250 mol Ag/L solution
0.0550 L
= 0.159 M
-X
([C₂H3O2-linit - X)
5. In each of your samples, you use your titration volume, V mL, to calculate the total moles of
Ag+ in the 20.0 mL sample of saturated solution you titrated by doing a stoichiometry
calculation using the chemical equation for the titration reaction.
Ag+ (aq) + SCN- (aq) -> AgSCN(s)
?mol
V mL
0.0500 M
V mL KSCN solution x 0.0500 mol KSCN x
1000 mL KSCN sol
6. Calculate [Ag+] at equilibrium
[Ag*lequil
1 mol SCN- x
1 mol KSCN
1 mol Ag+
1 mol SCN-
(5.00 x 10-5 x V) moles Ag+ in 20.0 mL
(5.00 x 10-5 x V moles Ag+) = (2.5 x 10-3 x V) mol/L
0.0200 L solution
Transcribed Image Text:C. Calculations 1. Assume that the total volume of each saturated solution is 55.00 mL. 2. When you prepare this equilibrium by mixing solutions, you are approaching this equilibrium from the right. Use an 'ICE Table' approach to getting the equilibrium concentrations [Ag leq and [C₂H302 leq in each of your six flasks. AgC₂H3O2 (s) Initial Change Equilibrium 3. The Initial [Aglinit and [C₂H3O₂ linit concentrations can be calculated from the molarities and volumes of solutions used. The initial concentration refers to the system after mixing but before precipitation. You will need to refer to the tables on pages 2 and 3 to get this information. example a: [Ag+]init Ag+ (aq) + C₂H3O₂ (aq) [C₂H3O2 linit [Ag+] init -X ([Ag+]init - X) total moles of Ag+ put into flask a total liters solution in flask a. in sample a. 4. The Change referred to in the' ICE table' occurs when precipitate forms after mixing. The equilibrium shifts to the left, reducing the silver and the acetate ion concentrations by X mol/L. = = 0.0350 L solutions x 0.250 mol Ag/L solution 0.0550 L = 0.159 M -X ([C₂H3O2-linit - X) 5. In each of your samples, you use your titration volume, V mL, to calculate the total moles of Ag+ in the 20.0 mL sample of saturated solution you titrated by doing a stoichiometry calculation using the chemical equation for the titration reaction. Ag+ (aq) + SCN- (aq) -> AgSCN(s) ?mol V mL 0.0500 M V mL KSCN solution x 0.0500 mol KSCN x 1000 mL KSCN sol 6. Calculate [Ag+] at equilibrium [Ag*lequil 1 mol SCN- x 1 mol KSCN 1 mol Ag+ 1 mol SCN- (5.00 x 10-5 x V) moles Ag+ in 20.0 mL (5.00 x 10-5 x V moles Ag+) = (2.5 x 10-3 x V) mol/L 0.0200 L solution
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